PRACTICE MATERIAL - CLASSICAL
MECHANICS - LAGRANGIAN MECHANICS
Problem 1: Classical Mechanics - Lagrangian Mechanics
Question: A particle of mass m is moving in a one-dimensional potential given by
V(x)=1
2kx2. Use the Lagrangian mechanics approach to find the equation of motion for
the particle.
Solution:
1. Write the Lagrangian:
The Lagrangian L is given by the difference between the kinetic energy T and the
potential energy V: L=T−V
For a particle of mass m moving with velocity x, the kinetic energy is:
T=1
2mx2
The potential energy given in the problem is:
V(x)=1
2kx2
So, the Lagrangian is: L=1
2mx2−1
2kx2
2. Apply the Euler-Lagrange equation:
The Euler-Lagrange equation is: d
dt(∂L
∂x)−∂L
∂x=0
Calculate ∂L
∂x: ∂L
∂x=mx
Then:
d
dt(mx)=mx
Now, calculate ∂L
∂x: ∂L
∂x=−kx
Substitute these into the Euler-Lagrange equation:
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law:
∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So: B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
mx+kx=0
3. Solve the equation of motion:
The equation of motion is a second-order differential equation:
x+k
mx=0
This is a simple harmonic oscillator equation, where ω2=k
m. Therefore, the general
solution is: x(t)=Acos(ωt)+Bsin(ωt)
where A and B are constants determined by initial conditions.
Problem 2: Electromagnetism - Maxwell’s Equations
Question: A plane electromagnetic wave is propagating in a vacuum. The electric field is
given by E=E0cos(kz−ωt)x. Use Maxwell’s equations to find the corresponding
magnetic field B.
Solution:
1. Start with Faraday’s Law: ∇×E=−∂B
∂t
The electric field is given by: E=E0cos(kz−ωt)x
Calculate the curl of E:
∇×E=
|
|
x y z
∂
∂x ∂
∂y ∂
∂z
E0cos(kz−ωt)0 0
|
|
=(0−0,0−0,0−∂
∂y(E0cos(kz−ωt)))=0
2. Use Maxwell-Ampere’s Law: ∇×B=μ0ϵ0∂E
∂t
Calculate the time derivative of E:
∂E
∂t=−ωE0sin(kz−ωt)x
The curl of B must be: ∇×B=−μ0ϵ0ωE0sin(kz−ωt)x
Since the wave is propagating in the z-direction, the magnetic field B must be in the y-
direction. Assume: B=B0cos(kz−ωt)y
Then: ∇×B=−kB0sin(kz−ωt)x
Equate the expressions:
−kB0sin(kz−ωt)=−μ0ϵ0ωE0sin(kz−ωt)
So:
B0=ωE0
kμ0ϵ0=E0
c
Therefore, the magnetic field is: B=E0
ccos(kz−ωt)y
