PHY332 - Electromagnetism Question Bank
Question 1
Problem: Two point charges, q1 = +3
µ
C and q2 = -5
µ
C, are placed 0.2
meters apart in vacuum. Calculate the magnitude of the electromagnetic force
between them and specify whether it is attractive or repulsive.
Solution:
Step 1: Identifying Given Values - Charge q1 = +3
µ
C = +3
Ö
10−6C−
Chargeq2 = −5C=−510−6C−Distancer = 0.2m
Step 2: Formula for Electric Force The electric force between two point
charges is given by Coulomb’s Law:
F=k|q1×q2|
r2
where: - Fis the magnitude of the force, - kis Coulomb’s constant (8.9875 ×
109N·m2/C2), - q1 and q2 are the magnitudes of the charges, - ris the distance
between the charges.
Step 3: Substituting the Values
F= (8.9875 ×109)|(+3 ×10−6)×(−5×10−6)|
(0.2)2
F= (8.9875 ×109)15 ×10−12
0.04
F= (8.9875 ×109)×0.375 ×10−9
F= 3.37 N
Step 4: Determining the Nature of the Force Since the charges are of opposite
signs (one positive and one negative), the force between them will be attractive.
Conclusion: The magnitude of the force between the charges is 3.37 Newtons,
and it is attractive. Question 1: Electromagnetic Force Between Two
Charged Particles
Problem: Two point charges, q1 = +3
µ
C and q2 = -5
µ
C, are
placed 0.2 meters apart in vacuum. Calculate the magnitude of the
electromagnetic force between them and specify whether it is attrac-
tive or repulsive.
Solution:
1
Step 1: Identifying Given Values - Charge q1 = +3
µ
C = +3
Ö
10−6C−Chargeq2 = −5C=−510−6C−Distancer = 0.2m
Step 2: Formula for Electric Force The electric force between two
point charges is given by Coulomb’s Law:
F=k|q1×q2|
r2
where: - Fis the magnitude of the force, - kis Coulomb’s constant
(8.9875 ×109N·m2/C2), - q1and q2are the magnitudes of the charges,
-ris the distance between the charges.
Step 3: Substituting the Values
F= (8.9875 ×109)|(+3 ×10−6)×(−5×10−6)|
(0.2)2
F= (8.9875 ×109)15 ×10−12
0.04
F= (8.9875 ×109)×0.375 ×10−9
F= 3.37 N
Step 4: Determining the Nature of the Force Since the charges are
of opposite signs (one positive and one negative), the force between
them will be attractive.
Conclusion: The magnitude of the force between the charges is
3.37 Newtons, and it is attractive.
Question 2
Problem Statement: A long solenoid has a total of 2500 turns
and carries a current of 3.0 A. The length of the solenoid is 0.75 m.
Calculate the magnitude of the magnetic field inside the solenoid.
Given: - Number of turns (N) = 2500 - Current (I) = 3.0 A -
Length of solenoid (L) = 0.75 m
Useful Formula: The magnetic field inside a long solenoid is given
by B=µ0
N
LI
where: - Bis the magnetic field, - µ0is the permeability of free
space (4π×10−7T
·
m/A), - Nis the number of turns, - Lis the length
of the solenoid, and - Iis the current.
Solution:
Step 1: Write down the formula.
B=µ0
N
LI
Step 2: Substitute the given values into the formula.
-µ0= 4π×10−7T
·
m/A, - N= 2500, - L= 0.75 m, - I= 3.0A.
2
B= (4π×10−7T
·
m/A)×2500
0.75 m×3.0A
Step 3: Calculate the value.
B= (4π×10−7T
·
m/A)×2500
0.75 ×3.0
B= (4π×10−7T
·
m/A)×3333.33 ×3.0
B= (4π×10−7T
·
m/A)×10000
B= 4π×10−3T
B≈0.012566 T
Step 4: Write the final answer. The magnitude of the magnetic
field inside the solenoid is approximately 0.0126 T (Tesla). Question
2: Calculating the Magnetic Field Strength Inside a Long Solenoid
Problem Statement: A long solenoid has a total of 2500 turns
and carries a current of 3.0 A. The length of the solenoid is 0.75 m.
Calculate the magnitude of the magnetic field inside the solenoid.
Given: - Number of turns (N) = 2500 - Current (I) = 3.0 A -
Length of solenoid (L) = 0.75 m
Useful Formula: The magnetic field inside a long solenoid is given
by B=µ0
N
LI
where: - Bis the magnetic field, - µ0is the permeability of free
space (4π×10−7T
·
m/A), - Nis the number of turns, - Lis the length
of the solenoid, and - Iis the current.
Solution:
Step 1: Write down the formula.
B=µ0
N
LI
Step 2: Substitute the given values into the formula.
-µ0= 4π×10−7T
·
m/A, - N= 2500, - L= 0.75 m, - I= 3.0A.
B= (4π×10−7T
·
m/A)×2500
0.75 m×3.0A
Step 3: Calculate the value.
B= (4π×10−7T
·
m/A)×2500
0.75 ×3.0
B= (4π×10−7T
·
m/A)×3333.33 ×3.0
B= (4π×10−7T
·
m/A)×10000
B= 4π×10−3T
B≈0.012566 T
Step 4: Write the final answer. The magnitude of the magnetic
field inside the solenoid is approximately 0.0126 T (Tesla).
3
Question 3
Problem:
A circular loop of wire with radius R= 0.25 m is placed in a mag-
netic field that changes with time. The magnetic field perpendicular
to the plane of the coil varies according to the equation B(t) = 0.02t2,
where Bis in Teslas and tis in seconds.
Calculate the magnitude of the induced emf in the loop at t=
5seconds.
Solution:
Step 1: Identify the given quantities Radius of the wire loop, R=
0.25 m Magnetic field variation, B(t)=0.02t2
Step 2: Calculate the area of the loop The area Aof a circle is
given by A=πr2. Substituting the given radius,
A=π(0.25)2=π(0.0625) m2≈0.196 m2
Step 3: Differentiate the magnetic field equation
The magnetic flux Φ = B(t)A, where B(t)is time-dependent. To
find the induced emf (E), use Faraday’s Law of Electromagnetic In-
duction which states:
E=−dΦ
dt
Now, differentiate Φwith respect to t:
dΦ
dt =d
dt(B(t)A) = AdB(t)
dt
Given B(t)=0.02t2, differentiate B(t)with respect to t:
dB
dt = 0.04t
Step 4: Calculate the rate of change of flux at t= 5 s
dB
dt
t=5 = 0.04 ×5 = 0.2T/s
Now, substitute back to find the rate of change of flux:
dΦ
dt = 0.196 ×0.2=0.0392 Weber/s
Step 5: Compute the induced emf The negative sign in Faraday’s
law indicates the direction of induced emf which opposes the change
in flux (Lenz’s Law), so:
E=−dΦ
dt =−0.0392 V≈ −0.04 V
4
Answer: The magnitude of the induced emf in the loop at t=
5seconds is about 0.04 V.
(Note: Using the absolute value since the question asks for mag-
nitude) Question 3:
Problem:
A circular loop of wire with radius R= 0.25 m is placed in a mag-
netic field that changes with time. The magnetic field perpendicular
to the plane of the coil varies according to the equation B(t) = 0.02t2,
where Bis in Teslas and tis in seconds.
Calculate the magnitude of the induced emf in the loop at t=
5seconds.
Solution:
Step 1: Identify the given quantities Radius of the wire loop, R=
0.25 m Magnetic field variation, B(t)=0.02t2
Step 2: Calculate the area of the loop The area Aof a circle is
given by A=πr2. Substituting the given radius,
A=π(0.25)2=π(0.0625) m2≈0.196 m2
Step 3: Differentiate the magnetic field equation
The magnetic flux Φ = B(t)A, where B(t)is time-dependent. To
find the induced emf (E), use Faraday’s Law of Electromagnetic In-
duction which states:
E=−dΦ
dt
Now, differentiate Φwith respect to t:
dΦ
dt =d
dt(B(t)A) = AdB(t)
dt
Given B(t)=0.02t2, differentiate B(t)with respect to t:
dB
dt = 0.04t
Step 4: Calculate the rate of change of flux at t= 5 s
dB
dt
t=5 = 0.04 ×5 = 0.2T/s
Now, substitute back to find the rate of change of flux:
dΦ
dt = 0.196 ×0.2=0.0392 Weber/s
Step 5: Compute the induced emf The negative sign in Faraday’s
law indicates the direction of induced emf which opposes the change
in flux (Lenz’s Law), so:
E=−dΦ
dt =−0.0392 V≈ −0.04 V
5
Answer: The magnitude of the induced emf in the loop at t=
5seconds is about 0.04 V.
(Note: Using the absolute value since the question asks for mag-
nitude)
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