PHYS 201 - Electromagnetism Question Bank
Question 1
Problem Statement: A point charge of +3
µ
C is located at the origin. De-
termine the electric field at a point P located at coordinates (4, 3, 0) meters.
Relevant Formula: The electric field Edue to a point charge Qat a distance
ris given by:
E=k|Q|
r2
where kis Coulomb’s constant (k≈8.988 ×109N m2/C2).
Step-by-Step Solution:
1. Identify the Charge and Distance: - Point charge Q= +3 µC = +3 ×
10−6C. - Coordinates of point P are (4, 3, 0) meters.
2. Calculate the Distance rfrom the Charge to Point P: - Use the distance
formula:
r=px2+y2+z2
- Plug in the coordinates x= 4, y= 3, z= 0:
r=p42+ 32+ 02=√16 + 9 = √25 = 5 meters
3. Calculate the Electric Field Eat Point P: - Plug rand Qinto the electric
field formula:
E=k|Q|
r2= 8.988 ×109×3×10−6
52
- Simplify the expression:
E= 8.988 ×109×3×10−6
25 = 8.988 ×109×0.12 ×10−6
E= 1.07856 ×103N/C
4. Direction of the Electric Field: - Since the charge is positive, the electric
field points away from the charge. - The direction vector from the origin (charge
location) to point P is (4,3,0). - Normalize this vector:
Direction unit vector = 4
5,3
5,0
1
- The electric field vector at P is:
E= 1.07856 ×1034
5,3
5,0N/C
E= (862.848,647.136,0) N/C
Answer: The electric field at point P is approximately 862.848 i+647.136 j N/C
pointing in the direction from the origin towards the point (4, 3, 0). Question
1: Understanding Electric Fields
Problem Statement: A point charge of +3
µ
C is located at the
origin. Determine the electric field at a point P located at coordinates
(4, 3, 0) meters.
Relevant Formula: The electric field Edue to a point charge Qat
a distance ris given by:
E=k|Q|
r2
where kis Coulomb’s constant (k≈8.988 ×109N m2/C2).
Step-by-Step Solution:
1. Identify the Charge and Distance: - Point charge Q= +3 µC =
+3 ×10−6C. - Coordinates of point P are (4, 3, 0) meters.
2. Calculate the Distance rfrom the Charge to Point P: - Use the
distance formula:
r=px2+y2+z2
- Plug in the coordinates x= 4,y= 3,z= 0:
r=p42+ 32+ 02=√16 + 9 = √25 = 5 meters
3. Calculate the Electric Field Eat Point P: - Plug rand Qinto
the electric field formula:
E=k|Q|
r2= 8.988 ×109×3×10−6
52
- Simplify the expression:
E= 8.988 ×109×3×10−6
25 = 8.988 ×109×0.12 ×10−6
E= 1.07856 ×103N/C
4. Direction of the Electric Field: - Since the charge is positive,
the electric field points away from the charge. - The direction vector
from the origin (charge location) to point P is (4,3,0). - Normalize
this vector:
Direction unit vector =4
5,3
5,0
2
- The electric field vector at P is:
E= 1.07856 ×1034
5,3
5,0N/C
E= (862.848,647.136,0) N/C
Answer: The electric field at point P is approximately 862.848 i+
647.136 j N/C pointing in the direction from the origin towards the
point (4, 3, 0).
Question 2
Problem: A long straight conductor carries a current of 5 amps.
Calculate the magnitude and direction of the magnetic field 10 cm
away from the wire.
Relevant Formula: The magnitude of the magnetic field produced
by a long straight conductor carrying a current Iat a distance rfrom
the conductor is given by Ampere’s Law:
B=µ0I
2πr
Where: - Bis the magnetic field, - µ0is the permeability of free
space (4π×10−7T·m/A), - Iis the current in amperes, - ris the
distance from the conductor in meters.
Step-by-Step Solution:
Step 1: Identify the given values. - Current, I= 5 A - Distance
from the conductor, r= 10 cm = 0.1 m
Step 2: Substitute the given values into the formula:
B=4π×10−7×5
2π×0.1
Step 3: Simplify the expression:
B=4π×10−7×5
2π×0.1=4π×10−7×5
0.2π
B=4×10−7×5
0.2= 10−5×10 = 10−4T
The magnetic field strength is 10−4Tesla.
Step 4: Determine the direction of the magnetic field using the
right-hand rule. - Point your thumb in the direction of the current
(upwards if the current flows upwards). - Curl your fingers around
the wire; where your fingers point is the direction of the magnetic
field. In this case, the magnetic field circles the wire in a clockwise
direction when viewed from above.
3
Conclusion: The magnitude of the magnetic field 10 cm away from
a long straight conductor carrying a 5 A current is 10−4Tesla. The
direction is circular around the wire, following the right-hand rule.
Question 2: Magnetic Field Due to a Long Straight Current-Carrying
Conductor
Problem: A long straight conductor carries a current of 5 amps.
Calculate the magnitude and direction of the magnetic field 10 cm
away from the wire.
Relevant Formula: The magnitude of the magnetic field produced
by a long straight conductor carrying a current Iat a distance rfrom
the conductor is given by Ampere’s Law:
B=µ0I
2πr
Where: - Bis the magnetic field, - µ0is the permeability of free
space (4π×10−7T·m/A), - Iis the current in amperes, - ris the
distance from the conductor in meters.
Step-by-Step Solution:
Step 1: Identify the given values. - Current, I= 5 A - Distance
from the conductor, r= 10 cm = 0.1 m
Step 2: Substitute the given values into the formula:
B=4π×10−7×5
2π×0.1
Step 3: Simplify the expression:
B=4π×10−7×5
2π×0.1=4π×10−7×5
0.2π
B=4×10−7×5
0.2= 10−5×10 = 10−4T
The magnetic field strength is 10−4Tesla.
Step 4: Determine the direction of the magnetic field using the
right-hand rule. - Point your thumb in the direction of the current
(upwards if the current flows upwards). - Curl your fingers around
the wire; where your fingers point is the direction of the magnetic
field. In this case, the magnetic field circles the wire in a clockwise
direction when viewed from above.
Conclusion: The magnitude of the magnetic field 10 cm away from
a long straight conductor carrying a 5 A current is 10−4Tesla. The
direction is circular around the wire, following the right-hand rule.
Question 3
Problem Statement: Two long, straight wires are parallel and 12
cm apart. They each carry a current of 5 A, flowing in the same
4
direction. Calculate the magnitude and direction of the force per
meter between the two wires.
Relevant Concepts: - Amp`ere’s force law for parallel currents
states that two parallel wires exert a force on each other. The force
per unit length (f) between two parallel current-carrying wires is
given by:
f=µ0·I1·I2
2πd
where: - µ0is the permeability of free space (4π×10−7T·m/A)
-I1and I2are the currents through the wires - dis the separation
between the wires
- The direction of the force depends on the direction of the cur-
rents: - If the currents are in the same direction, the force is attrac-
tive. - If the currents are in opposite directions, the force is repulsive.
Given Data: - I1=I2= 5 A - d= 12 cm = 0.12 m
Solution Steps:
1. Convert Units: Ensure all units are in meters and amperes.
2. Insert Values into Formula:
Substituting the given values in the formula,
f=4π×10−7·5·5
2π×0.12
3. Simplify the Expression:
Calculate the expression 4π×10−7·25 = 10−5,
f=10−5
2×0.12
f=10−5
0.24
f= 4.17 ×10−5N/m
4. Conclusion:
The magnitude of the force per meter between the wires is 4.17 ×
10−5N/m.
5. Direction of the Force:
Since both currents are flowing in the same direction, the force is
attractive. Thus, each wire will experience a force towards the other
wire.
Final Answer: The magnitude of the force per meter between the
two parallel wires is 4.17 ×10−5N/m and the direction of the force is
attractive. Question 3: Calculating the Force Between Two Parallel
Current-Carrying Wires
Problem Statement: Two long, straight wires are parallel and 12
cm apart. They each carry a current of 5 A, flowing in the same
5
direction. Calculate the magnitude and direction of the force per
meter between the two wires.
Relevant Concepts: - Amp`ere’s force law for parallel currents
states that two parallel wires exert a force on each other. The force
per unit length (f) between two parallel current-carrying wires is
given by:
f=µ0·I1·I2
2πd
where: - µ0is the permeability of free space (4π×10−7T·m/A)
-I1and I2are the currents through the wires - dis the separation
between the wires
- The direction of the force depends on the direction of the cur-
rents: - If the currents are in the same direction, the force is attrac-
tive. - If the currents are in opposite directions, the force is repulsive.
Given Data: - I1=I2= 5 A - d= 12 cm = 0.12 m
Solution Steps:
1. Convert Units: Ensure all units are in meters and amperes.
2. Insert Values into Formula:
Substituting the given values in the formula,
f=4π×10−7·5·5
2π×0.12
3. Simplify the Expression:
Calculate the expression 4π×10−7·25 = 10−5,
f=10−5
2×0.12
f=10−5
0.24
f= 4.17 ×10−5N/m
4. Conclusion:
The magnitude of the force per meter between the wires is 4.17 ×
10−5N/m.
5. Direction of the Force:
Since both currents are flowing in the same direction, the force is
attractive. Thus, each wire will experience a force towards the other
wire.
Final Answer: The magnitude of the force per meter between the
two parallel wires is 4.17 ×10−5N/m and the direction of the force is
attractive.
6
Question 4
Problem: A network consists of three capacitors: Capacitor C1=
4µF , Capacitor C2= 6 µF , and Capacitor C3= 12 µF . Capacitor C1
and Capacitor C2are connected in series, and this series combina-
tion is connected in parallel with Capacitor C3. Calculate the total
capacitance of the network.
Solution:
Step 1: Calculate the equivalent capacitance of C1and C2in series.
The formula for capacitors in series is:
1
Cseries
=1
C1
+1
C2
Substitute the values:
1
Cseries
=1
4µF +1
6µF =1
4+1
6=3
12 +2
12 =5
12
Thus,
Cseries =12
5µF = 2.4µF
Step 2: Calculate the total capacitance with C3in parallel.
The formula for capacitors in parallel is:
Cparallel =Cseries +C3
Substitute the values:
Cparallel = 2.4µF + 12 µF = 14.4µF
Answer: The total capacitance of the network is 14.4µF . Question
4: Capacitor Network
Problem: A network consists of three capacitors: Capacitor C1=
4µF , Capacitor C2= 6 µF , and Capacitor C3= 12 µF . Capacitor C1
and Capacitor C2are connected in series, and this series combina-
tion is connected in parallel with Capacitor C3. Calculate the total
capacitance of the network.
Solution:
Step 1: Calculate the equivalent capacitance of C1and C2in series.
The formula for capacitors in series is:
1
Cseries
=1
C1
+1
C2
Substitute the values:
1
Cseries
=1
4µF +1
6µF =1
4+1
6=3
12 +2
12 =5
12
7
Thus,
Cseries =12
5µF = 2.4µF
Step 2: Calculate the total capacitance with C3in parallel.
The formula for capacitors in parallel is:
Cparallel =Cseries +C3
Substitute the values:
Cparallel = 2.4µF + 12 µF = 14.4µF
Answer: The total capacitance of the network is 14.4µF .
Question 5
Problem Statement: Two long, straight wires are parallel to each
other and are separated by a distance of 0.2 meters. Wire A carries a
current of 5 A directed into the page, and Wire B carries a current of
8 A directed out of the page. Calculate the magnitude and direction
of the magnetic force per unit length between the two wires.
Relevant Concepts: The force per unit length between two parallel
currents I1and I2, separated by a distance r, is given by the formula:
F
L=µ0I1I2
2πr
where µ0(the magnetic constant) is 4π×10−7T
·
m/A.
Direction of the Force: The direction of force between two parallel
current-carrying wires depends on the directions of the currents: - If
the currents are in the same direction, the force is attractive. - If the
currents are in opposite directions, the force is repulsive.
Solution:
Step 1: Identify the currents and the distances: - Current in wire
A, IA= 5 A (into the page) - Current in wire B, IB= 8 A (out of the
page) - Distance between wires, r= 0.2m
Step 2: Apply the formula for the magnetic force per unit length:
F
L=4π×10−7×5×8
2π×0.2
F
L=20 ×10−7×40
0.4
F
L= 2 ×10−4N/m
8
Step 3: Determine the direction using the right-hand rule: - Since
currents are in opposite directions (one into the page and one out of
the page), the force between them is repulsive.
Step 4: Conclusion: The magnetic force per unit length between
the two wires is 2×10−4N/m, and the force is repulsive. Each wire
experiences a force pushing it away from the other. Question 5: Cal-
culating the Force between Two Parallel Current-Carrying Wires
Problem Statement: Two long, straight wires are parallel to each
other and are separated by a distance of 0.2 meters. Wire A carries a
current of 5 A directed into the page, and Wire B carries a current of
8 A directed out of the page. Calculate the magnitude and direction
of the magnetic force per unit length between the two wires.
Relevant Concepts: The force per unit length between two parallel
currents I1and I2, separated by a distance r, is given by the formula:
F
L=µ0I1I2
2πr
where µ0(the magnetic constant) is 4π×10−7T
·
m/A.
Direction of the Force: The direction of force between two parallel
current-carrying wires depends on the directions of the currents: - If
the currents are in the same direction, the force is attractive. - If the
currents are in opposite directions, the force is repulsive.
Solution:
Step 1: Identify the currents and the distances: - Current in wire
A, IA= 5 A (into the page) - Current in wire B, IB= 8 A (out of the
page) - Distance between wires, r= 0.2m
Step 2: Apply the formula for the magnetic force per unit length:
F
L=4π×10−7×5×8
2π×0.2
F
L=20 ×10−7×40
0.4
F
L= 2 ×10−4N/m
Step 3: Determine the direction using the right-hand rule: - Since
currents are in opposite directions (one into the page and one out of
the page), the force between them is repulsive.
Step 4: Conclusion: The magnetic force per unit length between
the two wires is 2×10−4N/m, and the force is repulsive. Each wire
experiences a force pushing it away from the other.
9
- The electric field vector at P is:
E= 1.07856 ×1034
5,3
5,0N/C
E= (862.848,647.136,0) N/C
Answer: The electric field at point P is approximately 862.848 i+
647.136 j N/C pointing in the direction from the origin towards the
point (4, 3, 0).
Question 2
Problem: A long straight conductor carries a current of 5 amps.
Calculate the magnitude and direction of the magnetic field 10 cm
away from the wire.
Relevant Formula: The magnitude of the magnetic field produced
by a long straight conductor carrying a current Iat a distance rfrom
the conductor is given by Ampere’s Law:
B=µ0I
2πr
Where: - Bis the magnetic field, - µ0is the permeability of free
space (4π×10−7T·m/A), - Iis the current in amperes, - ris the
distance from the conductor in meters.
Step-by-Step Solution:
Step 1: Identify the given values. - Current, I= 5 A - Distance
from the conductor, r= 10 cm = 0.1 m
Step 2: Substitute the given values into the formula:
B=4π×10−7×5
2π×0.1
Step 3: Simplify the expression:
B=4π×10−7×5
2π×0.1=4π×10−7×5
0.2π
B=4×10−7×5
0.2= 10−5×10 = 10−4T
The magnetic field strength is 10−4Tesla.
Step 4: Determine the direction of the magnetic field using the
right-hand rule. - Point your thumb in the direction of the current
(upwards if the current flows upwards). - Curl your fingers around
the wire; where your fingers point is the direction of the magnetic
field. In this case, the magnetic field circles the wire in a clockwise
direction when viewed from above.
3
Conclusion: The magnitude of the magnetic field 10 cm away from
a long straight conductor carrying a 5 A current is 10−4Tesla. The
direction is circular around the wire, following the right-hand rule.
Question 2: Magnetic Field Due to a Long Straight Current-Carrying
Conductor
Problem: A long straight conductor carries a current of 5 amps.
Calculate the magnitude and direction of the magnetic field 10 cm
away from the wire.
Relevant Formula: The magnitude of the magnetic field produced
by a long straight conductor carrying a current Iat a distance rfrom
the conductor is given by Ampere’s Law:
B=µ0I
2πr
Where: - Bis the magnetic field, - µ0is the permeability of free
space (4π×10−7T·m/A), - Iis the current in amperes, - ris the
distance from the conductor in meters.
Step-by-Step Solution:
Step 1: Identify the given values. - Current, I= 5 A - Distance
from the conductor, r= 10 cm = 0.1 m
Step 2: Substitute the given values into the formula:
B=4π×10−7×5
2π×0.1
Step 3: Simplify the expression:
B=4π×10−7×5
2π×0.1=4π×10−7×5
0.2π
B=4×10−7×5
0.2= 10−5×10 = 10−4T
The magnetic field strength is 10−4Tesla.
Step 4: Determine the direction of the magnetic field using the
right-hand rule. - Point your thumb in the direction of the current
(upwards if the current flows upwards). - Curl your fingers around
the wire; where your fingers point is the direction of the magnetic
field. In this case, the magnetic field circles the wire in a clockwise
direction when viewed from above.
Conclusion: The magnitude of the magnetic field 10 cm away from
a long straight conductor carrying a 5 A current is 10−4Tesla. The
direction is circular around the wire, following the right-hand rule.
Question 3
Problem Statement: Two long, straight wires are parallel and 12
cm apart. They each carry a current of 5 A, flowing in the same
4
direction. Calculate the magnitude and direction of the force per
meter between the two wires.
Relevant Concepts: - Amp`ere’s force law for parallel currents
states that two parallel wires exert a force on each other. The force
per unit length (f) between two parallel current-carrying wires is
given by:
f=µ0·I1·I2
2πd
where: - µ0is the permeability of free space (4π×10−7T·m/A)
-I1and I2are the currents through the wires - dis the separation
between the wires
- The direction of the force depends on the direction of the cur-
rents: - If the currents are in the same direction, the force is attrac-
tive. - If the currents are in opposite directions, the force is repulsive.
Given Data: - I1=I2= 5 A - d= 12 cm = 0.12 m
Solution Steps:
1. Convert Units: Ensure all units are in meters and amperes.
2. Insert Values into Formula:
Substituting the given values in the formula,
f=4π×10−7·5·5
2π×0.12
3. Simplify the Expression:
Calculate the expression 4π×10−7·25 = 10−5,
f=10−5
2×0.12
f=10−5
0.24
f= 4.17 ×10−5N/m
4. Conclusion:
The magnitude of the force per meter between the wires is 4.17 ×
10−5N/m.
5. Direction of the Force:
Since both currents are flowing in the same direction, the force is
attractive. Thus, each wire will experience a force towards the other
wire.
Final Answer: The magnitude of the force per meter between the
two parallel wires is 4.17 ×10−5N/m and the direction of the force is
attractive. Question 3: Calculating the Force Between Two Parallel
Current-Carrying Wires
Problem Statement: Two long, straight wires are parallel and 12
cm apart. They each carry a current of 5 A, flowing in the same
5
direction. Calculate the magnitude and direction of the force per
meter between the two wires.
Relevant Concepts: - Amp`ere’s force law for parallel currents
states that two parallel wires exert a force on each other. The force
per unit length (f) between two parallel current-carrying wires is
given by:
f=µ0·I1·I2
2πd
where: - µ0is the permeability of free space (4π×10−7T·m/A)
-I1and I2are the currents through the wires - dis the separation
between the wires
- The direction of the force depends on the direction of the cur-
rents: - If the currents are in the same direction, the force is attrac-
tive. - If the currents are in opposite directions, the force is repulsive.
Given Data: - I1=I2= 5 A - d= 12 cm = 0.12 m
Solution Steps:
1. Convert Units: Ensure all units are in meters and amperes.
2. Insert Values into Formula:
Substituting the given values in the formula,
f=4π×10−7·5·5
2π×0.12
3. Simplify the Expression:
Calculate the expression 4π×10−7·25 = 10−5,
f=10−5
2×0.12
f=10−5
0.24
f= 4.17 ×10−5N/m
4. Conclusion:
The magnitude of the force per meter between the wires is 4.17 ×
10−5N/m.
5. Direction of the Force:
Since both currents are flowing in the same direction, the force is
attractive. Thus, each wire will experience a force towards the other
wire.
Final Answer: The magnitude of the force per meter between the
two parallel wires is 4.17 ×10−5N/m and the direction of the force is
attractive.
6
Question 4
Problem: A network consists of three capacitors: Capacitor C1=
4µF , Capacitor C2= 6 µF , and Capacitor C3= 12 µF . Capacitor C1
and Capacitor C2are connected in series, and this series combina-
tion is connected in parallel with Capacitor C3. Calculate the total
capacitance of the network.
Solution:
Step 1: Calculate the equivalent capacitance of C1and C2in series.
The formula for capacitors in series is:
1
Cseries
=1
C1
+1
C2
Substitute the values:
1
Cseries
=1
4µF +1
6µF =1
4+1
6=3
12 +2
12 =5
12
Thus,
Cseries =12
5µF = 2.4µF
Step 2: Calculate the total capacitance with C3in parallel.
The formula for capacitors in parallel is:
Cparallel =Cseries +C3
Substitute the values:
Cparallel = 2.4µF + 12 µF = 14.4µF
Answer: The total capacitance of the network is 14.4µF . Question
4: Capacitor Network
Problem: A network consists of three capacitors: Capacitor C1=
4µF , Capacitor C2= 6 µF , and Capacitor C3= 12 µF . Capacitor C1
and Capacitor C2are connected in series, and this series combina-
tion is connected in parallel with Capacitor C3. Calculate the total
capacitance of the network.
Solution:
Step 1: Calculate the equivalent capacitance of C1and C2in series.
The formula for capacitors in series is:
1
Cseries
=1
C1
+1
C2
Substitute the values:
1
Cseries
=1
4µF +1
6µF =1
4+1
6=3
12 +2
12 =5
12
7
Thus,
Cseries =12
5µF = 2.4µF
Step 2: Calculate the total capacitance with C3in parallel.
The formula for capacitors in parallel is:
Cparallel =Cseries +C3
Substitute the values:
Cparallel = 2.4µF + 12 µF = 14.4µF
Answer: The total capacitance of the network is 14.4µF .
Question 5
Problem Statement: Two long, straight wires are parallel to each
other and are separated by a distance of 0.2 meters. Wire A carries a
current of 5 A directed into the page, and Wire B carries a current of
8 A directed out of the page. Calculate the magnitude and direction
of the magnetic force per unit length between the two wires.
Relevant Concepts: The force per unit length between two parallel
currents I1and I2, separated by a distance r, is given by the formula:
F
L=µ0I1I2
2πr
where µ0(the magnetic constant) is 4π×10−7T
·
m/A.
Direction of the Force: The direction of force between two parallel
current-carrying wires depends on the directions of the currents: - If
the currents are in the same direction, the force is attractive. - If the
currents are in opposite directions, the force is repulsive.
Solution:
Step 1: Identify the currents and the distances: - Current in wire
A, IA= 5 A (into the page) - Current in wire B, IB= 8 A (out of the
page) - Distance between wires, r= 0.2m
Step 2: Apply the formula for the magnetic force per unit length:
F
L=4π×10−7×5×8
2π×0.2
F
L=20 ×10−7×40
0.4
F
L= 2 ×10−4N/m
8
Step 3: Determine the direction using the right-hand rule: - Since
currents are in opposite directions (one into the page and one out of
the page), the force between them is repulsive.
Step 4: Conclusion: The magnetic force per unit length between
the two wires is 2×10−4N/m, and the force is repulsive. Each wire
experiences a force pushing it away from the other. Question 5: Cal-
culating the Force between Two Parallel Current-Carrying Wires
Problem Statement: Two long, straight wires are parallel to each
other and are separated by a distance of 0.2 meters. Wire A carries a
current of 5 A directed into the page, and Wire B carries a current of
8 A directed out of the page. Calculate the magnitude and direction
of the magnetic force per unit length between the two wires.
Relevant Concepts: The force per unit length between two parallel
currents I1and I2, separated by a distance r, is given by the formula:
F
L=µ0I1I2
2πr
where µ0(the magnetic constant) is 4π×10−7T
·
m/A.
Direction of the Force: The direction of force between two parallel
current-carrying wires depends on the directions of the currents: - If
the currents are in the same direction, the force is attractive. - If the
currents are in opposite directions, the force is repulsive.
Solution:
Step 1: Identify the currents and the distances: - Current in wire
A, IA= 5 A (into the page) - Current in wire B, IB= 8 A (out of the
page) - Distance between wires, r= 0.2m
Step 2: Apply the formula for the magnetic force per unit length:
F
L=4π×10−7×5×8
2π×0.2
F
L=20 ×10−7×40
0.4
F
L= 2 ×10−4N/m
Step 3: Determine the direction using the right-hand rule: - Since
currents are in opposite directions (one into the page and one out of
the page), the force between them is repulsive.
Step 4: Conclusion: The magnetic force per unit length between
the two wires is 2×10−4N/m, and the force is repulsive. Each wire
experiences a force pushing it away from the other.
9
- The electric field vector at P is:
E= 1.07856 ×1034
5,3
5,0N/C
E= (862.848,647.136,0) N/C
Answer: The electric field at point P is approximately 862.848 i+
647.136 j N/C pointing in the direction from the origin towards the
point (4, 3, 0).
Question 2
Problem: A long straight conductor carries a current of 5 amps.
Calculate the magnitude and direction of the magnetic field 10 cm
away from the wire.
Relevant Formula: The magnitude of the magnetic field produced
by a long straight conductor carrying a current Iat a distance rfrom
the conductor is given by Ampere’s Law:
B=µ0I
2πr
Where: - Bis the magnetic field, - µ0is the permeability of free
space (4π×10−7T·m/A), - Iis the current in amperes, - ris the
distance from the conductor in meters.
Step-by-Step Solution:
Step 1: Identify the given values. - Current, I= 5 A - Distance
from the conductor, r= 10 cm = 0.1 m
Step 2: Substitute the given values into the formula:
B=4π×10−7×5
2π×0.1
Step 3: Simplify the expression:
B=4π×10−7×5
2π×0.1=4π×10−7×5
0.2π
B=4×10−7×5
0.2= 10−5×10 = 10−4T
The magnetic field strength is 10−4Tesla.
Step 4: Determine the direction of the magnetic field using the
right-hand rule. - Point your thumb in the direction of the current
(upwards if the current flows upwards). - Curl your fingers around
the wire; where your fingers point is the direction of the magnetic
field. In this case, the magnetic field circles the wire in a clockwise
direction when viewed from above.
3
Conclusion: The magnitude of the magnetic field 10 cm away from
a long straight conductor carrying a 5 A current is 10−4Tesla. The
direction is circular around the wire, following the right-hand rule.
Question 2: Magnetic Field Due to a Long Straight Current-Carrying
Conductor
Problem: A long straight conductor carries a current of 5 amps.
Calculate the magnitude and direction of the magnetic field 10 cm
away from the wire.
Relevant Formula: The magnitude of the magnetic field produced
by a long straight conductor carrying a current Iat a distance rfrom
the conductor is given by Ampere’s Law:
B=µ0I
2πr
Where: - Bis the magnetic field, - µ0is the permeability of free
space (4π×10−7T·m/A), - Iis the current in amperes, - ris the
distance from the conductor in meters.
Step-by-Step Solution:
Step 1: Identify the given values. - Current, I= 5 A - Distance
from the conductor, r= 10 cm = 0.1 m
Step 2: Substitute the given values into the formula:
B=4π×10−7×5
2π×0.1
Step 3: Simplify the expression:
B=4π×10−7×5
2π×0.1=4π×10−7×5
0.2π
B=4×10−7×5
0.2= 10−5×10 = 10−4T
The magnetic field strength is 10−4Tesla.
Step 4: Determine the direction of the magnetic field using the
right-hand rule. - Point your thumb in the direction of the current
(upwards if the current flows upwards). - Curl your fingers around
the wire; where your fingers point is the direction of the magnetic
field. In this case, the magnetic field circles the wire in a clockwise
direction when viewed from above.
Conclusion: The magnitude of the magnetic field 10 cm away from
a long straight conductor carrying a 5 A current is 10−4Tesla. The
direction is circular around the wire, following the right-hand rule.
Question 3
Problem Statement: Two long, straight wires are parallel and 12
cm apart. They each carry a current of 5 A, flowing in the same
4
direction. Calculate the magnitude and direction of the force per
meter between the two wires.
Relevant Concepts: - Amp`ere’s force law for parallel currents
states that two parallel wires exert a force on each other. The force
per unit length (f) between two parallel current-carrying wires is
given by:
f=µ0·I1·I2
2πd
where: - µ0is the permeability of free space (4π×10−7T·m/A)
-I1and I2are the currents through the wires - dis the separation
between the wires
- The direction of the force depends on the direction of the cur-
rents: - If the currents are in the same direction, the force is attrac-
tive. - If the currents are in opposite directions, the force is repulsive.
Given Data: - I1=I2= 5 A - d= 12 cm = 0.12 m
Solution Steps:
1. Convert Units: Ensure all units are in meters and amperes.
2. Insert Values into Formula:
Substituting the given values in the formula,
f=4π×10−7·5·5
2π×0.12
3. Simplify the Expression:
Calculate the expression 4π×10−7·25 = 10−5,
f=10−5
2×0.12
f=10−5
0.24
f= 4.17 ×10−5N/m
4. Conclusion:
The magnitude of the force per meter between the wires is 4.17 ×
10−5N/m.
5. Direction of the Force:
Since both currents are flowing in the same direction, the force is
attractive. Thus, each wire will experience a force towards the other
wire.
Final Answer: The magnitude of the force per meter between the
two parallel wires is 4.17 ×10−5N/m and the direction of the force is
attractive. Question 3: Calculating the Force Between Two Parallel
Current-Carrying Wires
Problem Statement: Two long, straight wires are parallel and 12
cm apart. They each carry a current of 5 A, flowing in the same
5
direction. Calculate the magnitude and direction of the force per
meter between the two wires.
Relevant Concepts: - Amp`ere’s force law for parallel currents
states that two parallel wires exert a force on each other. The force
per unit length (f) between two parallel current-carrying wires is
given by:
f=µ0·I1·I2
2πd
where: - µ0is the permeability of free space (4π×10−7T·m/A)
-I1and I2are the currents through the wires - dis the separation
between the wires
- The direction of the force depends on the direction of the cur-
rents: - If the currents are in the same direction, the force is attrac-
tive. - If the currents are in opposite directions, the force is repulsive.
Given Data: - I1=I2= 5 A - d= 12 cm = 0.12 m
Solution Steps:
1. Convert Units: Ensure all units are in meters and amperes.
2. Insert Values into Formula:
Substituting the given values in the formula,
f=4π×10−7·5·5
2π×0.12
3. Simplify the Expression:
Calculate the expression 4π×10−7·25 = 10−5,
f=10−5
2×0.12
f=10−5
0.24
f= 4.17 ×10−5N/m
4. Conclusion:
The magnitude of the force per meter between the wires is 4.17 ×
10−5N/m.
5. Direction of the Force:
Since both currents are flowing in the same direction, the force is
attractive. Thus, each wire will experience a force towards the other
wire.
Final Answer: The magnitude of the force per meter between the
two parallel wires is 4.17 ×10−5N/m and the direction of the force is
attractive.
6
Question 4
Problem: A network consists of three capacitors: Capacitor C1=
4µF , Capacitor C2= 6 µF , and Capacitor C3= 12 µF . Capacitor C1
and Capacitor C2are connected in series, and this series combina-
tion is connected in parallel with Capacitor C3. Calculate the total
capacitance of the network.
Solution:
Step 1: Calculate the equivalent capacitance of C1and C2in series.
The formula for capacitors in series is:
1
Cseries
=1
C1
+1
C2
Substitute the values:
1
Cseries
=1
4µF +1
6µF =1
4+1
6=3
12 +2
12 =5
12
Thus,
Cseries =12
5µF = 2.4µF
Step 2: Calculate the total capacitance with C3in parallel.
The formula for capacitors in parallel is:
Cparallel =Cseries +C3
Substitute the values:
Cparallel = 2.4µF + 12 µF = 14.4µF
Answer: The total capacitance of the network is 14.4µF . Question
4: Capacitor Network
Problem: A network consists of three capacitors: Capacitor C1=
4µF , Capacitor C2= 6 µF , and Capacitor C3= 12 µF . Capacitor C1
and Capacitor C2are connected in series, and this series combina-
tion is connected in parallel with Capacitor C3. Calculate the total
capacitance of the network.
Solution:
Step 1: Calculate the equivalent capacitance of C1and C2in series.
The formula for capacitors in series is:
1
Cseries
=1
C1
+1
C2
Substitute the values:
1
Cseries
=1
4µF +1
6µF =1
4+1
6=3
12 +2
12 =5
12
7
Thus,
Cseries =12
5µF = 2.4µF
Step 2: Calculate the total capacitance with C3in parallel.
The formula for capacitors in parallel is:
Cparallel =Cseries +C3
Substitute the values:
Cparallel = 2.4µF + 12 µF = 14.4µF
Answer: The total capacitance of the network is 14.4µF .
Question 5
Problem Statement: Two long, straight wires are parallel to each
other and are separated by a distance of 0.2 meters. Wire A carries a
current of 5 A directed into the page, and Wire B carries a current of
8 A directed out of the page. Calculate the magnitude and direction
of the magnetic force per unit length between the two wires.
Relevant Concepts: The force per unit length between two parallel
currents I1and I2, separated by a distance r, is given by the formula:
F
L=µ0I1I2
2πr
where µ0(the magnetic constant) is 4π×10−7T
·
m/A.
Direction of the Force: The direction of force between two parallel
current-carrying wires depends on the directions of the currents: - If
the currents are in the same direction, the force is attractive. - If the
currents are in opposite directions, the force is repulsive.
Solution:
Step 1: Identify the currents and the distances: - Current in wire
A, IA= 5 A (into the page) - Current in wire B, IB= 8 A (out of the
page) - Distance between wires, r= 0.2m
Step 2: Apply the formula for the magnetic force per unit length:
F
L=4π×10−7×5×8
2π×0.2
F
L=20 ×10−7×40
0.4
F
L= 2 ×10−4N/m
8
Step 3: Determine the direction using the right-hand rule: - Since
currents are in opposite directions (one into the page and one out of
the page), the force between them is repulsive.
Step 4: Conclusion: The magnetic force per unit length between
the two wires is 2×10−4N/m, and the force is repulsive. Each wire
experiences a force pushing it away from the other. Question 5: Cal-
culating the Force between Two Parallel Current-Carrying Wires
Problem Statement: Two long, straight wires are parallel to each
other and are separated by a distance of 0.2 meters. Wire A carries a
current of 5 A directed into the page, and Wire B carries a current of
8 A directed out of the page. Calculate the magnitude and direction
of the magnetic force per unit length between the two wires.
Relevant Concepts: The force per unit length between two parallel
currents I1and I2, separated by a distance r, is given by the formula:
F
L=µ0I1I2
2πr
where µ0(the magnetic constant) is 4π×10−7T
·
m/A.
Direction of the Force: The direction of force between two parallel
current-carrying wires depends on the directions of the currents: - If
the currents are in the same direction, the force is attractive. - If the
currents are in opposite directions, the force is repulsive.
Solution:
Step 1: Identify the currents and the distances: - Current in wire
A, IA= 5 A (into the page) - Current in wire B, IB= 8 A (out of the
page) - Distance between wires, r= 0.2m
Step 2: Apply the formula for the magnetic force per unit length:
F
L=4π×10−7×5×8
2π×0.2
F
L=20 ×10−7×40
0.4
F
L= 2 ×10−4N/m
Step 3: Determine the direction using the right-hand rule: - Since
currents are in opposite directions (one into the page and one out of
the page), the force between them is repulsive.
Step 4: Conclusion: The magnetic force per unit length between
the two wires is 2×10−4N/m, and the force is repulsive. Each wire
experiences a force pushing it away from the other.
9
