JEE Main Previous Year Questions (2025): Electromagnetic Induction

 Page 1


JEE Main Previous Year Questions 
(2025): Electromagnetic Induction 
Q1: A rectangular metallic loop is moving out of a uniform magnetic field region to a 
field free region with a constant speed. When the loop is partially inside the magnate 
field, the plot of magnitude of induced emf (?? ) with time (?? ) is given by 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A.  
B.  
C.  
D. 
Ans: B 
Solution: 
 
Page 2


JEE Main Previous Year Questions 
(2025): Electromagnetic Induction 
Q1: A rectangular metallic loop is moving out of a uniform magnetic field region to a 
field free region with a constant speed. When the loop is partially inside the magnate 
field, the plot of magnitude of induced emf (?? ) with time (?? ) is given by 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A.  
B.  
C.  
D. 
Ans: B 
Solution: 
 
Motional emf : ?? = Blv = c constant 
 
Q2: Regarding self-inductance: 
A. The self-inductance of the coil depends on its geometry. 
B. Self-inductance does not depend on the permeability of the medium. 
C. Self-induced e.m.f. opposes any change in the current in a circuit. 
D. Self-inductance is electromagnetic analogue of mass in mechanics. 
E. Work needs to be done against self-induced e.m.f. in establishing the current. 
Choose the correct answer from the options given below: 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. A, B, C, E only 
B. A, B, C, D only 
C. A, C, D, E only 
D. B, C, D, E only 
Ans: C 
Solution: 
Let's analyze each statement: 
A. The self-inductance of the coil depends on its geometry. 
This is true because the self-inductance of a coil is given by formulas that include its geometrical 
parameters (number of turns, cross-sectional area, length, etc.). For example, for a solenoid, 
?? = ?? 0
?? ?? ?? 2
?? ?? , 
where ?? is the number of turns, ?? is the cross-sectional area, and ?? is the length. 
B. Self-inductance does not depend on the permeability of the medium. 
This is false. The permeability of the medium (represented by ?? = ?? 0
?? ?? ) directly influences the 
inductance, as seen in the formula above. So, any change in the medium's permeability will 
affect the inductance. 
C. Self-induced e.m.f. opposes any change in the current in a circuit. 
This is true, and it is a direct consequence of Lenz's law. The induced electromotive force (e.m.f.) 
always acts in a direction such that it opposes the change in current that produced it. 
D. Self-inductance is the electromagnetic analogue of mass in mechanics. 
This is true. Just as mass resists changes in velocity (inertia), inductance resists changes in 
current, which is why it is often compared to the inertial mass in mechanical systems. 
E. Work needs to be done against self-induced e.m.f. in establishing the current. 
This is true because, when you try to change the current, you must do work to overcome the 
opposing selfinduced e.m.f., storing energy in the magnetic field of the inductor. 
Based on the above reasoning, the true statements are A,C,D, and E . 
Thus, the correct answer is: 
Page 3


JEE Main Previous Year Questions 
(2025): Electromagnetic Induction 
Q1: A rectangular metallic loop is moving out of a uniform magnetic field region to a 
field free region with a constant speed. When the loop is partially inside the magnate 
field, the plot of magnitude of induced emf (?? ) with time (?? ) is given by 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A.  
B.  
C.  
D. 
Ans: B 
Solution: 
 
Motional emf : ?? = Blv = c constant 
 
Q2: Regarding self-inductance: 
A. The self-inductance of the coil depends on its geometry. 
B. Self-inductance does not depend on the permeability of the medium. 
C. Self-induced e.m.f. opposes any change in the current in a circuit. 
D. Self-inductance is electromagnetic analogue of mass in mechanics. 
E. Work needs to be done against self-induced e.m.f. in establishing the current. 
Choose the correct answer from the options given below: 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. A, B, C, E only 
B. A, B, C, D only 
C. A, C, D, E only 
D. B, C, D, E only 
Ans: C 
Solution: 
Let's analyze each statement: 
A. The self-inductance of the coil depends on its geometry. 
This is true because the self-inductance of a coil is given by formulas that include its geometrical 
parameters (number of turns, cross-sectional area, length, etc.). For example, for a solenoid, 
?? = ?? 0
?? ?? ?? 2
?? ?? , 
where ?? is the number of turns, ?? is the cross-sectional area, and ?? is the length. 
B. Self-inductance does not depend on the permeability of the medium. 
This is false. The permeability of the medium (represented by ?? = ?? 0
?? ?? ) directly influences the 
inductance, as seen in the formula above. So, any change in the medium's permeability will 
affect the inductance. 
C. Self-induced e.m.f. opposes any change in the current in a circuit. 
This is true, and it is a direct consequence of Lenz's law. The induced electromotive force (e.m.f.) 
always acts in a direction such that it opposes the change in current that produced it. 
D. Self-inductance is the electromagnetic analogue of mass in mechanics. 
This is true. Just as mass resists changes in velocity (inertia), inductance resists changes in 
current, which is why it is often compared to the inertial mass in mechanical systems. 
E. Work needs to be done against self-induced e.m.f. in establishing the current. 
This is true because, when you try to change the current, you must do work to overcome the 
opposing selfinduced e.m.f., storing energy in the magnetic field of the inductor. 
Based on the above reasoning, the true statements are A,C,D, and E . 
Thus, the correct answer is: 
Option C 
A, C, D, E only 
Q3: A uniform magnetic field of 0.4 T acts perpendicular to a circular copper disc 20 cm 
in radius. The disc is having a uniform angular velocity of ???? ?? ?????? ?? -?? about an axis 
through its centre and perpendicular to the disc. What is the potential difference 
developed between the axis of the disc and the rim? (?? = ?? .???? ) 
JEE Main 2025 (Online) 28th January Evening Shift 
Options: 
A. 0.5024 V 
B. 0.0628 V 
C. 0.2512 V 
D. 0.1256 V 
Ans: C 
Solution: 
To determine the potential difference between the center and the rim of the disc (often called a 
Faraday disc or unipolar generator), we use the formula for motional EMF in a rotating disc: 
?? =
1
2
???? ?? 2
 
where: 
?? is the magnetic field strength, 
?? is the angular velocity, 
?? is the radius of the disc. 
Given: 
?? = 0.4 T, 
?? = 10?? rad/s, 
?? = 20 cm = 0.2 m, 
we substitute these values into the formula: 
First calculate ?? 2
 : 
?? 2
= (0.2 m)
2
= 0.04 m
2
 
Substitute into the formula: 
?? =
1
2
× 0.4 × (10?? ) × 0.04 
Multiply step-by-step: 
0.4 × 10?? = 4?? , 
Then, 4?? × 0.04 = 0.16?? , 
Finally, multiply by 
1
2
 : 
?? =
1
2
× 0.16?? = 0.08?? 
Substitute ?? ˜ 3.14 : 
?? ˜ 0.08× 3.14 = 0.2512 V 
Thus, the potential difference developed between the axis and the rim is approximately 0.2512 
V , which corresponds to Option C. 
Page 4


JEE Main Previous Year Questions 
(2025): Electromagnetic Induction 
Q1: A rectangular metallic loop is moving out of a uniform magnetic field region to a 
field free region with a constant speed. When the loop is partially inside the magnate 
field, the plot of magnitude of induced emf (?? ) with time (?? ) is given by 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A.  
B.  
C.  
D. 
Ans: B 
Solution: 
 
Motional emf : ?? = Blv = c constant 
 
Q2: Regarding self-inductance: 
A. The self-inductance of the coil depends on its geometry. 
B. Self-inductance does not depend on the permeability of the medium. 
C. Self-induced e.m.f. opposes any change in the current in a circuit. 
D. Self-inductance is electromagnetic analogue of mass in mechanics. 
E. Work needs to be done against self-induced e.m.f. in establishing the current. 
Choose the correct answer from the options given below: 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. A, B, C, E only 
B. A, B, C, D only 
C. A, C, D, E only 
D. B, C, D, E only 
Ans: C 
Solution: 
Let's analyze each statement: 
A. The self-inductance of the coil depends on its geometry. 
This is true because the self-inductance of a coil is given by formulas that include its geometrical 
parameters (number of turns, cross-sectional area, length, etc.). For example, for a solenoid, 
?? = ?? 0
?? ?? ?? 2
?? ?? , 
where ?? is the number of turns, ?? is the cross-sectional area, and ?? is the length. 
B. Self-inductance does not depend on the permeability of the medium. 
This is false. The permeability of the medium (represented by ?? = ?? 0
?? ?? ) directly influences the 
inductance, as seen in the formula above. So, any change in the medium's permeability will 
affect the inductance. 
C. Self-induced e.m.f. opposes any change in the current in a circuit. 
This is true, and it is a direct consequence of Lenz's law. The induced electromotive force (e.m.f.) 
always acts in a direction such that it opposes the change in current that produced it. 
D. Self-inductance is the electromagnetic analogue of mass in mechanics. 
This is true. Just as mass resists changes in velocity (inertia), inductance resists changes in 
current, which is why it is often compared to the inertial mass in mechanical systems. 
E. Work needs to be done against self-induced e.m.f. in establishing the current. 
This is true because, when you try to change the current, you must do work to overcome the 
opposing selfinduced e.m.f., storing energy in the magnetic field of the inductor. 
Based on the above reasoning, the true statements are A,C,D, and E . 
Thus, the correct answer is: 
Option C 
A, C, D, E only 
Q3: A uniform magnetic field of 0.4 T acts perpendicular to a circular copper disc 20 cm 
in radius. The disc is having a uniform angular velocity of ???? ?? ?????? ?? -?? about an axis 
through its centre and perpendicular to the disc. What is the potential difference 
developed between the axis of the disc and the rim? (?? = ?? .???? ) 
JEE Main 2025 (Online) 28th January Evening Shift 
Options: 
A. 0.5024 V 
B. 0.0628 V 
C. 0.2512 V 
D. 0.1256 V 
Ans: C 
Solution: 
To determine the potential difference between the center and the rim of the disc (often called a 
Faraday disc or unipolar generator), we use the formula for motional EMF in a rotating disc: 
?? =
1
2
???? ?? 2
 
where: 
?? is the magnetic field strength, 
?? is the angular velocity, 
?? is the radius of the disc. 
Given: 
?? = 0.4 T, 
?? = 10?? rad/s, 
?? = 20 cm = 0.2 m, 
we substitute these values into the formula: 
First calculate ?? 2
 : 
?? 2
= (0.2 m)
2
= 0.04 m
2
 
Substitute into the formula: 
?? =
1
2
× 0.4 × (10?? ) × 0.04 
Multiply step-by-step: 
0.4 × 10?? = 4?? , 
Then, 4?? × 0.04 = 0.16?? , 
Finally, multiply by 
1
2
 : 
?? =
1
2
× 0.16?? = 0.08?? 
Substitute ?? ˜ 3.14 : 
?? ˜ 0.08× 3.14 = 0.2512 V 
Thus, the potential difference developed between the axis and the rim is approximately 0.2512 
V , which corresponds to Option C. 
Q4: Consider ?? ?? and ?? ?? are the currents flowing simultaneously in two nearby coils 
?? &?? , respectively. If ?? ?? = self inductance of coil ?? ,?? ????
= mutual inductance of coil 1 
with respect to coil 2 , then the value of induced emf in coil 1 will be : 
JEE Main 2025 (Online) 29th January Morning Shift 
Options: 
A. e
1
= -L
1
?? ?? 2
????
- M
12
?? ?? 1
????
 
B. e
1
= -L
1
?? ?? 1
????
+ M
12
?? ?? 2
????
 
C. e
1
= -L
1
?? ?? 1
????
- M
12
?? ?? 1
????
 
D. e
1
= -L
1
?? ?? 1
????
- M
12
?? ?? 2
????
 
Ans: D 
Solution: 
?? 1
= L
1
I
1
+ M
12
I
2
?? 1
= -
d?? 1
dt
= -L
1
dII
1
dt
- M
12
dI
dt
 
Q5: A coil of area ?? and ?? turns is rotating with angular velocity ?? in a uniform 
magnetic field ?? ?? 
 about an axis perpendicular to ?? ?? 
. Magnetic flux ?? and induced emf ?? 
across it, at an instant when ?? ?? 
 is parallel to the plane of coil, are : 
JEE Main 2025 (Online) 29th January Morning Shift 
Options: 
A. ?? = ???? ,?? = ???????? 
B. ?? = ???? ,?? = 0 
C. ?? = 0,?? = 0 
D. ?? = 0,?? = ?????? ?? 
Ans: D 
Solution: 
Page 5


JEE Main Previous Year Questions 
(2025): Electromagnetic Induction 
Q1: A rectangular metallic loop is moving out of a uniform magnetic field region to a 
field free region with a constant speed. When the loop is partially inside the magnate 
field, the plot of magnitude of induced emf (?? ) with time (?? ) is given by 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A.  
B.  
C.  
D. 
Ans: B 
Solution: 
 
Motional emf : ?? = Blv = c constant 
 
Q2: Regarding self-inductance: 
A. The self-inductance of the coil depends on its geometry. 
B. Self-inductance does not depend on the permeability of the medium. 
C. Self-induced e.m.f. opposes any change in the current in a circuit. 
D. Self-inductance is electromagnetic analogue of mass in mechanics. 
E. Work needs to be done against self-induced e.m.f. in establishing the current. 
Choose the correct answer from the options given below: 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. A, B, C, E only 
B. A, B, C, D only 
C. A, C, D, E only 
D. B, C, D, E only 
Ans: C 
Solution: 
Let's analyze each statement: 
A. The self-inductance of the coil depends on its geometry. 
This is true because the self-inductance of a coil is given by formulas that include its geometrical 
parameters (number of turns, cross-sectional area, length, etc.). For example, for a solenoid, 
?? = ?? 0
?? ?? ?? 2
?? ?? , 
where ?? is the number of turns, ?? is the cross-sectional area, and ?? is the length. 
B. Self-inductance does not depend on the permeability of the medium. 
This is false. The permeability of the medium (represented by ?? = ?? 0
?? ?? ) directly influences the 
inductance, as seen in the formula above. So, any change in the medium's permeability will 
affect the inductance. 
C. Self-induced e.m.f. opposes any change in the current in a circuit. 
This is true, and it is a direct consequence of Lenz's law. The induced electromotive force (e.m.f.) 
always acts in a direction such that it opposes the change in current that produced it. 
D. Self-inductance is the electromagnetic analogue of mass in mechanics. 
This is true. Just as mass resists changes in velocity (inertia), inductance resists changes in 
current, which is why it is often compared to the inertial mass in mechanical systems. 
E. Work needs to be done against self-induced e.m.f. in establishing the current. 
This is true because, when you try to change the current, you must do work to overcome the 
opposing selfinduced e.m.f., storing energy in the magnetic field of the inductor. 
Based on the above reasoning, the true statements are A,C,D, and E . 
Thus, the correct answer is: 
Option C 
A, C, D, E only 
Q3: A uniform magnetic field of 0.4 T acts perpendicular to a circular copper disc 20 cm 
in radius. The disc is having a uniform angular velocity of ???? ?? ?????? ?? -?? about an axis 
through its centre and perpendicular to the disc. What is the potential difference 
developed between the axis of the disc and the rim? (?? = ?? .???? ) 
JEE Main 2025 (Online) 28th January Evening Shift 
Options: 
A. 0.5024 V 
B. 0.0628 V 
C. 0.2512 V 
D. 0.1256 V 
Ans: C 
Solution: 
To determine the potential difference between the center and the rim of the disc (often called a 
Faraday disc or unipolar generator), we use the formula for motional EMF in a rotating disc: 
?? =
1
2
???? ?? 2
 
where: 
?? is the magnetic field strength, 
?? is the angular velocity, 
?? is the radius of the disc. 
Given: 
?? = 0.4 T, 
?? = 10?? rad/s, 
?? = 20 cm = 0.2 m, 
we substitute these values into the formula: 
First calculate ?? 2
 : 
?? 2
= (0.2 m)
2
= 0.04 m
2
 
Substitute into the formula: 
?? =
1
2
× 0.4 × (10?? ) × 0.04 
Multiply step-by-step: 
0.4 × 10?? = 4?? , 
Then, 4?? × 0.04 = 0.16?? , 
Finally, multiply by 
1
2
 : 
?? =
1
2
× 0.16?? = 0.08?? 
Substitute ?? ˜ 3.14 : 
?? ˜ 0.08× 3.14 = 0.2512 V 
Thus, the potential difference developed between the axis and the rim is approximately 0.2512 
V , which corresponds to Option C. 
Q4: Consider ?? ?? and ?? ?? are the currents flowing simultaneously in two nearby coils 
?? &?? , respectively. If ?? ?? = self inductance of coil ?? ,?? ????
= mutual inductance of coil 1 
with respect to coil 2 , then the value of induced emf in coil 1 will be : 
JEE Main 2025 (Online) 29th January Morning Shift 
Options: 
A. e
1
= -L
1
?? ?? 2
????
- M
12
?? ?? 1
????
 
B. e
1
= -L
1
?? ?? 1
????
+ M
12
?? ?? 2
????
 
C. e
1
= -L
1
?? ?? 1
????
- M
12
?? ?? 1
????
 
D. e
1
= -L
1
?? ?? 1
????
- M
12
?? ?? 2
????
 
Ans: D 
Solution: 
?? 1
= L
1
I
1
+ M
12
I
2
?? 1
= -
d?? 1
dt
= -L
1
dII
1
dt
- M
12
dI
dt
 
Q5: A coil of area ?? and ?? turns is rotating with angular velocity ?? in a uniform 
magnetic field ?? ?? 
 about an axis perpendicular to ?? ?? 
. Magnetic flux ?? and induced emf ?? 
across it, at an instant when ?? ?? 
 is parallel to the plane of coil, are : 
JEE Main 2025 (Online) 29th January Morning Shift 
Options: 
A. ?? = ???? ,?? = ???????? 
B. ?? = ???? ,?? = 0 
C. ?? = 0,?? = 0 
D. ?? = 0,?? = ?????? ?? 
Ans: D 
Solution: 
 
?? = BAN· cos? (?? t)
?? =
-d?? dt
= BA?? N · sin? (?? t)
 
When ?? is parallel to plane, ?? t =
?? 2
 
? ?? = 0,?? = BA?? N 
 
Q6:  
 
In the given circuit the sliding contact is pulled outwards such that electric current in 
the circuit changes at the rate of ?? ?? /?? . At an instant when ?? is ?????? , the value of the 
current in the circuit will be ____ A. 
JEE Main 2025 (Online) 23rd January Morning Shift 
Ans: 3 
Solution: