Electromagnetic Induction and Alternating Currents: JEE Main Previous Year Questions (2021-2026)

 Page 1


JEE Main Previous Year Questions (2021-2026): 
Electromagnetic Induction and Alternating 
Currents  
 
(January 2026) 
 
 
 
Electromagnetic Induction 
 
Q1: Match the List-I with List-II  
 
Choose the correct answer from the options given below: 
(a) A-III, B-IV, C-II, D-I 
(b) A-I, B-III, C-IV, D-II 
(c) A-IV, B-III, C-I, D-II 
(d) A-III, B-IV, C-I, D-II 
Ans: (d) 
Sol: 
The force on a current-carrying wire is given by F = BII sin  ?  
 
Dimensions of force [F] = [MLT
-2
]  
Dimensions of current [I] = [A]  
Dimension of length [l] = [L]  
Dimension of trigonometric function is [M
0
 L
0
 T
0
], so [sin ?] = [M
0
 L
0
 T
0
]  
So, the dimension of magnetic field is, 
Page 2


JEE Main Previous Year Questions (2021-2026): 
Electromagnetic Induction and Alternating 
Currents  
 
(January 2026) 
 
 
 
Electromagnetic Induction 
 
Q1: Match the List-I with List-II  
 
Choose the correct answer from the options given below: 
(a) A-III, B-IV, C-II, D-I 
(b) A-I, B-III, C-IV, D-II 
(c) A-IV, B-III, C-I, D-II 
(d) A-III, B-IV, C-I, D-II 
Ans: (d) 
Sol: 
The force on a current-carrying wire is given by F = BII sin  ?  
 
Dimensions of force [F] = [MLT
-2
]  
Dimensions of current [I] = [A]  
Dimension of length [l] = [L]  
Dimension of trigonometric function is [M
0
 L
0
 T
0
], so [sin ?] = [M
0
 L
0
 T
0
]  
So, the dimension of magnetic field is, 
 
Magnetic flux is the product of magnetic field and area: F = B · A  
Dimensions of magnetic induction [B] = [MT
-2 
A
-1
]  
Dimensions of area [A] = [L
2
]  
So, the dimension of magnetic flux is,  
 
From Ampere's Law or the force between two parallel wires : 
 
Dimensions of Force [F] = [MLT
-2
]  
Dimensions of distance [r] = [L]  
Dimensions of 2 and p is [M
0
 L
0
 T
0
]  
Dimensions of current [I] = [A]  
Dimensions of length [l] = [L]  
So, the dimensions of magnetic permeability is, 
 
The energy stored in an inductor is  
Page 3


JEE Main Previous Year Questions (2021-2026): 
Electromagnetic Induction and Alternating 
Currents  
 
(January 2026) 
 
 
 
Electromagnetic Induction 
 
Q1: Match the List-I with List-II  
 
Choose the correct answer from the options given below: 
(a) A-III, B-IV, C-II, D-I 
(b) A-I, B-III, C-IV, D-II 
(c) A-IV, B-III, C-I, D-II 
(d) A-III, B-IV, C-I, D-II 
Ans: (d) 
Sol: 
The force on a current-carrying wire is given by F = BII sin  ?  
 
Dimensions of force [F] = [MLT
-2
]  
Dimensions of current [I] = [A]  
Dimension of length [l] = [L]  
Dimension of trigonometric function is [M
0
 L
0
 T
0
], so [sin ?] = [M
0
 L
0
 T
0
]  
So, the dimension of magnetic field is, 
 
Magnetic flux is the product of magnetic field and area: F = B · A  
Dimensions of magnetic induction [B] = [MT
-2 
A
-1
]  
Dimensions of area [A] = [L
2
]  
So, the dimension of magnetic flux is,  
 
From Ampere's Law or the force between two parallel wires : 
 
Dimensions of Force [F] = [MLT
-2
]  
Dimensions of distance [r] = [L]  
Dimensions of 2 and p is [M
0
 L
0
 T
0
]  
Dimensions of current [I] = [A]  
Dimensions of length [l] = [L]  
So, the dimensions of magnetic permeability is, 
 
The energy stored in an inductor is  
 
Dimensions of energy [U] = [ML
2
 T
-2
]  
Dimensions of current [I] = [A]  
So, the dimensions of self-inductance is, 
 
Hence, the correct match is A-III, B-IV, C-I, D-II.  
Therefore, the correct option is (d). 
 
Q2: A circular loop of radius 7 cm is placed in uniform magnetic field of 0.2 T directed 
perpendicular to plane of loop. The loop is converted into a square loop in 0.5s. The EMF 
induced in the loop is ______ mV.  
(a) 13.2  
(b) 6.6  
(c) 1.32  
(d) 8.25  
Ans: (c) 
Sol: 
According to Faraday's Law of Electromagnetic Induction, the induced EMF (E) is equal to the 
rate of change of magnetic flux (?) through the loop.  
Initially circular loop is with radius r = 7 cm = 0.07 m.  
Finally a square loop formed from the same wire.  
Magnetic field B = 0.2 T (constant and perpendicular).  
Time interval = ? t = 0.5 s.  
Initial Area (A
1
)  
  
The perimeter of the square must be equal to the circumference of the circle because the wire 
length is constant. 
 
So, final area (A
2
)  
a
2
 = (0.11)
2
 = 0.0121 m
2
.  
Since the field is perpendicular, ? = B · A.  
So, the change in magnetic flux is, 
Page 4


JEE Main Previous Year Questions (2021-2026): 
Electromagnetic Induction and Alternating 
Currents  
 
(January 2026) 
 
 
 
Electromagnetic Induction 
 
Q1: Match the List-I with List-II  
 
Choose the correct answer from the options given below: 
(a) A-III, B-IV, C-II, D-I 
(b) A-I, B-III, C-IV, D-II 
(c) A-IV, B-III, C-I, D-II 
(d) A-III, B-IV, C-I, D-II 
Ans: (d) 
Sol: 
The force on a current-carrying wire is given by F = BII sin  ?  
 
Dimensions of force [F] = [MLT
-2
]  
Dimensions of current [I] = [A]  
Dimension of length [l] = [L]  
Dimension of trigonometric function is [M
0
 L
0
 T
0
], so [sin ?] = [M
0
 L
0
 T
0
]  
So, the dimension of magnetic field is, 
 
Magnetic flux is the product of magnetic field and area: F = B · A  
Dimensions of magnetic induction [B] = [MT
-2 
A
-1
]  
Dimensions of area [A] = [L
2
]  
So, the dimension of magnetic flux is,  
 
From Ampere's Law or the force between two parallel wires : 
 
Dimensions of Force [F] = [MLT
-2
]  
Dimensions of distance [r] = [L]  
Dimensions of 2 and p is [M
0
 L
0
 T
0
]  
Dimensions of current [I] = [A]  
Dimensions of length [l] = [L]  
So, the dimensions of magnetic permeability is, 
 
The energy stored in an inductor is  
 
Dimensions of energy [U] = [ML
2
 T
-2
]  
Dimensions of current [I] = [A]  
So, the dimensions of self-inductance is, 
 
Hence, the correct match is A-III, B-IV, C-I, D-II.  
Therefore, the correct option is (d). 
 
Q2: A circular loop of radius 7 cm is placed in uniform magnetic field of 0.2 T directed 
perpendicular to plane of loop. The loop is converted into a square loop in 0.5s. The EMF 
induced in the loop is ______ mV.  
(a) 13.2  
(b) 6.6  
(c) 1.32  
(d) 8.25  
Ans: (c) 
Sol: 
According to Faraday's Law of Electromagnetic Induction, the induced EMF (E) is equal to the 
rate of change of magnetic flux (?) through the loop.  
Initially circular loop is with radius r = 7 cm = 0.07 m.  
Finally a square loop formed from the same wire.  
Magnetic field B = 0.2 T (constant and perpendicular).  
Time interval = ? t = 0.5 s.  
Initial Area (A
1
)  
  
The perimeter of the square must be equal to the circumference of the circle because the wire 
length is constant. 
 
So, final area (A
2
)  
a
2
 = (0.11)
2
 = 0.0121 m
2
.  
Since the field is perpendicular, ? = B · A.  
So, the change in magnetic flux is, 
 
The magnitude of the induced EMF is : 
 
Therefore, the induced EMF is 1.32 mV . 
Hence, the correct option is (c). 
 
Q3: Suppose a long solenoid of 100 cm length, radius 2 cm having 500 turns per unit 
length, carries a current I = 10 sin  (?t) A, where ? = 1000 rad. /s. A circular conducting 
loop (B) of radius 1 cm coaxially slided through the solenoid at a speed v = 1 cm / s. The 
r.m.s. current through the loop when the coil B is inserted 10 cm inside the solenoid is 
 The value of a is ______.  
[Resistance of the loop = 10 O]  
(a) 80  
(b) 280  
(c) 100  
(d) 197 
Ans: (d) 
Sol: 
For a long solenoid, the magnetic field B inside is uniform and given by :  
B = µ
0
 nI  
Where :  
µ
0
 = 4p × 10
-7
 T · m / A  
n = 500 turns/unit length. 
I = 10 sin (1000t)A 
Page 5


JEE Main Previous Year Questions (2021-2026): 
Electromagnetic Induction and Alternating 
Currents  
 
(January 2026) 
 
 
 
Electromagnetic Induction 
 
Q1: Match the List-I with List-II  
 
Choose the correct answer from the options given below: 
(a) A-III, B-IV, C-II, D-I 
(b) A-I, B-III, C-IV, D-II 
(c) A-IV, B-III, C-I, D-II 
(d) A-III, B-IV, C-I, D-II 
Ans: (d) 
Sol: 
The force on a current-carrying wire is given by F = BII sin  ?  
 
Dimensions of force [F] = [MLT
-2
]  
Dimensions of current [I] = [A]  
Dimension of length [l] = [L]  
Dimension of trigonometric function is [M
0
 L
0
 T
0
], so [sin ?] = [M
0
 L
0
 T
0
]  
So, the dimension of magnetic field is, 
 
Magnetic flux is the product of magnetic field and area: F = B · A  
Dimensions of magnetic induction [B] = [MT
-2 
A
-1
]  
Dimensions of area [A] = [L
2
]  
So, the dimension of magnetic flux is,  
 
From Ampere's Law or the force between two parallel wires : 
 
Dimensions of Force [F] = [MLT
-2
]  
Dimensions of distance [r] = [L]  
Dimensions of 2 and p is [M
0
 L
0
 T
0
]  
Dimensions of current [I] = [A]  
Dimensions of length [l] = [L]  
So, the dimensions of magnetic permeability is, 
 
The energy stored in an inductor is  
 
Dimensions of energy [U] = [ML
2
 T
-2
]  
Dimensions of current [I] = [A]  
So, the dimensions of self-inductance is, 
 
Hence, the correct match is A-III, B-IV, C-I, D-II.  
Therefore, the correct option is (d). 
 
Q2: A circular loop of radius 7 cm is placed in uniform magnetic field of 0.2 T directed 
perpendicular to plane of loop. The loop is converted into a square loop in 0.5s. The EMF 
induced in the loop is ______ mV.  
(a) 13.2  
(b) 6.6  
(c) 1.32  
(d) 8.25  
Ans: (c) 
Sol: 
According to Faraday's Law of Electromagnetic Induction, the induced EMF (E) is equal to the 
rate of change of magnetic flux (?) through the loop.  
Initially circular loop is with radius r = 7 cm = 0.07 m.  
Finally a square loop formed from the same wire.  
Magnetic field B = 0.2 T (constant and perpendicular).  
Time interval = ? t = 0.5 s.  
Initial Area (A
1
)  
  
The perimeter of the square must be equal to the circumference of the circle because the wire 
length is constant. 
 
So, final area (A
2
)  
a
2
 = (0.11)
2
 = 0.0121 m
2
.  
Since the field is perpendicular, ? = B · A.  
So, the change in magnetic flux is, 
 
The magnitude of the induced EMF is : 
 
Therefore, the induced EMF is 1.32 mV . 
Hence, the correct option is (c). 
 
Q3: Suppose a long solenoid of 100 cm length, radius 2 cm having 500 turns per unit 
length, carries a current I = 10 sin  (?t) A, where ? = 1000 rad. /s. A circular conducting 
loop (B) of radius 1 cm coaxially slided through the solenoid at a speed v = 1 cm / s. The 
r.m.s. current through the loop when the coil B is inserted 10 cm inside the solenoid is 
 The value of a is ______.  
[Resistance of the loop = 10 O]  
(a) 80  
(b) 280  
(c) 100  
(d) 197 
Ans: (d) 
Sol: 
For a long solenoid, the magnetic field B inside is uniform and given by :  
B = µ
0
 nI  
Where :  
µ
0
 = 4p × 10
-7
 T · m / A  
n = 500 turns/unit length. 
I = 10 sin (1000t)A 
 
The loop is inside the solenoid, so it experiences this magnetic field. The flux through the 
circular loop of radius r = 1 cm = 0.01 m is: 
 
According to Faraday's Law, the induced EMF is   
Since the loop is already 10 cm inside a 100 cm long solenoid, it is in the region where the 
magnetic field is uniform. When the loop is inside the magnetic field then its motion along axis of 
solenoid will not cause any change in effective magnetic flux through loop because loop is 
co-axial with the solenoid.