COMEDK 2024 Shift 1 Paper

The magnetic field inside the solenoid is given by
B=µ0nI=µ0‌

N

L

I
where, N is the number of turns, L is the length of the solenoid, I is the current flowing through the solenoid, and µ0 is the permeability of free space.
The magnetic flux through the smaller coil is given by
φ=BA=µ0‌

N

L

IA
where A is the area of the smaller coil.
The induced emf in the smaller coil is given by Faraday's law of electromagnetic induction as
e=−‌

dφ

dt

Substituting the value of φ in the above equation, we get
e=−‌

d

dt

(µ0‌

N

L

IA)=−µ0‌

N

L

A‌

dI

dt

Given that the current is reversed in 0.2s, the rate of change of current is
‌

dI

dt

=‌

2−(−2)

0.2

=20A∕ s
Substituting the values, we get
Therefore, the induced emf produced is 1.32×10−1V.