The magnetic field inside the solenoid is given by
B=µ0nI=µ0‌Iwhere,
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=µ0‌IAwhere
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=−‌Substituting the value of
φ in the above equation, we get
e=−‌(µ0‌IA)=−µ0‌A‌ Given that the current is reversed in
0.2s, the rate of change of current is
‌=‌=20A∕ sSubstituting the values, we get
Therefore, the induced emf produced is
1.32×10−1V.