Question

A uniform magnetic field is restricted within a region of radius $r$. The magnetic field changes with time at a rate $\frac{d \vec{B}}{dt}$. Loop $1$ of radius $R > r$ enclosed the region $r$ and loop $2$ of radius $R$ is outside the region of magnetic field as shown in the figure below. Then the e.m.f. generated is

• A. Zero in loop $1$ and zero in loop $2$
• B. $- \frac{d \vec{B}}{dt} \pi r^2$ in loop $1$ and $- \frac{d \vec{B}}{dt} \pi r^2$ in loop $2$
• C. $- \frac{d \vec{B}}{dt} \pi R^2$ in loop $1$ and zero in loop $2$
• D. $- \frac{d \vec{B}}{dt} \pi r^2$ in loop $1$ and zero in loop $2$

Answer 1

The Correct Option is D

Induced Electromotive Force (e.m.f.) Calculation 

Let's calculate the induced e.m.f. in two different loops based on the magnetic field changes.

Induced e.m.f. in Loop (1)

The induced e.m.f. in loop (1) is given by Faraday's law of induction:

\(e = - \frac{d\phi}{dt} = - \frac{A dB}{dt}\)

Where: - \( e \) is the induced e.m.f., - \( \phi \) is the magnetic flux, - \( A \) is the area of the loop, - \( B \) is the magnetic field strength, and - \( \frac{dB}{dt} \) is the rate of change of the magnetic field.

For a circular loop of radius \( r \), the area \( A \) is given by \( A = \pi r^2 \). Substituting this into the equation, we get:

\(e = - \pi r^{2} \frac{dB}{dt}\)

Induced e.m.f. in Loop (2)

For loop (2), which is not in the magnetic field, there is no change in the magnetic field through the loop. Since there is no change in magnetic flux, the induced e.m.f. is zero:

\(Induced e.m.f. = 0\)

Conclusion:

Loop (1) experiences an induced e.m.f. based on the change in the magnetic field, while loop (2) does not experience any induced e.m.f. as it is not in the magnetic field.