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

The problem involves a changing magnetic field confined within a circular region of radius $r$. We need to find the e.m.f. induced in two loops, one inside and the other outside this region. Here’s how we approach it:

1. Understanding the Situation:

1. A uniform magnetic field $B(t)$ is within a circular region of radius $r$ and changes with time at a rate $\frac{d \vec{B}}{dt}$.
2. Loop 1 has a radius $R$ such that $R > r$ and encloses the entire magnetic region.
3. Loop 2 is entirely outside and does not encircle any changing magnetic field.

2. Applying Faraday's Law of Electromagnetic Induction:

According to Faraday's law, the induced e.m.f. ($\varepsilon$) is given by the rate of change of magnetic flux through the loop:

$$ \varepsilon = -\frac{d\Phi}{dt} $$

where $\Phi = B \cdot A$ is the magnetic flux through the area $A$.

3. Calculating e.m.f. for Loop 1:

Loop 1 encircles the entire region of changing magnetic field:

• Area affected by the changing field = $\pi r^2$ (since only this region has a changing field)
• Induced e.m.f: $$ \varepsilon_1 = -\frac{d \vec{B}}{dt} \cdot \pi r^2 $$

4. Calculating e.m.f. for Loop 2:

Loop 2 is outside the magnetic field region. Therefore, it does not enclose any magnetic flux, and hence:

• Induced e.m.f: $$ \varepsilon_2 = 0 $$

Conclusion:

For loop 1, the e.m.f is $- \frac{d \vec{B}}{dt} \pi r^2$, and for loop 2, the e.m.f is zero. Therefore, the correct answer is:

$- \frac{d \vec{B}}{dt} \pi r^2$ in loop $1$ and zero in loop $2$