Question

An elastic conducting material is stretched into a circular loop. It is placed with its plane perpendicular to uniform magnetic field \(\mathrm{B}=0.6 \mathrm{~T}\). When loop is released, the radius starts shrinking at constant rate \(3 \mathrm{~cm} / \mathrm{s}\). The induced e.m.f. in the loop at an instant when the radius of the loop is 10 cm will be (in mV )

• A. \(2.6\pi\)
• B. \(3.6\pi\)
• C. \(3.2\pi\)
• D. \(1.8\pi\)

Hint:

Induced emf equals B times the rate of change of loop area, which depends on the radius and its shrinking rate.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Flux \(\Phi = B\pi r^2\), so \(\varepsilon = \left|\dfrac{d\Phi}{dt}\right| = 2\pi B r\left|\dfrac{dr}{dt}\right|\).

Step 2: Substitute
\(B = 0.6\ \text{T}\), \(r = 0.1\ \text{m}\), \(\left|\dfrac{dr}{dt}\right| = 0.03\ \text{m/s}\)

Step 3: Calculation
\(\varepsilon = 2\pi\times0.6\times0.1\times0.03 = 3.6\times10^{-3}\pi\ \text{V}\)

Step 4: Result
\(\varepsilon = 3.6\pi\ \text{mV}\), option B.

Final Answer:
The induced emf is 3.6 pi millivolt. \[ \boxed{\text{(B) }3.6\pi} \]