Question

A square loop PQRS having 10 turns, area \(3.6 \times 10^{-3} \, \text{m}^2\), and resistance \(100 \, \Omega\) is slowly and uniformly being pulled out of a uniform magnetic field of magnitude \(B = 0.5 \, \text{T}\) as shown. Work done in pulling the loop out of the field in \(1.0 \, \text{s}\) is ______ \( \times 10^{-6} \, \text{J} \).

Answer 1

Correct Answer: 3

The emf induced in the loop is:
\[\mathcal{E} = N B v \ell,\]
where:
\[v = \frac{\ell}{t}.\]
The current induced in the loop is:
\[i = \frac{\mathcal{E}}{R} = \frac{N B \ell / t}{R}.\]
The force acting is:
\[F = N \cdot i \cdot B \cdot \ell = \frac{N^2 B^2 \ell^2}{R t}.\]
The work done is:
\[W = F \cdot \ell = \frac{N^2 B^2 \ell^2}{R t} \cdot \ell = \frac{N^2 B^2 \ell^3}{R t}.\]
Substitute values:
\[W = \frac{(10)^2 (0.5)^2 (3.6 \times 10^{-3})^2}{100 \cdot 1}.\]
\[W = 3.24 \times 10^{-6} \, \text{J}.\]
Final Answer: $3.24 \times 10^{-6} \, \text{J}$.