Question

A square loop of area \(25\text{ cm}^2\) has a resistance of \(10\) \(\Omega\). The loop is placed in uniform magnetic field of magnitude \(40\) T. The plane of the loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in one second will be

• A. \(2.5\times 10^{-3}\) J
• B. \(1.0\times 10^{-3}\) J
• C. \(1.0\times 10^{-4}\) J
• D. \(5\times 10^{-3}\) J

Hint:

Find emf $=\frac{BA}{t}$, then work $=\frac{\varepsilon^2}{R}t$.

Answer 1

The Correct Option is B

Step 1: Find the emf
\(A=25\times10^{-4}\) m\(^2\). Flux \(=BA=40\times25\times10^{-4}=0.1\) Wb. Removing it in \(1\) s gives \(\varepsilon=0.1\) V.

Step 2: Power and work
\(P=\frac{\varepsilon^2}{R}=\frac{0.01}{10}=10^{-3}\) W. Work \(=Pt=10^{-3}\times1=1.0\times10^{-3}\) J. Option (B).

Final Answer:
The work done is \(1.0\times10^{-3}\) J, option (B). \[ \boxed{\text{(B)}} \]