A square loop PQRS having 10 turns, area $3.6\times10^{-3}\ \text{m}^2$ 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 ______ $\times10^{-6}\ \text{J}$.
Numerical value type. Enter your answer.
Answer: 3.24
Side $a = \sqrt{3.6\times10^{-3}} = 0.06\ \text{m}$. The loop leaves the field in $1\ \text{s}$, so $v = 0.06\ \text{m s}^{-1}$.
$$\varepsilon = NBav = 10\times0.5\times0.06\times0.06 = 0.018\ \text{V}$$
All the work done appears as heat:
$$W = \frac{\varepsilon^2}{R}t = \frac{(0.018)^2}{100}\times1 = 3.24\times10^{-6}\ \text{J}$$
To the nearest integer, the answer is 3.
Solution by Sreeraj P, M.Sc Physics