Q 12-06-064JEE MainJEE Main 2023 (29 Jan, Shift 2)Medium
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 $40.0\ \text{T}$. The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in $1.0\ \text{s}$, will be
Answer: (B) $1.0\times10^{-3}\ \text{J}$
Flux change $\Delta\phi=BA=40\times25\times10^{-4}=0.1\ \text{Wb}$, removed uniformly in $1\ \text{s}$, so $\varepsilon=0.1\ \text{V}$.
The work done equals the heat produced:
$$W=\frac{\varepsilon^2}{R}t=\frac{0.01}{10}\times1=1.0\times10^{-3}\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics