Q 12-04-107JEE MainJEE Main 2023 (30 Jan, Shift 1)Medium
A massless square loop, of wire of resistance $10\ \Omega$, supporting a mass of $1\ \text{g}$, hangs vertically with one of its sides in a uniform magnetic field of $10^3\ \text{G}$, directed outwards in the shaded region. A dc voltage $V$ is applied to the loop. For what value of $V$, the magnetic force will exactly balance the weight of the supporting mass of $1\ \text{g}$? (If sides of the loop $=10\ \text{cm}$, $g=10\ \text{m s}^{-2}$)
Answer: (D) $10\ \text{V}$
Only the top side ($L=0.1\ \text{m}$) is in the field $B=10^3\ \text{G}=0.1\ \text{T}$. For balance $BIL=mg$:
$$I=\frac{mg}{BL}=\frac{10^{-3}\times10}{0.1\times0.1}=1\ \text{A}$$
$V=IR=1\times10=10\ \text{V}$.
Solution by Sreeraj P, M.Sc Physics