Q87 / 90 · JEE Main 2024 (05 Apr Shift 1) · Physics
A 2 A current carrying straight metal wire of resistance $1 \Omega$, resistivity $2 \times 10^{-6} \Omega \mathrm{m}$, area of cross-section $10 \mathrm{~mm}^2$ and mass $500 \mathrm{~g}$ is suspended horizontally in mid air by applying a uniform magnetic field $\vec{B}$. The magnitude of $B$ is _____$\times 10^{-1} \mathrm{~T}$ (given, $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ ).
Correct answer: 5
Solution
$\begin{aligned} & \mathrm{R}=\frac{\rho \ell}{\mathrm{A}} \Rightarrow \frac{2 \times 10^{-6} \times \ell}{10^{-5}}=1 \Rightarrow \ell=5 \\ & \mathrm{mg}=\mathrm{Bi} \ell\end{aligned}$
$\mathrm{B}=\frac{\mathrm{mg}}{\mathrm{i} \ell}=\frac{5}{2 \times 5}=0.5=5 \times 10^{-1} \mathrm{Tesla}$
Hence, the correct answer is 5.