A 2 A current carrying straight metal... |JEE Main 2024| EMF

JEE MAIN 2024

05-04-24 S1

Question

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 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, $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

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Solution

$\begin{aligned} & \mathrm{R}=\frac{\rho \ell}{\mathrm{A}} \Rightarrow \frac{2 \times 10^{-6} \times \ell}{10^{-5}}=1 \Rightarrow \ell=5 \\ & \mathrm{mg}=\mathrm{Bi} \ell \\ & \mathrm{B}=\frac{\mathrm{mg}}{\mathrm{i} \ell}=\frac{5}{2 \times 5}=0.5=5 \times 10^{-1} \text { Tesla }\end{aligned}$

Question Tags

JEE Main

Physics

Medium

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