A 2A 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, $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$).
Answer (integer)
5
Solution
<p>$$\begin{aligned}
& i L B=m g \text { and } L=\frac{A R}{\rho} \\
& \begin{aligned}
\therefore B & =\frac{m g \rho}{i A R} \\
& =\frac{0.5 \times 10 \times 2 \times 10^{-6}}{2 \times 10 \times 10^{-6} \times 1} \\
& =0.5 \mathrm{~T}
\end{aligned}
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Magnetic Effects of Current · Topic: Biot-Savart Law
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