JEE MAIN - Physics (2024 - 5th April Morning Shift - No. 30)
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
5
Explanation
$$\begin{aligned}
& i L B=m g \text { and } L=\frac{A R}{\rho} \\
& \begin{aligned}
\therefore B & =\frac{m g \rho}{i A R} \\
& =\frac{0.5 \times 10 \times 2 \times 10^{-6}}{2 \times 10 \times 10^{-6} \times 1} \\
& =0.5 \mathrm{~T}
\end{aligned}
\end{aligned}$$
Comments (0)
