A straight wire AB of mass 40 g and length 50 cm is suspended by a pair of flexible leads in uniform magnetic field of magnitude 0.40 T as shown in the figure. The magnitude of the current required in the wire to remove the tension in the supporting leads is ________A. (Take $\mathrm{g}=10 \mathrm{~ms}^{-2}$ ).
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Solution
For equilibrium
$$
\begin{aligned}
& M g=I \ell B \\
& I=\frac{m g}{\ell B}=\frac{40 \times 10^{-3} \times 10}{50 \times 10^{-2} \times 0.4}=2 \mathrm{~A}
\end{aligned}
$$
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