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Q.

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 magnetic of the current required in the wire to remove the tension in the supporting leads is _______A.
(Take g = 10ms^-2)

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answer is 2.

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Detailed Solution

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$mg = ilB$

$\frac{40 \times 10}{10^{3}} = l \times \frac{50}{100} \times 0.40$

$l = \frac{40}{50 \times 0.40}$
= 2A

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