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A wire of $62 c m$ length and $13 g$ mass is suspended by a pair of flexible leads in a uniform magnetic field of magnitude $0.440 T$ in figure. What are the magnitude and direction of current required to remove the tension in the supporting leads? Take $g = 10 m / s^{2}$ .
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$0.47 A$ from left to right

$0.47 A$ from right to left

$0.27 A$ from left to right

$0.27 A$ from right to left

answer is A.

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Detailed Solution

$W = F_{m}$ or $m g = i l B \sin 90 °$

$∴ i = \frac{m g}{B l} = \frac{(13 \times 10^{- 3}) (10)}{0.44 \times 0.62} = 0.47 A$

Magnetic force should be upwards to balance the weight. Hence, from Fleming's left hand rule we can see that direction of current should be from left to right.

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