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A $2.5 m$ long straight wire having mass of $500 g$ is suspended in mid air by a uniform horizontal magnetic field $B$ . If a current of $4 A$ is passing through the wire then the magnitude of the field is (Take $g = 10 m s^{- 2}$ ).

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$0.5 T$

$0.6 T$

$0.25 T$

$0.8 T$

answer is A.

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Detailed Solution

Here, $m = 500 g = 0.5 k g, I = 4 A, l = 2.5 m$

As $F = I l B \sin θ m g = I l B \sin 90 ° (∵ θ = 90 °, F = m g ∴ B = \frac{m g}{I l} = \frac{0.5 \times 10}{4 \times 2.5} = 0.5 T$

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