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A straight horizontal wire of mass 10 mg and length 1m carries a current 2A. What minimum magnetic field should be applied in the region so that the magnetic force on the wire may balance its weight

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$4.9 \times 10^{- 5} T$

$3.92 \times 10^{- 5} T$

$9.8 \times 10^{- 5} T$

$1.96 \times 10^{- 5} T$

answer is A.

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Detailed Solution

Given data

Mass $= 10 m g$

Length $L = 1 m$

Current $I = 2 A$

Concepts used moving charges and magnetism

Explanation

Required magnetic field

$F_{g} = F_{m}$

$M g = I B L \sin θ$

$B = \frac{M g}{I L}$

$B = \frac{10^{- 5} \times 9.8}{2 \times 1}$

By solving we get

$B = 4.9 \times 10^{- 5}$

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