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Q.

A horizontal long straight wire placed east -west carries a current of 3.6 A. The distance of the null point from the wire if the horizontal component of the earth's magnetic induction from south to north is $B_{H} = 3 .6 \times 10^{- 5} T$ is

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a

$2 \times 10^{- 2} cm$

b

$1 .2 \times 10^{- 2} m$

c

$1 \times 10^{- 2} cm$

d

$2 \times 10^{- 2} m$

answer is B.

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Detailed Solution

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$B = \frac{μ_{0} i}{2 πr} = B_{H}; 3 .6 \times 10^{- 5} = \frac{4 π \times 10^{- 7} \times 3 .6}{2 π \times r}$
$r = 2 \times 10^{- 2} m$

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A horizontal long straight wire placed east -west carries a current of 3.6 A. The distance of the null point from the wire if the horizontal component of the earth's magnetic induction from south to north is BH=3.6×10−5T is