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Magnetic Force on Current Carrying Wire

A horizontal ...

Question

A horizontal rod of 0.200 m long is mounted on a balance and carries a current. At the location of the rod a uniform horizontal magnetic field has magnitude 0.067 T and direction perpendicular to the rod.The magnetic force on the rod is measured by the balance and is found to be 0.13 N. The current is $970 \times 10^{- x} A$ . Find x.

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Solution

$F = i l B s i n 90^{o}$
$i = \frac{F}{l B} = \frac{0.13}{0.2 \times 0.067}$

$= 9.7 A$

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