A 4.0 cm long straight wire carrying a current of 8A is placed perpendicular to an uniform magnetic field of strength 0.15 T. The magnetic force on the wire is ______ mN.
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Correct Answer: 48
Solution and Explanation
$F = IlB$
$F = 8 \times \frac{4}{100} \times 0.15$
$F = \frac{48 \times 100}{10000} N$
$F = 48 \times 10^{-3} N$
$F = 48 \text{ mN}$
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