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JEE Main 2025 (3 Apr, Shift 1) · Force on Current-Carrying Conductors

Force on Current-Carrying Conductors question from JEE Main 2025 (3 Apr, Shift 1)

Q11JEE Main 2025 · 3 APR · SHIFT 1Easy

Force on Current-Carrying Conductors

A 4.0 cm long straight wire carrying a current of 8 A is placed perpendicular to a uniform magnetic field of strength 0.15 T. The magnetic force on the wire is ________mN.

Numerical answer type. Work it out, then open the answer.

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Answer: 48

$F = BIL\sin 90^\circ$
$F = 0.15 \times 8 \times 0.04 = 0.048\text{ N}$
So $F = 48\text{ mN}$ .

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