Magnetic Effects & Magnetism

JEE Main level, numerical answer, medium

186 of 276

Question 186 of 276

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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Answer: 48

F=IℓB=8×4100×0.15=48×10−3 N=48mN\begin{aligned} & F=I \ell B \\ & =8 \times \frac{4}{100} \times 0.15 \\ & =48 \times 10^{-3} \mathrm{~N}=48 \mathrm{mN}\end{aligned}

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