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JEE Mains · Physics · STD 12 - Moving charges and magnetism
  • JEE Mains 2025
  • Easy
  • Numerical

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 & Solution

Correct Answer

(C) 48

Step-by-step Solution

Detailed explanation

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