A 4.0 cm long straight wire carrying a current of 8A is placed perpendicular to an uniform magnetic field of…

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.

Solution

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

Asked in: JEE Main 2025 (03 Apr Shift 1)

Practice more Magnetic Fields due to Electric Current questions on Aicharya