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JEE Main 20253 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
3 April
Shift
Morning Shift
Subject
Physics
Chapter
Magnetic Effects of Current

JEE Main 2025 · 3 April · Morning Shift · Physics · Magnetic Effects of Current

A 4.0 cm long straight wire carrying a current of 8A is placed...

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.

Correct answer

0

Step-by-step solution

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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