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Physics · Moving Charges and Magnetic Field

JEE Main 2025 — 3 April, Morning Shift — Question 67

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 \qquad mN .

Answer: 48

Numerical answer — enter this value.

Step-by-step solution

F=i∣BF=i \mid B

=8×4×10−2×0.15 newton =8×4×10−2×0.15×103mN=32×1.5mN=48mN\begin{aligned} & =8 \times 4 \times 10^{-2} \times 0.15 \text { newton } \\ & =8 \times 4 \times 10^{-2} \times 0.15 \times 10^{3} \mathrm{mN} \\ & =32 \times 1.5 \mathrm{mN} \\ & =48 \mathrm{mN} \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
3 April, Morning Shift
Subject
Physics
Chapter
Moving Charges and Magnetic Field
Topic
Force and Torque on Wires and Loops, Magnetic Dipole Moment

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