Given:
q=4μC,v=4×106j^ms−1 and B=2k^T
Force acting on charged particle, F=q(v×B)
=4×10−6(4×106j^×2k^)
=4×10−6×8×106i^
⇒F=32i^N
⇒x=32
JEE Main 2024 — Physics Electromagnetism
A charge of 4.0μC is moving with a velocity of 4.0×106ms−1 along the positive y-axis under a magnetic field B of strength (2k^)T. The force acting on the charge is xi^N. The value of x is ______.
Held on 29 Jan 2024 · Verified 6 Jul 2026.
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