As we know, radius of the circular path is given by,
r=qBmv=qB2mK=qB2×m×qV
⇒r2=qB22mV
⇒m=2Vr2qB2=2×100(3×10−2)2×(2×10−6)×(4×10−3)2=144×10−18kg
JEE Main 2023 — Physics Electromagnetism
A charge particle of 2μC accelerated by a potential difference of 100V enters a region of uniform magnetic field of magnitude 4mT at right angle to the direction of field. The charge particle completes semicircle of radius 3cm inside magnetic field. The mass of the charge particle is ______ ×10−18kg.
Held on 1 Feb 2023 · Verified 6 Jul 2026.
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