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JEE Main 2026 · 8 Apr · Shift 2, Physics question 48

A 5 mg particle carrying a charge of 5π×10−65\pi\times10^{-6} C is moving with velocity of (3i^+2k^)×10−2\left(3\hat i+2\hat k\right)\times10^{-2} m/s in a region having magnetic field B⃗=0.1 k^\vec B=0.1\,\hat k Wb/m². It moves a distance of α\alpha meter along k^\hat k when it completes 5 revolutions. The value of α\alpha is ________.

Answer and worked solution

Official answer: 2

Period.
T=2πmqB=2π×5×10−65π×10−6×0.1=20 s.T=\frac{2\pi m}{qB}=\frac{2\pi\times5\times10^{-6}}{5\pi\times10^{-6}\times0.1}=20\text{ s}.

Advance along k^\hat k. v∥=0.02v_\parallel=0.02 m/s for 5 revolutions (100 s): 2 m.

Answer: 2

Topics:Force on a moving charge
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