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JEE MAIN 2026 · Magnetic Effects of Current and Magnetism

JEE Main 2026 Session 2 Question Paper: 8 Apr, Shift 2 (B.E./B.Tech), question 48

Updated
Numerical answer+4 / -1Answer matches NTA’s final key

A 5 mg particle carrying a charge of 5π×10−65\pi \times 10^{-6} C is moving with velocity of (3i^+2k^)×10−2\left(3\hat i + 2\hat k\right) \times 10^{-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 ______.

Your answer

Solution

Solution

The velocity component along B⃗\vec B (0.020.02 m/s along k^\hat k) is unaffected, so the path is a helix.

Period of one revolution, with m=5×10−6m = 5 \times 10^{-6} kg:

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

Pitch =0.02×20=0.4= 0.02 \times 20 = 0.4 m, so five revolutions cover 5×0.4=25 \times 0.4 = 2 m along k^\hat k.

Answer: α=2\alpha = 2.

Answer checked against the official answer key. The solution is our own working.

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