A 5 mg particle carrying a charge of 5π× 10^-6 C is moving with velocity of (3 i+2 k)× 10^-2 m/s in a region having magnetic field B = 0.1 k Wb/m^2.…
JEE Main 2026 — Physics Electromagnetism
2026integermedium
A 5 mg particle carrying a charge of 5π×10−6 C is moving with velocity of (3i^+2k^)×10−2 m/s in a region having magnetic field B=0.1k^ Wb/m2. It moves a distance of α meter along k^ when it completes 5 revolutions. The value of α is ________.
Official previous-year question
Held on 8 Apr 2026 · Verified 6 Jul 2026.
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Solution
Given:
Mass of the particle, m=5 mg=5×10−6 kg
Charge, q=5π×10−6 C
Velocity, v=(3i^+2k^)×10−2 m/s
Magnetic field, B=0.1k^ Wb/m2
The velocity component parallel to the magnetic field is v∥=2×10−2 m/s.
The time period of one revolution is given by:
T=qB2πm
Substituting the given values:
T=5π×10−6×0.12π×5×10−6=0.12=20 s
The time taken to complete 5 revolutions is:
t=5T=5×20=100 s
The distance moved along the k^ direction (pitch for 5 revolutions) is:
α=v∥×t
α=2×10−2×100=2 m
Answer: 2
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