Magnetic Effects & Magnetism

JEE Main level, numerical answer, medium

106 of 276

Question 106 of 276

A 55 mg particle carrying a charge of 5π×10−65\pi\times 10^{-6} C is moving with velocity of (3i^+2k^)×10−2(3\hat{i}+2\hat{k})\times 10^{-2} m/s in a region having magnetic field B⃗=0.1k^\vec{B} = 0.1\hat{k} Wb/m2^2. It moves a distance of α\alpha meter along k^\hat{k} when it completes 55 revolutions. The value of α\alpha is ________.

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Answer: 2

Given:
Mass of the particle, m=5 mg=5×10−6 kgm = 5 \text{ mg} = 5 \times 10^{-6} \text{ kg}
Charge, q=5π×10−6 Cq = 5\pi \times 10^{-6} \text{ C}
Velocity, v⃗=(3i^+2k^)×10−2 m/s\vec{v} = (3\hat{i} + 2\hat{k}) \times 10^{-2} \text{ m/s}
Magnetic field, B⃗=0.1k^ Wb/m2\vec{B} = 0.1\hat{k} \text{ Wb/m}^2

The velocity component parallel to the magnetic field is v∥=2×10−2 m/sv_{\parallel} = 2 \times 10^{-2} \text{ m/s}.

The time period of one revolution is given by:
T=2πmqBT = \dfrac{2\pi m}{qB}

Substituting the given values:
T=2π×5×10−65π×10−6×0.1=20.1=20 sT = \dfrac{2\pi \times 5 \times 10^{-6}}{5\pi \times 10^{-6} \times 0.1} = \dfrac{2}{0.1} = 20 \text{ s}

The time taken to complete 55 revolutions is:
t=5T=5×20=100 st = 5T = 5 \times 20 = 100 \text{ s}

The distance moved along the k^\hat{k} direction (pitch for 55 revolutions) is:
α=v∥×t\alpha = v_{\parallel} \times t
α=2×10−2×100=2 m\alpha = 2 \times 10^{-2} \times 100 = 2 \text{ m}

Answer: 22

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