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 ______。 Numerical answer type. Work it out, then open the answer.

Answer: 2. Only vz = 2 × 10⁻² m/s is along B; vx makes the circular motion. T = (2π m)/(qB) = 2π × 5 × 10⁻⁶5π × 10⁻⁶ × 0.1 = 20 s 5 revolutions take 100 s, so α = vz t = 2 × 10⁻² × 100 = 2 m So α = 2.

A 5 mg particle carrying a charge of 5π × 10⁻⁶ C is moving with… | JEE Main 2026 (8 Apr, Shift 2)

Q12JEE Main 2026 · 8 APR · SHIFT 2Medium

Moving Charge in a Magnetic Field

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

Numerical answer type. Work it out, then open the answer.

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Answer: 2

Only $v_z = 2 \times 10^{-2}$ m/s is along $\vec{B}$ ; $v_x$ makes the circular motion.
$T = \dfrac{2\pi m}{qB} = \dfrac{2\pi \times 5 \times 10^{-6}}{5\pi \times 10^{-6} \times 0.1} = 20$ s
5 revolutions take $100$ s, so $\alpha = v_z t = 2 \times 10^{-2} \times 100 = 2$ m
So $\alpha = 2$ .

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