What distance does particle travel along k̂ in 5 revolutions?

2m. T=20s (period). v_∥=2×10⁻² m/s. d=v_∥×5T=2×10⁻²×100=2m.

What is the period of revolution?

T=2πm/(qB)=2π×5×10⁻⁶/(5π×10⁻⁶×0.1)=20s.

Charged particle q=5π×10⁻⁶C m=5mg v=(3î+2k̂)×10⁻²m/s B=0.1k̂ — distance along k̂ in 5 revolutions

JEERANKERS

PhysicsMagnetism

QInteger TypeMagnetism

A 5 mg particle with charge $5\pi\times10^{-6}$ C moves with velocity $(3\hat{i}+2\hat{k})\times10^{-2}$ m/s in $\vec{B}=0.1\hat{k}$ T. Distance traveled along $\hat{k}$ in 5 revolutions is $\alpha$ m. Find $\alpha$.

✅ Correct Answer

α = 2

✓

Solution

Decompose velocity relative to B

$\vec{B}=0.1\hat{k}$. Perpendicular component: $v_\perp=3\times10^{-2}$ m/s ($\hat{i}$ direction). Parallel component: $v_\parallel=2\times10^{-2}$ m/s ($\hat{k}$ direction).

Find period of revolution

$T=\dfrac{2\pi m}{qB}=\dfrac{2\pi\times5\times10^{-6}}{5\pi\times10^{-6}\times0.1}=\dfrac{2\cancel{\pi}\times5\times10^{-6}}{5\cancel{\pi}\times10^{-7}}=20\text{ s}$

Distance along k̂ in 5 revolutions

$d=v_\parallel\times5T=2\times10^{-2}\times5\times20=\boxed{2\text{ m}},\quad\alpha=2$

📘 Helical Motion: v_∥ × nT gives distance along B

Particle in magnetic field with velocity component parallel to $\vec{B}$ follows a helix. Period $T=2\pi m/(qB)$ independent of speed. Distance along $\vec{B}$ = $v_\parallel\times nT$.

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Important Concepts

Helical Motionv_⊥ causes circular motion in xy-plane. v_∥=v_k causes uniform linear motion along k̂. Combined: helix.

Period FormulaT=2πm/(qB). Independent of speed! Here: T=2π×5e-6/(5π×1e-6×0.1). π cancels: T=20s.

5 RevolutionsTime for 5 revolutions=5T=100s. Distance along k̂=v_∥×100=2×10⁻²×100=2m.

α=2The distance along k̂ direction in 5 revolutions is 2 metres.

FAQs

Why does T not depend on speed?

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T=2πm/(qB) — no v in formula. This is why cyclotrons work: all particles complete one revolution in the same time.

What is v_⊥?

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v_⊥=3×10⁻² m/s (x-component). This determines the radius r=mv_⊥/(qB)=5e-6×3e-2/(5π×1e-6×0.1)≈0.0955m.

How does π cancel in T?

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T=2π×5×10⁻⁶/(5π×10⁻⁶×0.1)=(2π/5π)×(5/0.5)=(2/5)×10=4... wait: T=2π×5e-6/(5π×1e-7)=2×5e-6/(5×1e-7)=10e-6/5e-7=20s. Confirmed.

What is the pitch?

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Pitch=v_∥×T=2×10⁻²×20=0.4m per revolution. 5 revolutions: 5×0.4=2m ✓

Is this from JEE Main 2026?

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Yes, this appeared in JEE Main 2026 Section B.

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