HC Verma Chapter 34 Problem 2

Problem Statement

A particle of charge $q = 1.6\times10^{-19}\,\text{C}$ moves with velocity $v = 3\times10^6\,\text{m s}^{-1}$ perpendicular to a magnetic field $B = 0.3\,\text{T}$. Find the magnetic force.

Given Information

$q = 1.6\times10^{-19}\,\text{C}$
$v = 3\times10^6\,\text{m s}^{-1}$
$B = 0.3\,\text{T}$
$\theta = 90°$

Physical Concepts & Formulas

The Lorentz magnetic force $F = qvB\sin\theta$ acts perpendicular to both velocity and field. It does no work (always perpendicular to motion), only changes direction — causing circular motion.

$F = qvB\sin\theta$ — Lorentz force (max when $\theta=90°$)

Step-by-Step Solution

Step 1 — Apply Lorentz force (θ=90°): $$F = qvB$$

Step 2 — Substitute: $$F = 1.6\times10^{-19}\times3\times10^6\times0.3$$

Step 3 — Compute: $$F = 1.44e-13\,\text{N}$$

Worked Calculation

$$F = 1.6\times10^{-19}\times3\times10^6\times0.3 = 1.44e-13\,\text{N}$$

Answer

$$\boxed{F = 1.44e-13\,\text{N}}$$

The magnetic force 1.44e-13 N deflects the particle into circular motion. This is the principle behind cyclotrons, mass spectrometers, and the aurora borealis (charged particles spiraling in Earth’s field).