A particle carrying a charge of 1. 0 x 10-8C enters a magnetic field at 3.0x 10-2 ms -1 at right angles to the…

Analyze the question

A charged particle moves at right angles to a magnetic field. We are given the charge, speed and force, and we need the magnetic field strength B.

Concept

The force on a moving charge in a magnetic field is:

F = Bqv sin θ

Since the particle moves at right angles to the field, θ = 90° and sin 90° = 1, so:

F = Bqv, which gives B = F / (qv)

Solution

Given:

• q = 1.0 × 10⁻⁸ C
• v = 3.0 × 10⁻² m/s
• F = 1.8 × 10⁻⁸ N

B = (1.8 × 10⁻⁸) / (1.0 × 10⁻⁸ × 3.0 × 10⁻²)

Denominator: 1.0 × 10⁻⁸ × 3.0 × 10⁻² = 3.0 × 10⁻¹⁰

B = (1.8 × 10⁻⁸) / (3.0 × 10⁻¹⁰) = 0.6 × 10² = 60 T = 6.0 × 10¹ T

Identify the answer

B = 6.0 × 10 T, which matches option B.

(The numbers are unrealistic, since 60 T is a huge field, but the calculation is what the question asks for.)

Tip

Handle the powers of 10 separately from the numbers: 1.8/3.0 = 0.6, and 10⁻⁸ / 10⁻¹⁰ = 10². Always write the formula first, F = Bqv for perpendicular motion, and rearrange before substituting. Also check whether the exponents in the options are positive or negative, because that is where most mistakes happen.