A 10 C charge moves with a velocity of 1.0 x 105 ms^ -1 at right angle to a uniform magnetic field of flux…

Analyze the question: A charged particle moves perpendicular to a magnetic field. We need the magnetic force on it.

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

F = qvB sinθ

When the velocity is at right angles to the field, θ = 90° and sin 90° = 1, so F = qvB.

Given:

• q = 10 μC = 10 × 10⁻⁶ C = 1.0 × 10⁻⁵ C
• v = 1.0 × 10⁵ m/s
• B = 5.0 × 10⁻⁴ T

Solution:

F = (1.0 × 10⁻⁵)(1.0 × 10⁵)(5.0 × 10⁻⁴)

Combine the powers of 10 first: 10⁻⁵ × 10⁵ = 10⁰ = 1

F = 1 × 5.0 × 10⁻⁴ = 5.0 × 10⁻⁴ N

Tip: Always convert μC to coulombs first (1 μC = 10⁻⁶ C). The most common mistake is forgetting this, which gives an answer off by a factor of a million. Also remember that "at right angle" means sin θ = 1, so you can ignore the sine.