Question

7.55-μC charged particle with a speed of 30.50 m/s is found in a uniform magnetic field with magnitude 1.2 T. Solve for the magnitude of the magnetic force exerted on the charged particle if the particle is moving perpendicular to the field.

Answer 1

The Lorentz force is

$F = q[v \times B]$

If the particle is moving perpendicular to the field, $|v \times B| = vB$

$F = 7.55 \cdot 10^{-6} \cdot 30.50 \cdot 1.2 = 276.33 \cdot 10^{-6} = 0.276\; mN$

Answer: 0.276 mN