Question

What is the magnitude of the field if a charge of 10 C travelling at a speed of 8 × 106 m/s in a direction perpendicular to a magnetic field experiences a magnetic force of 8.7 × 10-3 N?

Answer 1

The magnetic force

$$F=qvB\sin\theta$$

Hence

$$B=\frac{F}{qv\sin\theta}\\ =\frac{8.7 \times10^{-3}}{10\times 8\times 10^6\times \sin90^\circ}=1.1\times 10^{-10}\:\rm T$$