Question

A charge of 8.4 × 10–4 C moves at an angle of 35° to a magnetic field that has a field strength of 6.7 × 10–3 T.

If the magnetic force is 3.5 × 10–2 N, how fast is the charge

Answer 1

Answer:10842.33m/s

Step-by-step explanation:

F=qvBsine

V=f/(qBsine)

V=(3.5×10^-2)÷(8.4×10^-4×6.7×10^-3×sin35)

V=10842.33m/s