Question

A charge of -5.92 μC is traveling at a speed of 5.16 × 106 m/s in a region of space where there is a magnetic field. The angle between the velocity of the charge and the field is 57.4o. A force of magnitude6.76 × 10-3 N acts on the charge. What is the magnitude of the magnetic field?

Answer 1

The magnetic force -

F=q(v x​​​​​​ B)

the magnitude-

F=qvBsin$\theta$ .............(1)

where $\theta$ is angle between velocity- of charge and the magnetic field=57.4^°

velocity (v)=5.16x10⁶m/s

charge (q)=-5.92μC=-5.92x10^-6C

magnitude of force (F)=6.76x10^-3N

using equation (1)-

6.76x10^-3 =5.92x10^-6 x 5.16x10⁶ xBsin(57.4°)

B=2.63x10^-4tesla Ans.