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?
The magnetic force -
F=q(v x B)
the magnitude-
F=qvBsin .............(1)
where is angle between velocity- of charge and the magnetic field=57.4^{°}
velocity (v)=5.16x10^{6}m/s
charge (q)=-5.92μC=-5.92x10^{-6}C
magnitude of force (F)=6.76x10^{-3}N
using equation (1)-
6.76x10^{-3} =5.92x10^{-6} x 5.16x10^{6} xBsin(57.4°)
B=2.63x10^{-4}tesla Ans.
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