Question

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20 Q.11 When at rest a proton experiences a net electromagnetic force of magnitude 8.0 x 10" N pointing in the positive x direction. When the proton moves with a speed of 1.5 x 10 m/s in the positive y direction, the net electromagnetic force on it decreases in magnitude to 7.5 x 10" N still pointing in the positive x direction Find the magnitude and direction of (a) the electric field and (b) the magnetic field.

Answer 1

a] when at rest, only electrostatic force will act

Force F = qE

or E = F/q = 8e-13/1.6e-19

E = 5000000 i^ N/C

b] when it is moving, qE + q v x B = 7.5e-13 i

8e-13 i + q v x B =  7.5e-13 i

1.6e-19*(1.5e6 j) x B = -0.5e-13 i

B = - 0.5e-13/(1.5e6*1.6e-19) k^

B = - 0.2083 k^ T