A proton at rest in some region of space where there is an e

A proton, at rest in some region of space where there is an electric and magnetic field, experiences a net electromagnetic force of 8.0 x 10-13 N pointing in the positive-x direction. If the same proton then moves with a speed of 1.5 x 106 m/s in the positive-y direction, the net electromagnetic force on it decreases in magnitude to 7.5 x 10-13 N, still pointing in the positive-x direction. Find the magnitude and direction of both the electric field and the magnetic field in which this proton is moving.

Solution

When at rest, the only force acting on the proton is the electric force

So, Eq = 8*10^-13 N

where E = eclectric field

q = 1.6*10^-19 C

So, E = 8*10^-13/(1.6*10^-19)

So, E = 5*10^6 N/C directed in positive x -direction

Now, when moving in +y direction,

Magnetic force acting on it, Fm = q*v*B

Now for decrease in net force, the magnetic force Fm must be directed in -ve x direction

So, for this to happen, magnetic field B must be directed in - z axis

So, -Fm + Eq = 7.5*10^-13

So, -qvB + 8*10^-13 = 7.5*10^-13

So, -1.6*10^-19*1.5*10^6*B + 8*10^-13 = 7.5*10^-13

So, B = 0.21 T

So, Electric field = 5*10^6 N/C directed in +x direction <------answer

Magnetic field = 0.21 T directed in -z direction <-------answer