When at rest, a proton experiences a net electromagnetic force of magnitude 8.1×

Question

When at rest, a proton experiences a net electromagnetic force of magnitude 8.1×1013 Npointing in the positive x direction. When the proton moves with a speed of 1.8×106 m/s in the positive y direction, the net electromagnetic force on it decreases in magnitude to 7.0×1013 N , still pointing in the positive x direction. A) Find the magnitude of the electric field. B) Find the direction of the electric field. C) Find the magnitude of the magnetic field. D) Find the direction of the magnetic field.

Explanation / Answer

force is F = 8.1*10^-13 N

speed v = 1.8 *10^6 m/s

the electric field is,

            E = F/q = 8.1*10^-13 N / 1.6*10^-19 C = 5.06x10^6 N/C

Direction: electric field is directed along the positivex-axis

(b).

the net force is,

    F'= 7.0×1013 N - 8.1×1013 = -1.1×1013 N

The magnetic field is,

   B = F/qv = -1.1×1013 N / 1.6*10^-19 C* 1.8×106 = -0.382 T

Magntiude: 0.382 T

direction: it is directed along the negative z-axis.

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