When at rest, a proton experiences a net electromagnetic force of magnitude 8.8×

Question

When at rest, a proton experiences a net electromagnetic force of magnitude 8.8×1013 Npointing in the positive x direction. When the proton moves with a speed of 1.6×106 m/s in the positive y direction, the net electromagnetic force on it decreases in magnitude to 7.3×1013 N , still pointing in the positive x direction.

A) Find the magnitude of the electric field.

E = _______ N/C

B) Find the direction of the electric field.

C) Find the magnitude of the magnetic field.

B = ______ T

D) Find the direction of the magnetic field.

positive x direction negative x direction positive y direction negative y direction positive z direction negative z direction

Explanation / Answer

a )

given

FB = 8.8 X 1013 N

V = 1.6 X 106 m/s

E = ( 8.8 X 1013 / 1.6 X 10-19 ) i

E = ( 5.5 X 106 N/C ) i

direction is n X - axis

or

positive x direction

b )

V = 1.6 X 106 m/s

F ' = 7.3 X 1013 N

FB - B V q = F '

8.8 X 1013 - B X 1.6 X 106 X 1.6 X 10-19 = 7.3 X 1013

B = ( 7.3 X 1013 - 8.8 X 1013 ) / 1.6 X 106 X 1.6 X 10-19

B = 0.585 T

the direction of magnetic field will be in negative Z - direction

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