A proton travels through uniform magnetic and electric fields. The magnetic fiel

Question

A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 2.07 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2180 m/s. At that instant, what is the magnitude of the net force acting on the proton if the electric field is (a) in the positive z direction and has a magnitude of 3.22 V/m, (b) in the negative z direction and has a magnitude of 3.22 V/m, and (c) in the positive x direction and has a magnitude of 3.22 V/m?

Explanation / Answer

Magnetic force acting on a charge (q) partical is given as

F=q*(V*B)

where
V=velocity of particle

B=magnetic field strength

Magnetic force

F = 1.6*10^-19*(2180*2.07*10^-3)

= 7.22*10^-19 N
  
and directed in the positive z direction

(a)Electric force = 3.22*1.6*10^-19

= 5.152*10^-19

since electric field is directed in positive direction so net force is

=5.152*10^-19 + 7.22*10^-19 N

= 12.372

(b)Electric force = 3.22*1.6*10^-19

= 5.152*10^-19

since electric field is directed in positive direction so net force is

- 5.152*10^-19 + 7.22*10^-19 N = 2.068

(c) Electric force = 3.22*1.6*10^-19

= 5.152*10^-19

Net force = 10^-19 *sqrt(5.152^2+7.22^2) = 8.869695* 10^-19 N

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