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 2.83 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2470 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 5.13 V/m, (b) in the negative z direction and has a magnitude of 5.13 V/m, and (c) in the positive x direction and has a magnitude of 5.13 V/m?

Explanation / Answer

B = 0.00283 T (-i)

v = 2470 m/s (j)

q = 1.6 x 10^-19 C

a) E = 5.13 V/m (k)

Fnet = Fe + Fm = qE + [ q (v X B) ]

= (1.6 x 10^-19 x 5.13k) + [1.6x10^-19 ( 2470j X -0.00283 i ]

= 1.6x 10^-19 ( 5.13k + 6.99k)

= 1.94 x 10^-18 N

b) E = - 5.13 V/m k

Fnet = = 1.6x 10^-19 (- 5.13k + 6.99k)

= 2.98 x 10^-19 N

c) E = 5.13 i

Fnet = 1.6 x 10^-19 ( 5.13i + 6.99k )

magnitude= 1.6 x 10^-19 sqrt(5.13^2 +6.99^2)

     = 1.39 x 10^-19 N

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