SHOW WORK 1 At t = 0, A proton is moving with speed v = 1.7 x 107 m/s along the

Question

SHOW WORK

1 At t = 0, A proton is moving with speed v = 1.7 x 107 m/s along the positive yaxis
in the presence of a magnetic field pointing in the negative x-axis of field
strength 1.0 T. What is the direction and magnitude of the magnetic force acting
on the proton at that instant?
a. 2.72e-12 N (positive z-direction) b. 1.6e-12 N (negative z-direction) c.
2.72e-12 N (positive x-direction) d. 1.6e-12N) (negative y-direction)

2) A proton is accelerated by a potential difference of 1500 V from rest to some
speed v. It enters with this speed in a uniform magnetic field perpendicular
everywhere to the velocity of the electron. Consequently, the electron gets
trapped. What is the electrons kinetic energy after 2.0 s while trapped in this
field?
a. 4.8e-16 J b. 2.4e-16 J c. 1.6e-19 J d. 1.6e-16 J e. 0.5 J

Explanation / Answer

1) Fb = q*v*B = 1.6e-19*1.7e7*1 = 2.72e-12 N (positive z-direction)

OPTION a

2) KE = W = dV*q

KE = 1500*1.6e-19 = 2.4e-16 J

option b

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