A particle with a mass of 20.0 times 10^-27 kg and charge of +e has velocity of

Question

A particle with a mass of 20.0 times 10^-27 kg and charge of +e has velocity of v as it enters the shaded region on the left. In this shaded region is a magnetic field B which bends the particles path into a averted of a circle with a radius of 5.60 cm before the particle exits. The particle then passes into the right shaded region where B_2 = 0.620 T and point up. In this region the particle passes through in a straight line because it also experiences as electric force with a magnitude of 9.00 times 10^-15 N which is equal and opposite to the magnetic force. a). what is the direction of the magnetic force to B_2. b) What is the velocity of the particle as it moves through B_2. c) What is the direction of the magnetic field in the shaded region B_1. d) What is the magnitude of the magnetic field in the shaded region B_1.

Explanation / Answer

a)

From right hand thumb rule the direction of magnetic force due to B2 is

out of the page

b)

the net force here is zero so

Fb = Fe

q * v * B = 9 * 10^-15

1.6 * 10^-19 * v * 0.62 = 9 * 10^-15

v = 9.07 * 10^5 m/s

c)

The direction of magnetic field B1 is into the page

d)

in region1

centripetal force = magnetic force

m * v^2 / r = q * v * B

B1 = (m * v) / (q * r)

B1 = (20 * 10^-27 * 9.07 * 10^5) / (1.6 * 10^-19 * 5.6 * 10^-2)

B1 = 2.02 T

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