Chapter 20 Magnetism In the diagram below a uniform magnetic field of magnitude

Question

Chapter 20 Magnetism

In the diagram below a uniform magnetic field of magnitude 0.035 Tesla (T) is directed to the right which is also away from an observer ("eyeball" in the diagram). A proton (1st proton) whose velocity vector is perpendicular to the field direction as shown has a speed of 5.3 times 105 m/s and is moving through the field. Draw a diagram that shows the motion of the proton from the point of view of the observer (recall how magnetic fields and other vectors directed into and out of the page can be represented). Show in the diagram the direction of the magnetic force on the proton. The motion of the proton is circular. From the point of view of the observer, is the proton orbiting in a clockwise or counterclockwise manner? Explain how you know. Determine the radius and frequency of the circular orbit. A second proton also moves through the field. It has the same speed as the first, however, its velocity vector makes a 37degree angle with the field direction. This proton follows a helical path through the field. From the point of view of the observer how does this second proton circulate, clockwise or counterclockwise? Explain how you know. Determine the radius and pitch (distance between loops) of the helix.

Explanation / Answer

1

a) F = q*v*B*sin(90)

= 1.6*10^-19*5.3*10^5*0.035*1

= 2.968*10^-15 N

This force acts out of the page

direction : clockwise

b) m*v^2/r = q*v*B*sin(90)

r = m*v/(B*q)

= 1.67*10^-27*5.3*10^5/(0.035*1.6*10^-19)

= 1.58*10^-1 m

= 0.158 m
c) it follows helical path and it rotates in closkwise direction.

d)
b) m*v^2/r = q*v*B*sin(37)

r = m*v/(B*q*sin(37))

= 1.67*10^-27*5.3*10^5/(0.035*1.6*10^-19*sin(37))

= 2.62*10^-1 m

= 0.2625 mm

time periode,T = 2*pi*m/B*q

= (2*pi*1.67*10^-27)/(0.035*1.6*10^-19)
= 1.874*10^-6 s

pitch = vox*T

= V*cos(37)*T

= 5.3*10^5*cos(37)*1.874*10^-6

= 0.793 m

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