As a part of an electron diffraction experiment, a magnetic field of strengthe 0

Question

As a part of an electron diffraction experiment, a magnetic field of strengthe 0.25 T is created towards South. Electrons are fired into this field. Find the direction and the magnitude of the force exerted by the field on an electron for each of the following electron velocities. 1 times 10^5 m/s, due northe 1 times 10^8 m/s, upward 1 times 10^8 m/s, downward A cosmic ray proton moving toward the Earthe a t 5 times 10^7 m is experiences a magnetic force of 1.70 times 10^16 N. What is the strengthe of the magnetic field if there is a 0 degree angle between the magnetic field and the proton's velocity? there is a 90 degree angle between the magnetic field and the proton's velocity? there is a 45 degree angle between the magnetic field and the proton's velocity?

Explanation / Answer

1) Force acting on moving charge with velocity v in a magnetic field B is F = q (V cross B) = qvB sin(theta)

a: the angle between v and B is 180.

   so, B =0.

b:direction of magnetic force : right

   magnitude F = q* v* B = 1.6 * 10-19 * 105 * 0.25 N = 4 * 10-13 N

C: direction : left

     magnitude = 1.6 * 10-19 * 105 * 0.25 N = 4 * 10-13 N

2)

a) 0 (sin(theta)=0 ).

b) F = qvB

    B = F/qv

c) F = qvB sin45

   B = F/ qvsin45

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