Aproton moving in the positive x direction with a speed of 9.4 105 m/s experiences zero magnetic force. when it moves in the positive y direction it experiences a force of 2.2 10-13 n that points in the positive z direction. determine the magnitude and direction of the magnetic field.

a) 1.46 T and directed to -x.

Explanation: In order to explain the above problem we have to consider the Lorentz force given by:

F=q*v ⊕ B ; ⊕  is the vector product.

The magnitude of B can be obtained as follow:

B=F/(q*v) where q and v are the charge and velocity of the proton.

B=2.2 *10^-13/(9.4*10^5*1.6*10^-19)=1.46 T

The direction of B is directed to  x axis because along this direction the proton experiences zero magnetic force.

Knowing that the magnetic force for the proton is directed to positive z direction and by inspection of the vector product bewteen v ( in +y direction) and B ( in x axis, due to zero line forze), we can conclude that the direction of B is to minus x axis.

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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