A group of particles is traveling in a magnetic field of unknown magnitude and d

Question

A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.70 km/s in the +x-direction experiences a force of 2.10×1016 N in the +y-direction, and an electron moving at 4.50 km/s in the z-direction experiences a force of 8.50×1016 N in the +y-direction.

a. What is the magnitude of the magnetic field?

b. What is the direction of the magnetic field? (in the xz-plane)

c. What is the magnitude of the magnetic force on an electron moving in the y-direction at 3.60 km/s ?

d. What is the direction of this the magnetic force? (in the xz-plane)

Explanation / Answer

Let the magnetic field be vector-B = B1*i +B2*k and magnitude of charge on proton or electron be q
We have F=qvB

q*1700*iX(vector-B) = (2.1*10^-16)*j

-1700*q*B2*j = 2.1*10^-16*j or
1700*q*B2 = -2.1*10^-16 ----------------------------------- 1
So B2 = (-2.1*10^-16)/(1700*1.6*10^-19) = -0.772

Also similarly for negatively charged electron
we have -q*4500*kX(vector-B) = (8.50*10^-16)*j or
-4500*q*B1*j = 8.50*10^-16*j or
4500*q*B1 = -8.50*10^-16 -----------------------------------2 or
B1 = (-8.50*10^-16)/(4500*1.6*10^-19) = -1.181 or
B = sq rt[B1^2 +b2^2] = 1.41 T
b)angle with x axis = tan^-1[B2/B1] = 33degree or 57 degree with z axis in negative XZ plane

C) vector force = -1.6*10^-19*3600(-j) X (B1*i +B2*k) =
= (-B1*k+B2*i)*1.6*10^-19*3600
F = 1.41*1.6*3.6*10^-16 N = 8.12*10^-16 N

d)Force will be in positive x and negative z plane making an angle 57 degrees with x axis.

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