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 1.60 km/s in the +x-direction experiences a force of 2.06×1016 N in the +y-direction, and an electron moving at 4.40 km/s in the z-direction experiences a force of 8.60×1016 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 3.20 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

so

q*1600*iX(vector-B) = (2.06*10^-16)*j or
-1600*q*B2*j = 2.06*10^-16*j or
1600*q*B2 = -2.06*10^-16 ----------------------------------- 1
So B2 = (-2.06*10^-16)/(1600*1.6*10^-19) = -0.80468

for negatively charged electron

we have -q*4400*kX(vector-B) = (8.60*10^-16)*j or
-4400*q*B1*j = 8.60*10^-16*j or
4400*q*B1 = -8.60*10^-16 -----------------------------------2 or
B1 = (-8.60*10^-16)/(4400*1.6*10^-19) = - 1.2215

B = sq rt[B1^2 +b2^2] =

angle with x axis = tan^-1[B2/B1] = tan^-1[-2.308]= 57.71 degree or 32.19 degree with z axis in negative XZ plane

magnetic force on an electron moving in the y-direction at 3.20 km/s is
= -1.6*10^-19*3200(-j) X (B1*i +B2*k) =
= (-B1*k+B2*i)*1.6*10^-19*3200
Force will be in positive x and negative z plane making an angle 57.71 degrees with x axis and its magnitude F will be given by
F = 1.5*1.6*3.2*10^-16 N = 7.68*10^-16 N

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