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 1.50 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 8.60×1016 N in the +y-direction.

Part A

What is the magnitude of the magnetic field?

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Part B

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

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Part C

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

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Part D

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

A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.50 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 8.60×1016 N in the +y-direction.

Part A

What is the magnitude of the magnetic field?

B =   T  

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Part B

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

=    from the z-direction

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Part C

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

F =   N  

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Part D

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

Explanation / Answer

given data

speed =1.50km/s=1.50*10^3 m/s

force in y- direction =F =2.10*10^-16 N

the electron moving in z-direction . with a speed of = 4.50km/s = 4.50*10^3 m/s

force= 8.60*10^-16 N

The magnitude of the magnetic feild is given by

Let the magnetic field be vector-B = B1*i +B2*k and magnitude of charge on proton or electron be q
we have q*1500*iX(vector-B) = (2.10*10^-16)*j or
-1500*q*B2*j = 2.10*10^-16*j or
1500*q*B2 = -2.10*10^-16 ----------------------------------- 1
So B2 = (-2.10*10^-16)/(1500*1.6*10^-19) = -0.875 Tesla

Also similarly for negatively charged electron
we have -q*4500*kX(vector-B) = (8.60*10^-16)*j or
-4500*q*B1*j = 8.60*10^-16*j or
4400*q*B1 = -8.60*10^-16 -----------------------------------2 or
B1 = (-8.60*10^-16)/(4500*1.6*10^-19) = -1.1944 or
B = sq rt[B1^2 +b2^2] =sqrt(1.1944^2+0.875^2) =1.48 T
angle with x axis = tan^-1[B2/B1] = 36.22degree or 53.78 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
Force will be in positive x and negative z plane making an angle 53.78 degrees with x axis and its magnitude F will be given by
F = 1.48*1.6*3.6*10^-16 N = 8.5248*10^-16 N

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