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 4.70 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?

(I did this one)

Answer I got is B = 1.38 T

Part B

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

=   ?    from the z -direction

(Answer should be 304 from the z -direction for Part B but I dont know how or why)

Part C

What is the magnitude of the magnetic force on an electron moving in the y -direction at 3.70 km/s ?

F = ? N

Part D

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

=   ?    from the z -direction

Please show work. Thanks.

Explanation / Answer

part a:

magnetic force=charge*cross product of velocity and magnetic field

let magnetic field=Bx i + By j + Bz k

where i,j and k are unit vectors aong +ve x, +ve y and +ve z axis

for the proton:

velocity=(1700 i) m/s

cross product of velocity and magnetic field

=1700*By k -1700*Bz j

charge=1.6*10^(-19) C

then magnetic force=1.6*10^(-19)*(1700*By k -1700*Bz j)

as given magnetic force is along +ve y direction,

By=0

1.6*10^(-19)*(-1700*Bz)=2.1*10^(-16)

==>Bz=-0.77206 T

for the electron:

velocity=(-4700 k) m/s

cross product of velocity and magnetic field=-4700*Bx j

charge=-1.6*10^(-19) C

force =-1.6*10^(-19)*(-4700*Bx) =8.6*10^(-16)

==>Bx=8.6*10^(-16)/(1.6*10^(-19)*4700)=1.1436 T

so magnetic field=1.1436 i -0.77206 k

angle with -ve z direction=360-arctan(1.1436/0.77206)=304 degrees

part C:

velocity=(-3700 j) m/s

force=charge*cross product of velocity and magnetic field

=-4.57*10^(-16) i -6.77*10^(-16) k

force magnitude=sqrt(4.57^2+6.77^2)*10^(-16)=8.1681*10^(-16) N

part d:

direction of magnetic force from -ve z axis=arctan(6.77/4.57)=55.98 degrees

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