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 1.30 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.

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)

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A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.30 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.

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)

=    from the x-direction

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Explanation / Answer

The z component of the magnetic field is
-2.1x10^(-16)N/(1.602x10(-19)C x 1.30 x 10^3 m/s) = 1.008 T

The x component of the magnetic field is
8.5 x 10^(-16)N / (1.602x10^(-19)C x 4.5 x 10^3 m/s) = 1.18 T,
but we can't know whether this is positive or negative, because you didn't say whether the electron is deflected in the +y direction or the -y direction.
If I assume that the force on the electron also acts in the +y direction,
then the x component of the magnetic field is also positive,
since (-)(-k) x (i) = +j.
I conclude that the magnetic field has magnitude
sqrt(1.18^2 + 1.008^2) T = 1.55 T
and its direction is in the x-z plane at
arctan (1.008/1.18) = 40.5 degrees away from the + x-axis
and 49.5 degrees away from the + z-axis.

(b) In doing this part, I will again assume that the electron moving in the -z direction
was deflected in the +y direction.
F = qv x B
= (-1.602 x 10^(-19)C)(-3.6 x 10^3 m/s j) x (1.18 i + 1.008 k) T
= (-6.8 k + 5.8 i) x 10^(-16) N
The magnitude of this force is
sqrt(6.8^2+5.8^2) x 10^(-16) N = 8.94 x 10^(-16) N
and its direction is in the x-z plane, perpendicular to the magnetic field,
so 40.5 degrees away from the negative z axis
and 49.5 degrees away from the positive x axis

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