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 2.10×10?16 N in the +y -direction, and an electron moving at 4.30 km/s in the ?z -direction experiences a force of 8.50×10?16 N in the +y -direction

What is the magnitude of the magnetic field? (B=______T)

What is the magnitude of the magnetic force on an electron moving in the ?y -direction at 3.50 km/s ?         (F=______N)

Explanation / Answer

The z component of the magnetic field is
-2.10*10^(-16)N/(1.602x10(-19)C x 1.60 x 10^3 m/s) = 0.8192 T

The x component of the magnetic field is
8.5 x 10^(-16)N / (1.602x10^(-19)C x 4.30 x 10^3 m/s) = 1.234 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(1.234^2 + 0.8192^2) T = 1.48 T
and its direction is in the x-z plane at
arctan (0.8192/1.234) = 33.57 degrees away from the + x-axis
and 56.43 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)(-3.5 x 10^3 m/s j) x (1.234 i + 0.8192 k) T
= (-6.91 k + 4.593 i) x 10^(-16) N
The magnitude of this force is
sqrt(6.91^2+4.593^2) x 10^(-16) N = 8.29x10^(-16) N
and its direction is in the x-z plane,
perpendicular to the magnetic field,
so 33.57 degrees away from the negative z axis
and 56.43 degrees away from the positive x axis.

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