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 mov- ing at 1.50 km/s in the +x-direction experiences a force of 2.25 x 10 16 N in the direction, and an electron moving at 4.75 km/s in the -z-direction experiences a force of 8.50 x 10 16 N in the +y. direction. (a) What are the magnitude and direction of the magnetic field? (b) What are the magnitude and direction of the magnetic force on an electron moving in the direction at 3.20 km/s?

Explanation / Answer

The z component of the magnetic field is
-2.25x10^(-16)N/(1.602x10(-19)C x 1.50 x 10^3 m/s) = 0.9363 T

The A component of the magnetic field is
8.5 x 10^(-16)N / (1.602x10^(-19)C x 4.75 x 10^3 m/s) = 1.117 T,
whether this is positive or negative,
whether the electron is deflected
in the +y direction or the -y direction.

Assume the force on the electron also acts in the +vdirection,
then the A component of the magnetic field is also positive,
since (-)(-k) A (i) = +j.

The magnetic field has magnitude
sqrt(1.117^2 + 0.9363^2) T = 1.46 T
and its direction is in the A-z plane at
arctan (0.9363/1.117) = 39.97 degrees away from the + A-axis
and 50.03 degrees away from the + z-axis.

(b) Assume that the electron moving in the -z direction
was deflected in the +v direction.
F = qv x B
= (-1.602 x 10^(-19)C)(-3.2 x 10^3 m/s j) x (1.117 i + 0.9363 k) T
= (-5.726 k + 4.800 i) x 10^(-16) N
The magnitude of this force is
sqrt(5.726^2+4.8^2) x 10^(-16) N = 7.28 x 10^(-16) N
and its direction is in the A-z plane,
perpendicular to the magnetic field,
so 39.97 degrees away from the negative z axis
and 50.03 degrees away from the positive A axis.

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