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 2.10×10⁻¹⁶ N in the +y -direction, and an electron moving at 4.30 km/s in the −z -direction experiences a force of 8.20×10⁻¹⁶ N in the +y -direction.

A.What is the magnitude of the magnetic field?   

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

C. What is the magnitude of the magnetic force on an electron moving in the −y -direction at 3.50 km/s ?        

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

Answer 1

The z component of the magnetic field is
-2.15x10^(-16)N/(1.602x10(-19)C x 1.70 x 10^3 m/s) = 0.78945 T

The x component of the magnetic field is
8.2 x 10^(-16)N / (1.602x10^(-19)C x 4.30 x 10^3 m/s) = 1.1903 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(1.1903^2 + 0.78945^2) T = 1.428 T
and its direction is in the x-z plane at
arctan (0.78945/1.1903) = 33.55 degrees away from the + x-axis
and 56.45 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 (1.1903 i + 0.78945 k) T
= (-6.674 k + 4.4264 i) x 10^(-16) N
The magnitude of this force is
sqrt(4.4264^2+6.674^2) x 10^(-16) N = 8.008 x 10^(-16) N
and its direction is in the x-z plane,
perpendicular to the magnetic field,
so 33.55 degrees away from the negative z axis
and 56.45 degrees away from the positive x axis.