A group of particles is A group of particles is traveling in a magnetic field of

Question

A group of particles is A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that the proton moving at 1.50 km/s in the +x direction experiences a force of 2.25 * 10^-16 N in the +y direction, and an electron moving at 4.75 km/s in the -z direction experiences a force of 8.50 * 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 -y direction at 3.20 km/s?

Explanation / Answer

given that

v1 =1.50 km/s

v2 = 4.75 km/s

F1 = 2.25*10^(-16) N in +y direction

F2 = 8.50*10^(-16) N in+y direction.

part(a)

The z component of the magnetic field is

Bz = -2.25*10^(-16) / (1.602*10^(-19)* 1.50 *10^3 ) = 0.9363 T

The x component of the magnetic field is

Bx = 8.5*10^(-16) / (1.602*10^(-19) * 4.75*10^3 ) = 1.117 T,

given 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.

the magnetic field has magnitude

B = sqrt(1.117^2 + 0.936^2) T = 1.46 T

and its direction is in the x-z plane at

tan^-1 (0.936/1.117) = 39.97 degrees away from the + x-axis

and 50.03 degrees away from the + z-axis

part(b)

given that the electron moving in the -z direction was deflected in the +y direction.

F = q*v x B

F = (-1.602*10^(-19))*(-3.2 *10^3 j) x (1.117 i + 0.9363 k)

F = (-5.72 k + 4.80 i) * 10^(-16) N

The magnitude of this force is

|F| = sqrt(5.72^2+4.80^2) * 10^(-16) N = 7.28 * 10^(-16) N

and its direction is in the x-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 x axis.

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