What is the magnitude of the magnetic field? What is the direction of the magnet

Question

What is the magnitude of the magnetic field?

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

What is the magnitude of the magnetic force on an electron moving in the y-direction at 3.70 km/s ?

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

A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.40 km/s in the +r-direction experiences a force of 2.06x10-16 N in the +y-direction, and an electron moving at 4.50 km/s in the -z-direction experiences a force of 8.20x10 N in the +y-direction. -16

Explanation / Answer

The z component of the magnetic field is

F1 = -2.06 x10^-16 N / (1.602 x 10-19 C x 1.40 x 10^3 m/s) = -0.9185 T

The x component of the magnetic field is

F2 = 8.20 x 10^-16 N / (1.602 x 10^-19 C x 4.50 x 10^3 m/s) = 1.1375 T

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.

IFI = sqrt(1.1375^2 + 0.9185^2) T

IFI = 1.46 T

its direction is in the x-z plane at tan^-1(0.9185/1.1375) = 38.9 degrees away from the + x-axis and 51.08 degrees

away from the + z-axis

(b)

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

F = qv x B

F = [(-1.602 x 10^-19) x (-3.7 x 10^3 m/s j)] x (1.1375 i + 0.9185 k) T

F = (-6.7424 k + 5.444 i) x 10^-16 N

The magnitude of this force is

F = sqrt(6.7424^2 + 5.444^2) x 10^-16 N

F = 8.666 x 10^(-16) N

its direction is in the x-z plane, perpendicular to the magnetic field, so 38.9 degrees away from the negative z axis

and 51.08 degrees away from the positive x axis.

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