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 1.40 km/s in the +x-direction experiences a force of 2.10×1016 N in the +y-direction, and an electron moving at 4.20 km/s in the z-direction experiences a force of 8.20×1016 N in the +y-direction.

A-What is the magnitude of the magnetic field?..............B=???T

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

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

D-What is the magnitude of the magnetic force on an electron moving in the y-direction at 3.70 km/s ?............ =???degree (from -x direction)

Explanation / Answer

F = q(vxB)

F is perpendicular to both v and B since the force on the proton is in the +y direction

By = 0

B = Bxi + Bzk for the proton

for proton

Fp = q(vp x B)

Fpj = evpj x (Bxi + Bzk) = -e*vp*Bz*j

Bz = -Fp/evp = 2.10 x 10^-16/(1.6x10^-19 x 1400 ) = -0.9375 T

for electron Fe = -e(ve xB)

Bx = Fe/eve

ve = 4200 m/s

Bx = 1.22T

B = 1.22i - 0.9375k

|B| = sqrt(Bx^2+Bz^2)

|B| = 1.54 T

theta = tan^-1Bz/Bx = -37.54 degree

part c )

F = q(vxB) = (-e)[ 3700m/s * (-j) x (Bxi + Bzk)] = e(3700)[Bx(-k)+Bzi]

F =e(3700)[-1.22k - 0.9375i] = -7.2224 x 10^-16k - 5.55 x 10^-16i

F = sqrt(Fx^2+Fz^2)

F = 9.1 x 10^-16 N

theta = tan^-1(Fz/Fx)52.5 degree =

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