A particle with a charge of -2.05x10^-8C is moving with an instantaneous velocit

Question

A particle with a charge of -2.05x10^-8C is moving with an instantaneous velocity of magnitude 40.5 km/s in the xy -plane at an angle of 50.0 degrees counterclockwise from the +x axis.

What is the magnitude of the force exerted on this particle by a magnetic field with magnitude 2.00T in the -x direction?

What is the direction of the force exerted on this particle by a magnetic field with magnitude 2.00T in the +z direction?

Please, enter your answer as a counterclockwise angle from the +y direction to the direction of the force.

What is the magnitude of the force exerted on this particle by a magnetic field with magnitude 2.00T in the +z direction?

Explanation / Answer

V = Vxi + Vyj = 40.5*10^3*cos50 i + 40.5*10^3*cos50 j

a) B = -Bx i = -2i T

F = q* ( V x B ) = q* ((Vxi + Vy j ) x (Bx i)) = q*(Vy*Bx - k) = -q*Vy*Bx k

F = 2.05*10^8*40.5*10^3*cos50*2 = 10.67e+12 N

b) B = +Bz k = -2k T

F = q* ( V x B ) = q* ((Vxi + Vy j ) x (-Bz k)) = q*(-Vy*Bz i +Vx*Bzj ) = -q*( -62.05e+3 i + 52.07e+3 j



direction = tan^-1(62.05e+3/52.07e+3) = 50


c) B = Bz i = 2i T

F = q* ( V x B ) = q* ((Vxi + Vy j ) x (-Bz k)) = q*(-Vy*Bz i +Vx*Bzj ) = -q*( -62.05e+3 i + 52.07e+3 j) = -12.72e+12 i + 10.67e+12 j

F = sqrt(12.72e+12^2 + 10.67e+12^2) = 16.6e+12 N

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