A particle with a charge of -2.5010-8 C is moving with an instantaneous velocity

Question

A particle with a charge of -2.5010-8 C is moving with an instantaneous velocity of magnitude 40.0 km/s in the xy-plane at an angle of 50.0 degree counterclockwise from the +x axis.

Part B

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

Part C

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.

Explanation / Answer

we know magnetic force on moving charged particle, F = q*(v cross B)

= q*v*B*sin(theta)

q --> charge

v ---> Velocity of charged particle

B --> magnetic field

theta ---> angle between v and B

B) F = 2.5*10^-8*40*10^3*2*sin(50)

= 1.53*10^-3 N

Direction: towards +Z axis

C) F = 2.5*10^-8*40*10^3*2*sin(90)

= 2*10^-3 N

Direction: towards +Y axis

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