A particle with a charge of 2.50×108 C is moving with an instantaneous velocity

Question

A particle with a charge of 2.50×108 C is moving with an instantaneous velocity of magnitude 40.0 km/s in the xy-plane at an angle of 50.0 degrees counterclockwise from the +x axis.

PART C: What is the direction of the force exerted on this particle by a magnetic field with magnitude 2.00 T in the +z direction?
Please, enter your answer as a counterclockwise angle from the +y direction to the direction of the force. [=o?]

I'm confused about this question...because in the initial part of it, which asked for both of these questions but with the magnitude and direction of B in the -x direction, I can't picture what's going on in the +z direction. Can you please show me all the steps and help me understand this question? Thanks!!!

Explanation / Answer

given
charge = q = -2.50*10-8C
velocity = v = 40.0km/s = 40*103 m/s
angle = = 500
magnetic field magnitude = B = 2.00T
force F = Bqv sin
plug in values and do caliculations F =----------N

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