A magnetic field is directed as shown in the diagram to the right. The magnetic

Question

A magnetic field is directed as shown in the diagram to the right. The magnetic field has a strength of B = 0.78 Tesla. A positively charged particle of 1.20 mu C enters the field from the top of the paper as shown with a velocity of V = 550 m/s. a. What will be the direction of the resultant magnetic force? b. What will be the magnitude of the magnetic force acting on this positive particle? c. What will be the direction of the force if the particle is negative? d. What will be the direction of the force if the negative particle is moving from left to right?

Explanation / Answer

a.) F =q(v xB) .............cross product

direction into the plane, by right-hand thumb rule

b.) F = 1.2 x 10-6 x 550 x 0.78 N

= 514.8 x 10-6 N

c.) out of the plane....opposite to when charge is positive

d.) F =0 ..............cross product is zero when angle b/w vectors is 180 degrees

directionless

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