A point charge Q = +1.15 C moving in a uniform magnetic field of B = 1.08 T has
ID: 1461474 • Letter: A
Question
A point charge Q = +1.15 C moving in a uniform magnetic field of B = 1.08 T has a velocity of v = 4.87×105 m/s that is perpendicular to the magnetic field. If the magnetic field points out of the page and the charge moves to the West, then what is the direction of the resulting force? (In this problem we use the points of the compass for directions on the paper with North pointing to the top of the page, East pointing to the right, and 'into' and 'out of' to indicate directions with respect to the page.)
Out of the page
Into the page
East
West
South
North
What is the magnitude of the force?
(in N)
Explanation / Answer
Apply Fleming left hand rule:
The direction of force is North
Use the following formula to find force
F=QvB sin
Substitute given values
F= ( 1.15 x 10-6 N ) ( 4.85 x105 ) ( 1.08 T ) sin900
= 6.05 x 10-1 N , north
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