A point charge Q = +1.44 C moving in a uniform magnetic field of B = 1.39 T has

Question

A point charge Q = +1.44 C moving in a uniform magnetic field of B = 1.39 T has a velocity of v = 3.44×105 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?

either (South,East, Into the page, Out of the page, North, or West)

(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.)

Explanation / Answer

Magnetic field B = 1.39 T

Charge q = 1.44 x10 -6 C

Velocity v = 3.44 x10 5 m/s

Force on the point charge F = Bvq

                                         = (1.39)( 3.44 x10 5 )(1.44 x10 -6 )

                                        = 0.688 N

According to Fleming left hand rule the direction of force F is along north direction.

According to Fleming's left hand rule, if the thumb, fore-finger and middle finger of left hand are stretched perpendicular to each other , and if fore finger represents the direction of magnetic field, the middle finger represents the direction of current, then the thumb represents the direction of force.

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