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A particle of mass 0.195 g carries a charge of -2.50 x 10^-8 C. The particle...

A particle of mass 0.195 g carries a charge of -2.50 x 10^-8 C. The particle is given an initial horizontal velocity that is due north and has magnitude 4.00 x 10^4 m/s. What are the magnitude and direction of the minimum mgnetic field that will keep the particle moving in the earth's gravitational field in the same horizontal, northward direction? Conceptually explain why the B-field is in this direction.

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Answer #1

The formula for magnetic force on a moving charge is

F = Bqv

the formula for earth's gravitational field is

F = mg

Bqv = mg

B = mg/ qv

    =(0.195 * 10 ^-3 kg) (9.8)/ -2.50 x 10^-8 C. ( 4.00 x 10^4 m/s. )

= 1.91 T

Since the charge is negative, the magnetic force is opposite to the
right-hand rule direction. The minimum magnetic field is when the field is perpendicular to v! . The force is also
perpendicular toB
soB is either eastward or westward.

The magnetic field could also have a component along the north-south direction, that would not
contribute to the force, but then the field wouldnt have minimum magnitude

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