Question

A particle with a mass of 1.98×10^−4 kg carries a negative charge of -3.50×10^−8 C. The particle is given an initial horizontal velocity that is due north and has a magnitude of 3.78×10^4 m/s.
a) What is the magnitude of the minimum magnetic field that will keep the particle moving in the earth's gravitational field in the same horizontal, northward direction?
b) What is the direction of the minimum magnetic field?

Answer 1

(a) The magnitude of the minimum magnetic field is 1.467 T.

(b) The direction of the minimum magnetic field will be eastwards.

Minimum magnetic field

F = qvB

mg = qvB

B = mg/qv

where;

• B is magnetic field
• v is velocity
• m is mass of the particle
• q is charge of the particle

B = (1.98 x 10⁻⁴ x 9.8) / (3.5 x 10⁻⁸ x 3.78 x 10⁴)

B = 1.467 T

Direction of the magnetic field

The magnetic field will be directed towards the negative charge. Thus, the direction of the minimum magnetic field will be eastwards.

Thus, the magnitude of the minimum magnetic field is 1.467 T and the direction is eastwards.

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