A particle with a mass of 1.98×10 4 kg carries a negative charge of - 3.70×10 8

Question

A particle with a mass of 1.98×104 kg carries a negative charge of - 3.70×108 C . The particle is given an initial horizontal velocity that is due north and has a magnitude of 3.67×104 m/s .

Part 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?

Part B

What is the direction of the minimum magnetic field?

What is the direction of the minimum magnetic field?

Explanation / Answer

A)given values: m=1.98×104 kg q=3.70×108 C,v= 3.67×104 m/s

F_lorentz = W
B v q = m g
B =mg/vq=((1.98*10^-4)*(9.81))/((3.67*10^4)*(3.70*10-8))=1.82*10^-9 T

B)Force, F = q(vXB) = mg. Applying Flemings left hand rule or by applying right hand cock-screw rule for vector product keeping in mind that the negative charge moving towards north constitutes conventional current towards south the magnetic field has to be eastwards so that the force due to applied magnetic field acts in the opposite direction of the gravity of earth. Also note that the earth's gravitational field will be ineffective in the beginning because the charge moves along the direction of magnetic field due to earth.

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