Aparticle with a mass of 2.00×10−4 kg carries a negative charge of - 3.80×10−8 c . the particle is given an initial horizontal velocity that is due north and has a magnitude of 4.21×104 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? (west, east, north, south)

A. The Magnitude is 1.2251 TESLA which keeps the particle moving in the earth's gravitational field in the same horizontal, northward direction.

B. The direction of “minimum magnetic field” is East.

Explanation:

(A) To calculate Magnitude:  

From gravitational force (mg) = magnetic force (qvB).

qvB = mg

Given that,

Substitute the given values in the formula,

B = 1.2251 TESLA

(B) Direction of minimum magnetic field:

The direction is east. Force, F = q(vXB) = mg.  

Applying “Fleming's left hand rule” for vector product we know that the “negative charge” moving towards “north” establishes “conventional current” towards “south” the “magnetic field” has to be “eastwards” so, that the force due to applied “magnetic field” acts in the reverse direction of the “gravity of earth”. The “earth's gravitational field” will be ineffective in the commencement because the charge moves same the direction of “magnetic field” due to earth. To balance “gravity” the “magnetic force” has to point upward and have the “same magnitude” as the “gravitational force”. From “right-hand rule” take that the particle has "negative" charge then the field has to point to the "East" to create upward force.