a. what magnitude force would be exerted on a q=100mc charged dust particle at t

Question

a. what magnitude force would be exerted on a q=100mc charged dust particle at the first point alpha if q is moving at 10 m/s due east?

b. what would be the direction of this force?

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Note A6 = 4 × 10-7 Tm2/C. Suppose that a long straight wire carries an electrical current horizontally due East through a room that is shielded from the Earth's magnetic field. At a point a, 2.0em due North of the center of the wire, the strength of the magnetic field the current induces is found to be 8x10-3 T.

Explanation / Answer

a) the force would be, F= qvB*sin(90)

[B is in upwards direction and v is due east, which gives a 90 degree angleb/w them ]

F= 100*10-3*10*8*10-3*1

F= 8*10-3 N

b) from Right hand rule, the force will be in south direction (towards the wire)

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