Astraight wire carries a current of 3 a which is in the plane of this page, pointed toward the top of the page. a particle of charge q0 = +6.5 × 10−6 c is moving parallel to the wire and in the same direction as the current at a distance of r = 0.05 m to the right of the wire. the speed of the particle is v = 280 m/s. determine the magnitude and direction of the magnetic force exerted on the moving charge by the current in the wire.

The magnitude of the magnetic force exerted on the moving charge by the current in the wire is 2.18 x N

The direction of the magnetic force exerted on the moving charge by the current in the wire is radially inward

Explanation:

given information:

current, I = 3 A

= +6.5 x C

r = 0.05 m

v = 280 m/s

and direction of the magnetic force exerted on the moving charge by the current in the wire, we can use the following formula:

F = qvB sin θ

where

F = magnetic force (N)

q = electric charge (C)

v = velocity (m/s)

θ = the angle between the velocity and magnetic field

to find B we use

B = μI/2πr

μ = 4π x or 1.26 x N/ , thus

B = 4π x x 3 / 2π(0.05)

  = 1.2  x   T

Now, we can calculate the magnitude force

F = qvB sin θ

θ = 90°, because the speed and magnetic are perpendicular

F = 6.5 x x 280  x 1.2 x   sin 90°

  = 2.18 x N

Using the hand law, the magnetic direction is radially inward

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