Question

A straight 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.

Answer 1

Answer:

The magnitude of the magnetic force exerted on the moving charge by the current in the wire is 2.18 x $10^{-8}$ 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

$q_{0}$ = +6.5 x $10^{-6}$ 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 = μ$_{0}$I/2πr

μ$_{0}$ = 4π x $10^{-7}$ or 1.26 x $10^{-6}$ N/$A^{2}$ , thus

B = 4π x $10^{-7}$ x 3 / 2π(0.05)

  = 1.2  x  $10^{-5}$ 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 $10^{-6}$ x 280  x 1.2 x  $10^{-5}$ sin 90°

  = 2.18 x $10^{-8}$ N

Using the hand law, the magnetic direction is radially inward