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

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

Step-by-step 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