Question

the wire is 5.20 cm from the charge and carries a current of 68.5 A in a direction opposite to that of the moving charge. Calculate the magnitude of the force on the charge.

Answer 1

Complete Question

A 6.00-µc charge is moving with a speed of 7.50x10^4 m/s parallel to a very long, straight wire. the wire is 5.20 cm from the charge and carries a current of 68.5 A in a direction opposite to that of the moving charge. Calculate the magnitude of the force on the charge.

Answer:

The value is
`F =1.20 *10^(-4) \ N`

Step-by-step explanation:

From the question we are told that

The magnitude of the charge is
`Q = 6.00\muC = 6.00 *10^(-6) \ C`

The speed is
`v = 7.50 *10^(4) \ m/s`

The distance of the wire from the charge is
`d = 5.20 \ cm = 0.0520 \ m`

The current flowing through the wire in the opposite direction to the charge is
`I _2 = 68.5 \ A`

Gnerally the magnitude of force on that charge is mathematically represented as

`F = qvB`

Here B is magnetic field which is mathematically represented as

`B = (\mu_o * I)/(2\pi d)`

Here
`\mu_o` is the permeability of free space with value
`\mu_o = 4\pi * 10^(-7) N/A^2`

So

`F = qv[(\mu_o * I)/(2\pi d)]`

=>
`F = 6.0*10^(-6) * (7.50 *10^(4))[(4\pi * 10^(-7) * 68.5 )/(2* 3.142 * 0.0520 )]`

=>
`F =1.20 *10^(-4) \ N`