Question

A long, straight wire carries a current of 4.23 A. An electron travels at 47100 m/s parallel to the wire, 61.1 cm from the wire. The permeability of free space is 1.25664 × 10−6 N/A 2 and the charge on an electron is 1.6 × 10−19 C. What force does the magnetic field of the current exert on the moving electron? Answer in units of N.

Answer 1

`F_e = 1.043*10^(-20) N`

Step-by-step explanation:

To give a proper solution to this problem we need to apply the concept about Force on electron, which is given by,

`F_e = BeV`

Where,

`F_e`= Perpendicular Force to the direction of magnetic field

B = Magnetic field

V = Velocity (Perpendicular also)

We know that B is given by

`B = (\mu_0I)/(2\pi d)`

Where
`\mu` is the permeability constant I is the current and d the distance

Replacing in the Force equation,

`F_e = (\mu_0I)/(2\pi d) eV`

`F_e = \frac{(4\pi*10^(-7))(4.23)} {2\pi 0.611}*1.6*10^(-19)*47100`

`F_e = 1.043*10^(-20) N`