Question

A long straight wire carries a 40.0 A current in the +x direction. At a particular instant, an electron moving at 1.0  107 m/s in the +y direction is 0.10 m from the wire. The charge on the electron is –1.6  10–19 C. What is the force on the electron at this instant?

Answer 1

`F=-1.28* 10^(-16)\ N`

Step-by-step explanation:

Given that,

Current flowing in the wire, I = 40 A (+x direction)

Speed of the electron,
`v=10^7\ m/s` (+y direction)

Distance from the wire, r = 0.1 m

Let F is the electric force on the electron. It is given by :

`F=qvB\ sin\theta`

Here,
`\theta=90`

`F=qvB`

Here,
`B=(\mu_oI)/(2\pi r)`

`F=qv(\mu_oI)/(2\pi r)`

`F=-1.6* 10^(-19)* 10^7* (4\pi* 10^(-7)* 40)/(2\pi * 0.1)`

`F=-1.28* 10^(-16)\ N`

So, the force on the electron at this instant is
`-1.28* 10^(-16)\ N`. Hence, this is the required solution.