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

Answer:

$F=-1.28\times 10^{-16}\ N$

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, $B=\dfrac{\mu_oI}{2\pi r}$

$F=qv\dfrac{\mu_oI}{2\pi r}$

$F=-1.6\times 10^{-19}\times 10^7\times \dfrac{4\pi\times 10^{-7}\times 40}{2\pi \times 0.1}$

$F=-1.28\times 10^{-16}\ N$

So, the force on the electron at this instant is $-1.28\times 10^{-16}\ N$. Hence, this is the required solution.