A straight wire carries a current of 10 \mathrm{~A}. An electron moving at 10^{7} \mathrm{~m} / \mathrm{s} is at distance 2.0 \mathrm{~cm} from the wire. Find the force acting on the electron if its velocity is directed towards the wire.
A straight wire carries a current of 10 \mathrm{~A}. An electron moving at 10^{7} \mathrm{~m} / \mathrm{s} is at distance 2.0 \mathrm{~cm} from the wire. Find the force acting on the electron if its velocity is directed towards the wire.

Ans: We are given the current through the straight wire to be, I=10 \mathrm{~A}
Speed of the electron, \mathrm{v}=10^{7} \mathrm{~m} / \mathrm{s}
Distance of electron from the wire, \mathrm{R}=2.0 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m},
Force acting on moving electron would be, \mathrm{F}=\mathrm{qVB} \sin \theta
We have the expression for magnetic field as,

    \[\mathrm{B}=\frac{\mu_{0}}{4 \pi} \frac{2 I}{r}\]

Substituting the given values,
\mathrm{B}=\frac{10^{-7} \times 2 \times 10}{2 \times 10^{-2}}=10^{-4} \mathrm{~T} and it is given to be \perp to the plane of paper and directed towards.
Now, force acting on the electron could be given by,

    \[\begin{array}{l} \mathrm{F}=4 \mathrm{~V} \mathrm{~B} \sin \theta \\ \Rightarrow \mathrm{F}=1.6 \times 10^{-16} \mathrm{~N} \end{array}\]

Therefore, we find the force to be, \mathrm{F}=1.6 \times 10^{-16} \mathrm{~N}.