An infinitely long straight conductor carries a current of 5 A as shown. An electron is moving with a speed of 10⁵ m/s parallel to the conductor. The perpendicular distance between the electron and the conductor is 20 cm at an instant. Calculate the magnitude of the force experienced by the electron at that instant.

CONCEPT:

To calculate the magnitude of the force experienced by the electron in the current having charge q moving with velocity v calculated with the help of Lorentz force. It is defined as the magnetic field exerting the force on the object which is having charge and velocity where force F is perpendicular to the magnetic field B. Therefore, it is written as;

\(\overrightarrow F = q\left( {\overrightarrow v × \overrightarrow B } \right)\)

Here q is the charge, \(\overrightarrow B \)  is the magnetic field, \(\overrightarrow v\) is the velocity of the electron, and \(\overrightarrow F\) is the force.

CALCULATION:

Given: In the given figure we have a magnetic field produced due to the current-carrying wire at the point "A" where

velocity, \(\overrightarrow v = {10^5} \:m/s\)

distance, \(r = 20\:cm = 20 × {{10}^{ - 2}} \:m\)

current, \(I = 5\:A\)

Now we use Lorentz force which is \(\overrightarrow F = q\left( {\overrightarrow v × \overrightarrow B } \right)=qvB\:---(1)\)

 Here, we first have to calculate the magnetic field \(\overrightarrow B \)

A magnetic field of the infinite long wire is written as;

\(B = \frac{{{μ _0}I}}{{2\pi r}} \) where,
\(μ _0\) is the permeability of free space which is 4π×10⁻⁷ Tm/A

Now, on putting the given values in the equation for B.

\(B = \frac{{{{4 \pi \times10}^{ - 7}} × 5}}{{2\pi \times20 × {{ 10}^{ - 2}}}} = \frac{1}{2} × {10^{ - 5}}\) T

Putting this value in equation (1) we get;​​

\(\left| {\vec F} \right| = 1.6 \times {10^{ - 19}} \times {10^5} \times \frac{1}{2} \times {10^{ - 5}} \)

       = 0.8 ×10-19 N​​​​​

​\(\left| {\vec F} \right| \) = 8 × 10-20 N

 Hence, Option 1) is the correct answer.