Question

An infinitely long straight conductor carries a current of $5\text{A}$ as shown. An electron is moving with a speed of ${10}^5\text{m/s}$ parallel to the conductor. The perpendicular distance between the electron and the conductor is $20\text{cm}$ at an instant. Calculate the magnitude of the force experienced by the electron at that instant.

• A. $8\times {10}^{-20}\text{N}$
• B. $4\times {10}^{-20}\text{N}$
• C. $8\pi \times {10}^{-20}\text{N}$
• D. $4\pi \times {10}^{-20}\text{N}$

Answer 1

The correct option is A $8\times {10}^{-20}\text{N}$


Magnetic field produced due to long current carrying wire at point $\text{A}$ is

$B=\frac{{\mu }_0}{4\pi }\frac{2I}{r}$

$B=\frac{{10}^{-7}\times 2\times 5}{20\times {10}^{-2}}=\frac{1}{2}\times {10}^{-5}\text{T}\odot$

Direction of magnetic field will be outward to the plane of paper.

Now, force acting on the electron due to this field,

$\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$

$|\overrightarrow{F}|=1.6\times {10}^{-19}\times {10}^5\times \frac{1}{2}\times {10}^{-5}=0.8\times {10}^{-19}$

$\therefore |\overrightarrow{F}|=8\times {10}^{-20}\text{N}$

Hence, $\left(\text{A}\right)$ is the correct option.