Question

An electron and a proton are revolving around common centre $O$ in two coplanar circular paths as shown in figure with time period of rotation $1s$ and $2s$ respectively. The net magnetic field at $O$ will be:

• A. $\frac{{\mu }_{0}e}{2}tesla$
• B. $\frac{{\mu }_{0}e}{2\pi }tesla$
• C. $2{\mu }_{0}etesla$
• D. $Zero$

Answer 1

The correct option is A $\frac{{\mu }_{0}e}{2}tesla$

Current is defined as $I=\frac{e}{t}$

For proton, t = 2s and r = 1m

$I=\frac{e}{2}$

So, magnetic field due to proton is

$B_1=\frac{{\mu }_{0}I}{2r}$

$B_1=\frac{{\mu }_{0}e}{2\times 2\times 1}$

$B_1=\frac{{\mu }_{0}e}{4}............\left(1\right)$

For electron, t = 1s and r = 2m

$I=\frac{-e}{1}=-e$

So, magnetic field due to electron is

$B_2=-\frac{{\mu }_{0}e}{2\times 2}$

$B_2=-\frac{{\mu }_{0}e}{4}..........\left(2\right)$

Net magnetic field

$B=B_2-B_1$

$=-\frac{{\mu }_{0}e}{4}-\frac{{\mu }_{0}e}{4}$

$=-\frac{{\mu }_{0}e}{2}$