Question

An electron revolves in an orbit of hydrogen atom with the speed of ${10}^{5}m/s$. The magnetic field produced at the nucleus due to the motion of electron is (Given, radius of orbit $=0.2\stackrel{o}{A},{\mu }_0=4\pi \times {10}^{-7}H/m$).

• A. $4$T
• B. $2$T
• C. $1$T
• D. $3$T

Answer 1

Answer: (A)

$4$T

Given, $r=0.2\stackrel{o}{A}=2\times {10}^{-11}m,v={10}^{5}m/s$
$e=1.6\times {10}^{-19}C$
$B=?$
$\because I=\frac{q}{t}=\frac{e}{T}=e\left(\frac{v}{2\pi r}\right)$
Also, $B=\frac{{\mu }_{0}I}{2r}=\frac{{\mu }_0}{2r}2\frac{v}{2\pi r}$
$=\frac{{\mu }_0}{4\pi }\times \frac{ev}{r^2}$
$={10}^{-7}\times \frac{1.6\times {10}^{-19}\times {10}^5}{(2\times {10}^{-11}{)}^2}$
$=\frac{1.6\times {10}^{-21}}{4\times {10}^{-22}}=4T$.