Question

A helium nucleus makes a full revolution in a circle of radius 0.8 m is 2 s. The magnetic field at the center of the circle will be-

• A. ${10}^{-19/}\mu 0$
• B. ${10}^{-19}\mu 0$
• C. $2\times {10}^{-19}\mu 0$
• D. $2\times {10}^{-19/}\mu 0$

Answer 1

The correct option is A ${10}^{-19/}\mu 0$

$Given:$ Radius= 0.8m ; time= 2s

$Solution:$ The current due to the revolution of the He

nucleus is $I=\frac{q\omega }{2\pi }=\frac{2\times 1.6\times {10}^{-19}}{2}=1.6\times$

${10}^{-19}A$

Magnetic field at the center $B=\frac{{\mu }_{0}I}{2r}=$

$\frac{{\mu }_0\times 1.6\times {10}^{-19}}{2\times 0.8}={10}^{-19}{\mu }_0$

$So,the$ $correct$ $option:A$