Question

A helium nucleus makes a full rotation in a circle of radius 0.8 m in 2s. The value of the magnetic field at the centre of the circle will be

• A. ${\mu }_0\times {10}^{-19}T$
• B. $1.6\times {\mu }_0\times {10}^{-19}T$
• C. $3{\mu }_0\times {10}^{-19}T$
• D. $2{\mu }_0\times {10}^{-19}T$

Answer 1

Answer: (A)

The magnetic field at centre
$B=\frac{{\mu }_{0}I}{2r}=\frac{{\mu }_0(q-t)}{2r}$
$=\frac{{\mu }_0\times 2\times 1.6\times {10}^{-19}}{2\times 0.8\times 2}$
$B={\mu }_0\times {10}^{-19}T$