Question

A helium nucleus makes a full rotation in a circle of radius $0.8$ meter in $2 \sec$. The value of the magnetic field induction $B$ in tesla at the centre of circle will be

• A. $2 \times 10^{-19} \mu_0$
• B. $10^{-19} / \mu_0$
• C. $10^{-19} \mu_0$
• D. $2 \times 10^{-20} / \mu_0$

Answer 1

Answer: (C)

$B =\frac{\mu_0}{4 \pi} \frac{2 \pi i }{ r }$ where
$i =\frac{2 e }{ t }=\frac{2 \times 1.6 \times 10^{-19}}{2}=1.6 \times 10^{-19} A$
$\therefore B=\frac{\mu_0 i}{2 r}=\frac{\mu_0 \times 1.6 \times 10^{-19}}{2 \times 0.8}=\mu_0 \times 10^{-19} T$