When a helium nucleus makes a full rotation of a circle of radius $0.8 \mathrm{~m}$ in $2.5 \mathrm{~s}$,…
When a helium nucleus makes a full rotation of a circle of radius $0.8 \mathrm{~m}$ in $2.5 \mathrm{~s}$, then the value of magnetic field $B$ at the centre of the circle will be
$4 \pi \times 10^{-25} \mathrm{~T}$
$2 \pi \times 10^{-26} T$
$4 \pi \times 10^{-26} \mathrm{~T}$
$2 \pi \times 10^{-25} \mathrm{~T}$
Solution
Charge on helium nucleus,
$
\begin{aligned}
q & =2 e=2 \times 1.6 \times 10^{-19} \\
& =3.2 \times 10^{-19} \mathrm{C} \\
r & =0.8 \mathrm{~m}, \mathrm{~T}=2.5 \mathrm{~s}
\end{aligned}
$
Current associated due to one rotation of helium nucleus,
$
I=\frac{q}{T}=\frac{3.2 \times 10^{-19}}{2.5}=1.28 \times 10^{-19} \mathrm{~A}
$
$\therefore$ Magnetic field at the centre of the circle formed by helium nucleus,
$
\begin{aligned}
B & =\frac{\mu_0 I}{2 r}=\frac{4 \pi \times 10^{-7} \times 1.28 \times 10^{-19}}{2 \times 0.8} \\
& =3.2 \pi \times 10^{-26} \mathrm{~T}
\end{aligned}
$
which is nearest to $4 \pi \times 10^{-26} \mathrm{~T}$