Question

A helium nucleus makes full rotation in a circle of
radius 0.8 min 2 s. The value of'magnetic field B at the
centre of the circle, will be $(\mu _0=$ permeability constant)

• A. $\frac {2 \times 10^{-19}}{\mu _0} $
• B. $2 \times 10^{-19} \mu _0 $
• C. $10^{-19} \mu _0 $
• D. $\frac {10^{-19}}{\mu_0} $
• E. None of these

Answer 1

Answer: (C)

The magnetic field at the centre of a circle is given by
$ B=\frac {\mu _0i}{2r} $
where, i is current and r the radius of circle.
Also, $ i= \frac {q}{t}$
For helium nucleus, q = 2e
$\therefore i= \frac {2e}{t} $
So, $ B= \frac {\mu _0.2e}{2rt} $
$ = \frac {\mu _0 \times 2 \times 1.6 \times 10^{-19}}{2 \times 0.8 \times 2} $
$ = 10^{-19}\mu _0 $