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. $\frac {10^{-19}}{\mu_0}$

Answer 1

Answer: (C)

$10^{-19} \mu _0$

Answer 2

The magnetic field at the centre of a circle is given by \hspace15mm B=\frac {\mu _0i}{2r} where, i is current and r the radius of circle. Also, \hspace15mm i= \frac {q}{t} For helium nucleus, q = 2e \therefore \hspace15mm i= \frac {2e}{t} So, \hspace15mm B= \frac {\mu _0.2e}{2rt} \hspace15mm = \frac {\mu _0 \times 2 \times 1.6 \times 10^{-19}}{2 \times 0.8 \times 2} \hspace15mm = 10^{-19}\mu _0