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.
$$\frac{{{{10}^{ - 19}}}}{{{\mu _0}}}$$
C.
$${10^{ - 19}}{\mu _0}$$
D.
$$\frac{{2 \times {{10}^{ - 20}}}}{{{\mu _0}}}$$
Answer :
$${10^{ - 19}}{\mu _0}$$
Solution :
$$\eqalign{
& B = \frac{{{\mu _0}}}{{4\pi }}\frac{{2\pi i}}{r}\,\,{\text{where}} \cr
& i = \frac{{2e}}{t} = \frac{{2 \times 1.6 \times {{10}^{ - 19}}}}{2} = 1.6 \times {10^{ - 19}}\,A \cr
& \therefore B = \frac{{{\mu _0}i}}{{2r}} = \frac{{{\mu _0} \times 1.6 \times {{10}^{ - 19}}}}{{2 \times 0.8}} \cr
& = {\mu _0} \times {10^{ - 19}}\,T \cr} $$