A helium nucleus makes a full rotation in a circle of radius $0.8$ metre in two seconds. The value of the magnetic field $B$ at the centre of the circle will be
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(b) $i = \frac{q}{T} = \frac{{2 \times 1.6 \times {{10}^{ - 19}}}}{2} = 1.6 \times {10^{ - 19}}A$
$\therefore \,B = \frac{{{\mu _o}i}}{{2r}} = \frac{{{\mu _o} \times 1.6 \times {{10}^{ - 19}}}}{{2 \times 0.8}} = {\mu _o} \times {10^{ - 19}}$
$\therefore \,B = \frac{{{\mu _o}i}}{{2r}} = \frac{{{\mu _o} \times 1.6 \times {{10}^{ - 19}}}}{{2 \times 0.8}} = {\mu _o} \times {10^{ - 19}}$
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