A beam of ions with velocity $2 \times {10^5}\,m/s$ enters normally into a uniform magnetic field of $4 \times {10^{ - 2}}\,tesla$. If the specific charge of the ion is $5 \times {10^7}\,C/kg$, then the radius of the circular path described will be.......$m$
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(a) $r = \frac{{mv}}{{Bq}}$$ = \frac{v}{{(q/m)B}} = \frac{{2 \times {{10}^5}}}{{5 \times {{10}^7} \times 4 \times {{10}^{ - 2}}}} = 0.1\,m$
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