and magnetic induction $( B )=10^{-4} Wb / m ^{2}$
Therefore $\frac{1}{2}\left(9.1 \times 10^{-31}\right) v^{2}=10 \times\left(1.6 \times 10^{-19}\right)$
$v^{2}=\frac{2 \times 10 \times\left(1.6 \times 10^{-19}\right)}{9.1 \times 10^{-31}}=3.52 \times 10^{12}$
$v=1.876 \times 10^{6} m$.
Centripetal force $=\frac{m v^{2}}{r}=B e v$.
$r=\frac{m v}{B e}=\frac{\left(9.1 \times 10^{-31}\right) \times\left(1.876 \times 10^{6}\right)}{10^{-4} \times\left(1.6 \times 10^{-19}\right)}$
$=11 \times 10^{-2} m=11 cm$
$1^{st}$ Way : By using spring of smaller torsion constant.
$2^{nd}$ Way : By using a smaller coil
$3^{rd}$ Way : By using a stronger magnet
$4^{th}$ Way : By using a coil having fewer number of turns.
