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Q. A beam of $\alpha$ particles having specific charge $2.5\times10^7\,C\,kg^{-1}$ moves with a speed of $2\times10^5\,ms^{-1}$ in a magnetic field of $0.05\, T$. The radius of the circular path described by the $\alpha$-particle is

Solution:

R = $\frac{M \, v}{B_{q0}} = \frac{v}{B} ( \frac{q_0}{M}) $
$= 2 \times 10^5 / 0.05 \times 2.5 \times 10^7$ = 0.16 m = 16 cm.