A beam of ions with velocit5r 2x10⁵ m/sec enters normally into a uniform magnetic field of 4 x10^-2tesla. If the specific charge of the ion is 5 x 10⁷ C/kg, the radius of the circular path described will be

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0.16m

0.20m

0.25m

0.10 m

answer is A.

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Detailed Solution

$\begin{array}{l} q_{0} Br = mv or r = \frac{mv}{q_{0} B} \\ ∴ r = \frac{V}{B (q_{0} / m)} = \frac{2 \times 10^{- 5}}{(4 \times 10^{- 2}) (5 \times 10^{7})} = 0 \cdot 10 m \end{array}$

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