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Circular Kinematics

A beam of ion...

Question

A beam of ions with velocity $2 \times 10^{5} m / s$ enters normally into a uniform magnetic field of $0.04 T$ . If the specific charge of ion $(\frac{q}{m} = 5 \times 10^{6} C / k g)$ . Find the radius (in metre) of the circular path.

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Solution

Radius of circular path is given as $r = \frac{m v}{q B} = \frac{v}{(q / m) B}$
putting as it is values we get $r = 1 m e t e r$

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