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A particle of mass ‘m’ carrying a charge ‘q’ enter normally into uniform magnetic induction field ‘B’ . The time taken by the particle to traverse $90^{0}$ arc is

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$\frac{2 π m}{B q}$

$\frac{π m}{B q}$

$\frac{π m}{2 B q}$

$\frac{π m}{4 B q}$

answer is C.

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Detailed Solution

$B q v = m r ω^{2}$

$360^{0} - T$

$90^{0} - ?$

$t = \frac{90 T}{360} = \frac{T}{4}$ But $T = \frac{2 π}{ω} = \frac{2 π m}{B q}$

$t = \frac{2 π m}{4 B q} = \frac{π m}{2 B q}$

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