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Q.

A charged particle enters a uniform magnetic field with velocity vector at an angle $45 °$ with the magnetic field. The pitch of the helical path followed by the particle is ‘P’, the radius of helix is

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a

\frac{P}{\sqrt{2} π}

b

\sqrt{2 P}

c

\frac{\sqrt{2} P}{π}

d

\frac{P}{2 π}

answer is C.

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Detailed Solution

P = (v \cos 45 °) T

$R = \frac{mV \sin 45 °}{Bq} \Rightarrow R = \frac{P}{2 π}$

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