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**Hint:**Every helical path has three distinct characteristics as radius, time period, and pitch. The helix pitch is the height of one complete helix turn, measured parallel to the helix axis. A double helix consists of two helices with the same axis (typically congruent), differentiated by a translation along the axis.

Formula Used: The radius of the helix is given by the following formula

\[{\text{r}} = \dfrac{{{\text{mv}}\sin \theta }}{{{\text{qB}}}}\]

Where

$\theta $ is the angle at which a charged particle enters into a uniform magnetic field with velocity

\[{\text{m}}\] is the mass of the particle

\[{\text{v}}\] is the velocity of the particle

\[{\text{q}}\] is the electric charge

\[{\text{B}}\] is the magnetic field

**Complete Step-by-Step Solution:**

According to the question, the following information is provided to us

The pitch of the helical path followed by the particle is ${\text{p}}$

The angle at which a charged particle enters into a uniform magnetic field with velocity, \[\theta = {45^ \circ }\]

The pitch is given by the equation

\[p = \dfrac{{2\pi mv}}{{qB}}\cos \theta \]

Which can be rewritten as

\[p = \dfrac{{2\pi P}}{{qB}}\cos \theta \]

Where

\[P\] is the momentum \[ = mv\]

Now, we will put the value of \[\theta = {45^ \circ }\] in the above equation to get

\[p = \dfrac{{2\pi P}}{{qB}}\cos {45^ \circ }\]

Now, the radius of the helix is given by

\[{\text{r}} = \dfrac{{{\text{mv}}\sin \theta }}{{{\text{qB}}}}\]

Which can be rewritten as

\[{\text{r}} = \dfrac{{{\text{P}}\sin \theta }}{{{\text{qB}}}}\]

Now, we will put the value of \[\theta = {45^ \circ }\] in the above equation to get

\[{\text{r}} = \dfrac{{{\text{P}}\sin {{45}^ \circ }}}{{{\text{qB}}}}\]

Also, \[\sin {45^ \circ } = \cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }}\]

Upon comparing the final results of pitch and radius of helix, we can conclude that

\[p = 2\pi r\]

So, we get

\[\therefore r = \dfrac{p}{{2\pi }}\]

**Hence, the correct option is (C.)**

**Note:**When a velocity component is present along the direction of magnetic field, its magnitude remains unchanged throughout the motion, as there is no effect of a magnetic field on it. The movement is also circular in nature because of the perpendicular velocity component.

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