Question

A charged particle enters into a uniform magnetic field with a velocity vector at an angle of ${45^\circ }$ with the magnetic field. The pitch of the helical path followed by the particle is ${\text{p}}$. The radius of the helix will be
(A) $\dfrac{{\text{p}}}{{\sqrt 2 \pi }}$
(B) $\sqrt 2 {\text{p}}$
(C) $\dfrac{{\text{p}}}{{2\pi }}$
(D) $\dfrac{{\sqrt {2{\text{p}}} }}{\pi }$

Answer 1

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.