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# 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) $\dfrac{{\sqrt {2{\text{p}}} }}{\pi }$

Last updated date: 25th Jul 2024
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Answer
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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.