Solveeit Logo

Question

Question: A charged particle enters into a uniform magnetic field with a velocity vector at an angle of \({45^...

A charged particle enters into a uniform magnetic field with a velocity vector at an angle of 45{45^\circ } with the magnetic field. The pitch of the helical path followed by the particle is p{\text{p}}. The radius of the helix will be
(A) p2π\dfrac{{\text{p}}}{{\sqrt 2 \pi }}
(B) 2p\sqrt 2 {\text{p}}
(C) p2π\dfrac{{\text{p}}}{{2\pi }}
(D) 2pπ\dfrac{{\sqrt {2{\text{p}}} }}{\pi }

Explanation

Solution

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
r=mvsinθqB{\text{r}} = \dfrac{{{\text{mv}}\sin \theta }}{{{\text{qB}}}}
Where
θ\theta is the angle at which a charged particle enters into a uniform magnetic field with velocity
m{\text{m}} is the mass of the particle
v{\text{v}} is the velocity of the particle
q{\text{q}} is the electric charge
B{\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 p{\text{p}}
The angle at which a charged particle enters into a uniform magnetic field with velocity, θ=45\theta = {45^ \circ }
The pitch is given by the equation
p=2πmvqBcosθp = \dfrac{{2\pi mv}}{{qB}}\cos \theta
Which can be rewritten as
p=2πPqBcosθp = \dfrac{{2\pi P}}{{qB}}\cos \theta
Where
PP is the momentum =mv = mv
Now, we will put the value of θ=45\theta = {45^ \circ } in the above equation to get
p=2πPqBcos45p = \dfrac{{2\pi P}}{{qB}}\cos {45^ \circ }
Now, the radius of the helix is given by
r=mvsinθqB{\text{r}} = \dfrac{{{\text{mv}}\sin \theta }}{{{\text{qB}}}}
Which can be rewritten as
r=PsinθqB{\text{r}} = \dfrac{{{\text{P}}\sin \theta }}{{{\text{qB}}}}
Now, we will put the value of θ=45\theta = {45^ \circ } in the above equation to get
r=Psin45qB{\text{r}} = \dfrac{{{\text{P}}\sin {{45}^ \circ }}}{{{\text{qB}}}}
Also, sin45=cos45=12\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=2πrp = 2\pi r
So, we get
r=p2π\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.