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) ${\displaystyle \frac{\text{p}}{\sqrt{2}\pi }}$
(B) $\sqrt{2}\text{p}$
(C) ${\displaystyle \frac{\text{p}}{2\pi }}$
(D) ${\displaystyle \frac{\sqrt{2\text{p}}}{\pi }}$

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
$$\text{r}={\displaystyle \frac{\text{mv}\sin \theta }{\text{qB}}}$$
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
$$p={\displaystyle \frac{2\pi mv}{qB}}\cos \theta$$
Which can be rewritten as
$$p={\displaystyle \frac{2\pi P}{qB}}\cos \theta$$
Where
$$P$$ is the momentum $$=mv$$
Now, we will put the value of $$\theta ={45}^{\circ }$$ in the above equation to get
$$p={\displaystyle \frac{2\pi P}{qB}}\cos {45}^{\circ }$$
Now, the radius of the helix is given by
$$\text{r}={\displaystyle \frac{\text{mv}\sin \theta }{\text{qB}}}$$
Which can be rewritten as
$$\text{r}={\displaystyle \frac{\text{P}\sin \theta }{\text{qB}}}$$
Now, we will put the value of $$\theta ={45}^{\circ }$$ in the above equation to get
$$\text{r}={\displaystyle \frac{\text{P}\sin {45}^{\circ }}{\text{qB}}}$$
Also, $$\sin {45}^{\circ }=\cos {45}^{\circ }={\displaystyle \frac{1}{\sqrt{2}}}$$
Upon comparing the final results of pitch and radius of helix, we can conclude that
$$p=2\pi r$$
So, we get
$$\therefore r={\displaystyle \frac{p}{2\pi }}$$

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.