Question

A particle of mass ‘m’ and charge ‘q’ is projected into a region having a perpendicular magnetic field B. Find the angle of deviation (figure) of the particle as it comes out of the magnetic field if the width ‘d’ of the region is very slightly smaller than

1.$\dfrac{mv}{qB}$.
2. $\dfrac{mv}{\sqrt{2}qB}$
3. $\dfrac{3mv}{qB}$

Answer 1

We need to understand the relation between the deviation caused to an electric charge that is moving perpendicular to a magnetic field and the radius of the circular path the particle will be travelling to solve the given problem in different situations.

Complete answer:
We are given a situation in which an electric charge ‘q’ is moving into a magnetic field in the perpendicular direction. We know that the charge will get deviated by the magnetic field and will describe a circular path.

We know that the charge when encountering a magnetic field will describe a circular path depending on the perpendicular component of the velocity with which the particle enters the field. We can derive the radius from the centripetal force and the magnetic force as –

& \dfrac{mv}{2qB} \\\ & {{F}_{C}}={{F}_{B}} \\\ & \Rightarrow \dfrac{m{{v}_{\bot }}^{2}}{r}=q{{v}_{\bot }}B \\\ & \therefore r=\dfrac{mv\sin \theta }{qB}\text{ --(1)} \\\ \end{aligned}$$ Now, we are given the radius for three situations. We can find the angle of deviation using the above relation. 1\. $$r=\dfrac{mv}{qB}$$: We can equate the equation (1) to get the angle of deviation as – $$\begin{aligned} & \dfrac{mv}{qB}=\dfrac{mv\sin \theta }{qB} \\\ & \Rightarrow \sin \theta =1 \\\ & \therefore \theta =\dfrac{\pi }{2} \\\ \end{aligned}$$ 2\. $$r=\dfrac{mv}{\sqrt{2}qB}$$: We can equate the given radius with (1) to get the angle of deviation as – $$\begin{aligned} & \dfrac{mv}{\sqrt{2}qB}=\dfrac{mv\sin \theta }{qB} \\\ & \Rightarrow \sin \theta =\dfrac{1}{\sqrt{2}} \\\ & \therefore \theta =\dfrac{\pi }{4} \\\ \end{aligned}$$ 3\. $$r=\dfrac{mv}{2qB}$$: We can find the angle of deviation for this case also similarly as – $$\begin{aligned} & \dfrac{mv}{2qB}=\dfrac{mv\sin \theta }{qB} \\\ & \Rightarrow \sin \theta =\dfrac{1}{2} \\\ & \therefore \theta =\dfrac{\pi }{6} \\\ \end{aligned}$$ We get the angles of deviation for all the given cases of radius as – 1\. $$\theta =\dfrac{\pi }{2}$$ 2\. $$\theta =\dfrac{\pi }{3}$$ 3\. $$\theta =\dfrac{\pi }{6}$$ These are the required angles for the three given cases. **Note:** The velocity component of the charged particle along the perpendicular direction as that of the magnetic field determines the radius of the circular motion of the charged particle for a given mass and charge of the particle kept as a constant.