Question

A particle of mass m and charge q is projected into a region that has 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
(a) $\frac{mv}{qB}$ (b) $\frac{mv}{2qB}$ (c) $\frac{2mv}{qB}$.

Answer 1

Given:
Mass of the particle = m
Charge of the particle = q
Magnetic field = B
As per the question, the particle is projected into a perpendicular magnetic field.

(a) When the width, d = $\frac{mv}{qB}$:
d is equal to the radius and θ is the angle between the radius and tangent, which is equal to $\frac{\mathrm{\pi }}{2}$.


(b) When the width, d = $\frac{mv}{2qB}$
Width of the region in which a magnetic field is applied is half of the radius of the circular path described by the particle.
As the magnetic force is acting only along the y direction, the velocity of the particle will remain constant along the x direction. So, if d is the distance travelled along the x axis, then
d = v_xt
$\mathrm{t}=\frac{d}{{\mathrm{v}}_{\mathrm{x}}}...\left(\mathrm{i}\right)$
(i) (ii)
The acceleration along the x direction is zero. The force will act only along the y direction.
Using the equation of motion for motion along the y axis:
v_y = u_y + a_yt
$$\begin{gathered} v_{\mathrm{y}}=0+\frac{q{u}_{\mathrm{x}}Bt}{m} \\ =\frac{q{u}_{\mathrm{x}}Bt}{m} \end{gathered}$$
Putting the value of t from equation (i), we get:
$\frac{q{u}_{\mathrm{x}}Bd}{m{v}_{\mathrm{x}}}$
We know:
$$\begin{gathered} \mathrm{tan\theta }=\frac{v_{\mathrm{y}}}{v_{\mathrm{x}}} \\ \frac{qBd}{m{v}_{\mathrm{x}}}=\frac{qBm{v}_{\mathrm{x}}}{2qBm{v}_{\mathrm{x}}}=\frac{1}{2} \\ \Rightarrow \theta ={\tan }^{-1}\left(\frac{1}{2}\right) \\ =26.4=30°=\frac{\mathrm{\pi }}{6} \end{gathered}$$
(c) When the width, d = $\frac{2mv}{qB}$
From the figure, it can be seen that the angle between the initial direction and final direction of velocity is π.