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) mvqB (b) mv2qB (c) 2mvqB.
Figure

Answer 1

Explanation:

Particle's mass = m          

Charge = q                

Magnetic field = B

According to the information provided, the projection of the particle into a perpendicular magnetic field is given by

(a) With width, d = mvqB:

θ is the angle between the tangent and radius, d is equal to the radius and , which is identical to $\frac{\pi }{2}$.

(b) With width, d = mv2qB

The region’s width in which there is an application of magnetic field is half of the circular path radius explained by the particle.

Only along the y direction, the magnetic force is acting, the particle’s velocity will stay constant along the x direction. So, along the x axis, if d is the travelled distance, then

Along the x direction, the acceleration is zero. Along the y direction, the force will act.

Along the y axis, using the equation of motion:

$\begin{array}{l}{\mathrm{t}=\frac{\mathrm{d}}{\mathrm{V}_{\mathrm{X}}} \quad \ldots(1)} \\{\mathrm{V}_{\mathrm{Y}}=\mathrm{u}_{\mathrm{Y}}+\mathrm{a}_{\mathrm{Y}} \mathrm{t}=\frac{0+\mathrm{qu}_{\mathrm{X}} \mathrm{Bt}}{\mathrm{m}}=\frac{\mathrm{qu}_{\mathrm{X} \mathrm{Bt}}}{\mathrm{m}}}\end{array}$

We understand:

$\tan \theta=\frac{V_{Y}}{V_{X}}=\frac{q B d}{m V_{X}}=\frac{q B m V_{X}}{2 q B m V_{X}}=\frac{1}{2}$

$\Rightarrow \theta=\tan ^{-1}\left(\frac{1}{2}\right)=26.4 \approx 30^{\circ}=\pi / 6$

(c) With width, d = 2mvqB

From the figure, the angle between the final direction and initial direction of velocity is π.