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 34.E14) of the particle as it comes out of the magnetic field if the width d of the region is very slightly smaller than

$\left(a\right)\frac{mv}{qB}\left(b\right)\frac{mv}{2qB}\left(c\right)\frac{2mv}{qB}$

Answer 1

Mass of the particle = m;

charge = q,

width = d

(a) If $d=mV/qB$

then d is equal the radius q is the angle between the radius and tangent which is equal to $\frac{\pi }{2}$

(b) If d = distance travelled

$=\frac{1}{2}$ of radius along x-distance

$d=V_{x}t$

i.e. $t=\frac{d}{Vx}...\left(1\right)$

$v_y=u_y+a_{y}t$

From (1) putting the value of t,

$tan\theta =\frac{1}{2}$

$\Rightarrow \theta =ta{n}^{-1}\left(\frac{1}{2}\right)$

$=26.4={30}^{\circ }=\frac{\pi }{6}$

(c) The angle between the initial direction and final direction of velocity is $\pi .$