A slightly divergent beam of non-relativistic charged particles accelerated by a potential difference $V$ propagates from a point $A$ along the axis of a straight solenoid. The beam is brought into focus at a distance $A$ from the point $A$ at two successive values of magnetic induction $B_{1}$ and $B_{2}$ . Find the specific charge $q / m$ of the particles.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{2 π^{2} V}{l^{2} {(B_{2} - B_{1})}^{2}}$

$\frac{8 π^{2} V}{l^{2} {(B_{1} - B_{2})}^{2}}$

$\frac{8 π^{2} V}{l^{2} {(B_{2} - B_{1})}^{2}}$

$\frac{4 π^{2} m}{l^{2} {(B_{2} - B_{1})}^{2}}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Suppose $v_{0}$ is the velocity of the particle, then

$\frac{1}{2} m v_{0}^{2} = q V \Rightarrow v_{0} = \sqrt{\frac{2 q V}{m}}$ …………………..(i)

For slightly divergent particle, $θ \to 0 \Rightarrow v_{0} \cos θ ≅ v_{0}$

If $t$ is the time to travel a distance $l$ , then $t = \frac{l}{v_{0}}$ . v A In this duration, let particle completes $n$ circles for $B_{1}$ and $(n + 1)$ for $B_{2}$ , then

$\frac{l}{v_{0}} = n \frac{2 π m}{q B_{1}} = (n + 1) \frac{2 π m}{q B_{2}}$ ………………….(ii)

On solving $n = \frac{B_{1}}{B_{2} - B_{1}}$

Substituting this values of $v_{0}$ and $n$ in equation (ii) and solving, we get

$\frac{l}{\sqrt{\frac{2 q V}{m}}} = \frac{B_{1}}{B_{2} - B_{1}} \frac{2 π m}{q B_{1}} o r \frac{q}{m} = \frac{8 π^{2} V}{l^{2} {(B_{2} - B_{1})}^{2}}$