Question

Consider a bob of mass $m$ and having charge $q$ attached with light string of length $l$ and pivoted at point $O$. It is released at rest at ${60}^{\circ }$ with vertical. There are two regions- Region- $I$ (left of line $PQ$) has a uniform and constant magnetic field $B$ directed into plane of paper. Region-$II$ (right of line $PQ$) has a constant and uniform electric field $E$ directed vertically up as shown. Consider no effect of gravity in both the regions.

• A. time taken by particle to cross region -I for $1^{st}$ time is $\pi \sqrt{\frac{2ml}{5qE}}$
• B. time taken by particle to cross region -I for $2^{nd}$ time is $\pi \sqrt{\frac{2ml}{13qE}}$
• C. Angular speed of $1^{st}$ revolution in magnetic field is $\sqrt{\frac{5qE}{2ml}}$
• D. Angular speed of $2^{nd}$ revolution in magnetic field is $\sqrt{\frac{13qE}{ml}}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A time taken by particle to cross region -I for $1^{st}$ time is $\pi \sqrt{\frac{2ml}{5qE}}$
B time taken by particle to cross region -I for $2^{nd}$ time is $\pi \sqrt{\frac{2ml}{13qE}}$
C Angular speed of $1^{st}$ revolution in magnetic field is $\sqrt{\frac{5qE}{2ml}}$

Using work-energy theorem till string become taut
Work done by $qE$ = change in K.E
$\Rightarrow qEl=\frac{1}{2}m{v}^2-0$
$\Rightarrow v=\sqrt{\frac{2qEl}{m}}$
This speed $v$ is vertically upwards.




When string becomes taut an impulse is received and speed (tangential) of the bob becomes
$v\cos {30}^{\circ }=\sqrt{\frac{3qEl}{2m}}$
Further using work energy theorem
Work done by $qE$ = Change in K.E,
$qE\frac{l}{2}=\frac{1}{2}m{v}^{\prime 2}-\frac{1}{2}m(v\cos {30}^{\circ }{)}^2$
$qE\frac{l}{2}=\frac{1}{2}m{v}^{\prime 2}-\frac{1}{2}m\frac{3qEl}{2m}$
$qE\frac{l}{2}=\frac{1}{2}m{v}^{\prime 2}-\frac{1}{2}\frac{3qEl}{2}$
$m{v}^{\prime 2}=\frac{5qEl}{2}$
speed of the bob as it enters magnetic field
$v^{\prime }=\sqrt{\frac{5qEl}{2m}}$
For the first half revolution in the magnetic field particle will move with same constant speed $v^{\prime }$
$\therefore$ time period to cross region $I$ for first time =$\frac{\pi l}{v^{\prime }}=\pi \sqrt{\frac{2ml}{5qE}}$
After $1^{st}$ half revolution in magnetic field, particle enters the region of electric field. Due to work done by electric field there is change in KE.
Let $v_1$ be velocity of particle as it enters region-$I$ for the second time.



Work done by $qE$ = Change in K.E, from lowermost to uppermost point.
$qE.2l=\frac{m{v}_1^2}{2}-\frac{m}{2}\frac{5qEl}{2m}$
$v_1=\sqrt{\frac{13qEl}{2m}}$
time period to cross region $I$ for second time = $\frac{\pi l}{v_1}=\pi \sqrt{\frac{2ml}{13qE}}$
Angular speed of $1^{st}$ half revolution in region $I$= $\frac{v^{\prime }}{l}=\sqrt{\frac{5qE}{2ml}}$
Angular speed of $2^{nd}$ half revolution in region = $\frac{v_1}{l}=\sqrt{\frac{13qE}{2ml}}$