Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

A light charged particle is revolving in a circle of radius r in electrostatic attraction of a static heavy particle with opposite charge. How does the magnetic field B at the centre of the circle due to the moving charge depend on r ?
Option: 1
$B \propto \frac{1}{r}$

Option: 2
$B \propto \frac{1}{r^2}$

Option: 3
$B \propto \frac{1}{r^{3/2}}$

Option: 4
$B \propto \frac{1}{{r}^{5 / 2}}$

Answers (1)

best_answer

Let a light particle of mass m and charge $q_2$ revolve around a heavy particle of charge $+q_1$ in a circular path of radius r with a velocity v

Now, $\frac{m v^2}{r}=\frac{1}{4 \pi \epsilon_0} \frac{\left(+q_1\right) \times\left(-q_2\right)}{r^2}$

$\begin{aligned} & \text { or, } r^2=\frac{1}{4 \pi E_0} \cdot \frac{-q_1 q_2}{m r} \\ & \therefore \quad V \propto \frac{1}{\sqrt{r}} \end{aligned}$

Time period of the charge in circular orbit. $T=\frac{2 \pi r}{v}$

$\therefore$ Current due to the revolving charge

$I=\frac{-q_2}{T}=\frac{q_2 \times v}{2 \pi r}$

$\therefore I \propto \frac{v}{r} \text { or } I \propto \frac{1}{r^3 / 2}\left[\because V \propto \frac{1}{\sqrt{r}}\right]$

So, magnetic field at the centre of the circle due to the moving charge $+q_2$ ,

$B=\frac{\mu_0 I}{2 r} \text { or }, B \propto \frac{I}{r}$

$\text { or } B \propto \frac{1}{r^{3 / 2} \times r} \text { or, } B \propto \frac{1}{r^{5 / 2}}$

Posted by

qnaprep

View full answer

NEET 2024 Most scoring concepts

Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

Want a career in medicine without NEET? Consider paramedical courses after Class 12

anam-khan

AFMC hikes MBBS seat leaving bond penalty, mandates 7 years of commission service

anam-khan

‘Patience and not backing down’: Kabir Paharia’s SC fight eases MBBS admissions for disabled candidates

Latest Question