Question

An $n$ sided regular polygon is inscribed in a circular region of radius $R$ using a wire of cross sectional radius $r$ and resistivity $\rho$. If the magnetic field in the circular region has a time rate of change given by $B$ then the current in the wire loop at any time is given by

• A. $\frac{\pi RB{r}^2\cos \left(\pi /n\right)}{2\rho }$
• B. $\frac{\pi RB{r}^2\cos \left(\pi /n\right)}{\sqrt{2}\rho }$
• C. $\frac{\pi RB{r}^2\sin \left(\pi /n\right)}{2\rho }$
• D. $\frac{\pi RB{r}^2\cos \left(\pi /n\right)}{3\rho }$

Answer 1

The correct option is A $\frac{\pi RB{r}^2\cos \left(\pi /n\right)}{2\rho }$

Side length of rectangular side polygon =$2R\sin \frac{\pi }{n}$

Area of polygon $=\frac{1}{2}n{R}^2\sin \frac{\pi }{n}$

Now flux $\varphi$ passes through $=\overrightarrow{B}.ds$

emf induced $d\epsilon =-\frac{d\varphi }{dt}=-\frac{d\overrightarrow{B}.ds}{dt}$

$\epsilon =-\frac{d\overrightarrow{B}\times \frac{1}{2}n{R}^2\sin \left(\frac{2\pi }{n}\right)}{dt}=\frac{n{R}^2\sin \left(\frac{2\pi }{n}\right)}{2}\frac{d\overrightarrow{B}}{dt}$

$\left|\epsilon \right|=\frac{n}{2}{R}^2\sin \left(\frac{2\pi }{n}\right)\times B(Given\frac{d\overrightarrow{B}}{dt}=B)$

$\int ds\to \frac{n}{2}{R}^2\sin \left(\frac{2\pi }{n}\right)$

Current $i$ through loop at any time t will be,

$i=\frac{\epsilon }{R}$

$R=\rho \frac{l}{A}$

where, $\rho$=restivity

$l$=length of wire

$A$=area of cross section of area=$\pi r^2$

$l=2R\sin \frac{\pi }{n}$

$i=\frac{R^2\frac{1}{2}nB\sin \left(\frac{2\pi }{n}\right)}{\rho \times \left(\frac{2R\sin \frac{\pi }{n}}{\pi r^2}\right)}$

$i=\frac{\pi RB{r}^2\cos \frac{\pi }{n}}{2\rho }$