Two long current carrying conductors are placed parallel to each other at a distance of 8 cm between them. The magnitude of magnetic field produced at mid-point between the two conductors due to current flowing in them is 300 $\mu T$ . the equal current flowing in the two conductors is:

When the ring rotates about its axis with a uniform frequency fHz, the current flowing in the ring is

I=q/T=qf

Magnetic field at the centre of the ring is

for $\pi R^{2}Area\to \text{'}l\text{'}$ current\

1 unit Area ® $\frac{l}{\pi R^2}$  

              $\pi r^2\to \frac{l}{\pi R^2}\times \pi r^2$  

$i=\frac{l{r}^2}{R^2}$              

 

Now, consider Amperian loop of radius small ‘r’ ln Amperian loop magnetic field will be tangential to the amperian loop.

${\displaystyle ?\overrightarrow{B}.d\overrightarrow{i}={\mu }_0{l}_{enclosed}}$        

${\stackrel{?}{\mathrm{B}}}_{\mathrm{P}}={\stackrel{?}{\mathrm{B}}}_{\text{upper wire}}\otimes +{\stackrel{?}{\mathrm{B}}}_{\text{semi-circle}}?+{\stackrel{?}{\mathrm{B}}}_{\text{lower-wire}}\otimes$

$B_P=-\frac{{\mu }_{0}i}{4\pi R}+\frac{{\mu }_{0}i}{4R}-\frac{{\mu }_{0}i}{4\pi R}=\frac{{\mu }_0\mathrm{i}}{4\mathrm{R}}\left(1-\frac{2}{\pi }\right)$ pointing away from the page

In external magnetic field, a magnetic force acts on every small part of the loop in direction perpendicular to the wire. Thus, loop assumes a shape (circular) in which it covers maximum area