AB µ r² long straight wire with a circular cross-section having radius R, is carrying a steady current I: The current I is uniformly distributed across this cross-section. Then the variation of magnetic field due to current I with distance r(r < R) from its centre will be:

If currents are flowing in same direction, magnetic field will cancel each other, so the currents must flowing in opposite direction

$B_P=\frac{{\mu }_{0}I}{2\pi r}\times 2$

$\Rightarrow 300\times 10^{-6}=\frac{4\pi \times 10^{-7}}{2\pi \times 4\times 10^{-2}}\times 2$                                               &

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

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

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