Determine the magnetic field B at the centre of a rectangular loop...

The magnetic field at the centre of the rectangular loop is the sum of the magnetic fields produced by its four sides.

The magnetic field due to a finite straight wire of length $L$ at a perpendicular distance $d$ from its centre is given by:

$B' = \dfrac{\mu_0 i}{4 \pi d} (\sin \theta_1 + \sin \theta_2)$

For the two sides of length $l$, the perpendicular distance to the centre is $d = \dfrac{b}{2}$. The angles are $\sin \theta_1 = \sin \theta_2 = \dfrac{l}{\sqrt{l^2 + b^2}}$.

The magnetic field due to these two sides is:

$B_1 = 2 \times \dfrac{\mu_0 i}{4 \pi \left(\dfrac{b}{2}\right)} \left( \dfrac{2l}{\sqrt{l^2 + b^2}} \right) = \dfrac{2 \mu_0 i l}{\pi b \sqrt{l^2 + b^2}}$

For the two sides of length $b$, the perpendicular distance to the centre is $d = \dfrac{l}{2}$. The angles are $\sin \theta_1 = \sin \theta_2 = \dfrac{b}{\sqrt{l^2 + b^2}}$.

The magnetic field due to these two sides is:

$B_2 = 2 \times \dfrac{\mu_0 i}{4 \pi \left(\dfrac{l}{2}\right)} \left( \dfrac{2b}{\sqrt{l^2 + b^2}} \right) = \dfrac{2 \mu_0 i b}{\pi l \sqrt{l^2 + b^2}}$

The total magnetic field at the centre is the sum of $B_1 and$B_2$as they point in the same direction:<br><br>$B = B_1 + B_2 = \dfrac{2 \mu_0 i}{\pi \sqrt{l^2 + b^2}} \left(\dfrac{l}{b} + \dfrac{b}{l} \right)

$B =$\dfrac{2 \mu_0 i}{\pi \sqrt{l^2 + b^2}} \left( \dfrac{l^2 + b^2}{l b} \right) = \dfrac{2 \mu_0 i \sqrt{l^2 + b^2}}{\pi l b}