Question

Find the magnetic field B at the centre of a rectangular loop of length l and width b, carrying a current i.

Answer 1

Let the angles made by points A and B with point O be \[\theta_1   \text{ and }    \theta_2\] , respectively.

And ,  

\[\theta_1 = \theta_2 = \theta\]

Separation of the point from the wire, d = b/2

\[\text{ In } ∆ AOM, \]
\[\text{ AO }= \sqrt{\left( \frac{b}{2} \right)^2 + \left( \frac{l}{2} \right)^2} = \frac{1}{2}\sqrt{b^2 + l^2}\]

And,

\[\sin \theta = \frac{\frac{l}{2}}{\frac{1}{2}1\sqrt{b^2 + l^2}} = \frac{l}{\sqrt{b^2 + l^2}}\]

Thus, the magnetic field due to current in wire AB is given by 

 

\[B = \frac{\mu_0 i}{4\pi d}(\sin   \theta_1  + \sin   \theta_2 )\] 

\[ \Rightarrow B = \frac{\mu_0 i}{4\pi d} \times 2\sin  \theta    \]

\[\Rightarrow B = \frac{\mu_0 i}{4\pi\left( \frac{b}{2} \right)} \times \frac{2l}{\sqrt{l^2 + b^2}}\]
\[ = \frac{\mu_0 i}{\pi b}\frac{l}{\sqrt{l^2 + b^2}}\]

Similarly, the magnetic field due to current in wire BC is given by

\[B' = \frac{\mu_0 i}{\pi l}\frac{b}{\sqrt{l^2 + b^2}}\]

Now,
Magnetic field due to current in wire CD = Magnetic field due to current in wire AB = B
And,
Magnetic field due to current in wire DA = Magnetic field due to current in wire BC = B'

\[\therefore B_{net} = 2\left[ B + B' \right]\]
\[ = 2\left[ \frac{\mu_0 i}{\pi b}\frac{l}{\sqrt{l^2 + b^2}} + \frac{\mu_0 i}{\pi l}\frac{b}{\sqrt{l^2 + b^2}} \right]\]
\[ = \frac{2 \mu_0 i}{\pi\sqrt{l^2 + b^2}}\left[ \frac{l}{b} + \frac{b}{l} \right]\]
\[ = \frac{2 \mu_0 i\sqrt{b^2 + l^2}}{\pi bl}\]