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

Explanation:

Find the magnetic field B at the centre of a rectangular loop of length l and width b, carrying a current i. =2μoi√b2+l2πbl. CBSE to introduce major changes in exam pattern of Class 10 and Class 12 board classes by 2023. Get here complete updates on the CBSE 2020 exam.

Answer 2

The magnetic field B at the centre of a rectangular loop of length l and width b, carrying a current i is given by $\frac{2 \mu_{0} i \sqrt{b^{2}+l^{2}}}{\pi l b}$ .

Explanation:

Step 1:

Let $\theta_1$ and $\theta_2$ , respectively, be the angles of points A and B with point O.

$\theta_1 = \theta_2$ = θ  

Divide the point from wire, d = $\frac{b}{2}$

In case of  ∆AOM,

$A O=\sqrt{\left(\frac{b}{2}\right)^{2}+\left(\frac{l}{2}\right)^{2}}=\frac{1}{2} \sqrt{b^{2}+l^{2}}$

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

Step 2:

Therefore, the current magnetic field in wire AB is given by

$B=\frac{\mu_{0} i}{4 \pi d}\left(\sin \theta_{1}+\sin \theta_{2}\right)$

$B=\frac{\mu_{0} t}{4 \pi d} \times 2 \sin \theta$

$B=\frac{\mu_{0} i}{4 \pi \frac{b}{2}} \times \frac{2 l}{\sqrt{b^{2}+l^{2}}}$

  $=\frac{\mu_{0} i}{\pi b} \times \frac{l}{\sqrt{b^{2}+l^{2}}}$

Step 3:

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

$b^{\prime}=\frac{\mu_{0} i}{\pi l} \times \frac{b}{\sqrt{b^{2}+l^{2}}}$

Current magnetic field in wire CD = Magnetic field due to current in wire AB = B  

And,  

Current in wire DA = Magnetic field due to current in wire BC = B '

$b^{\prime}=\frac{\mu_{0} i}{\pi l} \times \frac{b}{\sqrt{b^{2}+l^{2}}}$

  $=2\left[\frac{\mu_{0} i}{\pi b} \times \frac{l}{\sqrt{b^{2}+l^{2}}}+\frac{\mu_{0} i}{\pi l} \times \frac{b}{\sqrt{b^{2}+l^{2}}}\right]$

   $=2 \frac{\mu_{0} i}{\pi \sqrt{b^{2}+l^{2}}} \times\left[\frac{l}{b}+\frac{b}{l}\right]$

  $=\frac{2 \mu_{0} i \sqrt{b^{2}+l^{2}}}{\pi l b}$