Question

Figure shows a square loop of edge a made of a uniform wire. A current i enters the loop at the point A and leaves it at the point C. Find the magnetic field at the point P which is on the perpendicular bisector of AB at a distance a/4 from it.
Figure

Answer 1

Explanation:

To find : The magnetic field at the point P which is on the perpendicular bisector of AB at a distance a/4 from it.

Step 1:

B at P since AD $=\frac{\mu_{0}}{4 \pi} \cdot \frac{i}{2} \cdot \frac{4}{d^{2}} \cdot \frac{4}{d^{2}} \cdot\left[\frac{\frac{a}{a}}{\sqrt{\left(\frac{a}{2}\right)^{2}+\left(\frac{a}{4}\right)^{2}}}+\frac{\frac{a}{2}}{\sqrt{\left(\frac{a}{2}\right)^{2}+\left(\frac{3 a}{4}\right)^{2}}}\right] \frac{\mu_{0} i}{4 \pi a}\left[\frac{\frac{a}{2}}{\sqrt{\left(\frac{a}{2}\right)^{2}+\left(\frac{a}{4}\right)^{2}}}+\frac{\frac{a}{2}}{\sqrt{\left(\frac{a}{2}\right)^{2}+\left(\frac{3 a}{2}\right)^{2}}}\right]$

$B \text { at } P \text { due to } A C=\frac{\mu_{0}}{4 \pi} \cdot \frac{i}{2} | \frac{16}{9 a^{2}} \cdot a \cdot \frac{\frac{3 a}{4}}{\sqrt{\left(\frac{a}{2}\right)^{2}+\left(\frac{3 a}{4}\right)^{2}}}$

$=\frac{4 \mu_{0} i}{9 \pi a} \frac{\frac{3 a}{4}}{\sqrt{\left(\frac{a}{2}\right)^{2}+\left(\frac{3 a}{4}\right)^{2}}}$

$\text { B at } P \text { due to } A B=\frac{\mu_{0}}{4 \pi} \cdot \frac{i}{2} \cdot \frac{16}{9 a^{2}} \cdot a \cdot \frac{\frac{3 a}{4}}{\sqrt{\left(\frac{a}{2}\right)^{2}+\left(\frac{3 a}{4}\right)^{2}}}$

$\text { B at } P \text { since } B C=\frac{\mu_{0}}{4 \pi} \cdot \frac{i}{2} \cdot \frac{4}{a^{2}} \cdot \frac{4}{a^{2}} \cdot a\left[\frac{\frac{a}{2}}{\sqrt{\left(\frac{a}{2}\right)^{2}+\left(\frac{a}{4}\right)^{2}}}+\frac{\frac{a}{2}}{\sqrt{\left(\frac{a}{2}\right)^{2}+\left(\frac{3 a}{4}\right)^{2}}}\right]$

$=\frac{\mu_{0} i}{2 \pi a}\left[\frac{\frac{a}{2}}{\sqrt{\left(\frac{a}{2}\right)^{2}+\left(\frac{a}{4}\right)^{2}}}+\frac{\frac{a}{2}}{\sqrt{\left(\frac{a}{2}\right)^{2}+\left(\frac{3 a}{4}\right)^{2}}}\right]$

Step 2:

So,at point P, net magnetic field is

$B=\frac{4 \mu_{0} t}{\pi a}\left[\frac{\frac{a}{4}}{\sqrt{\left(\frac{a}{2}\right)^{2}+\left(\frac{a}{4}\right)^{2}}}\right]-\frac{4 \mu_{0} i}{9 \pi a}\left[\frac{\frac{3 a}{4}}{\sqrt{\left(\frac{a}{2}\right)^{2}+\left(\frac{3 a}{4}\right)^{2}}}\right]$

$B=\frac{4 \mu_{0} i}{4 \pi a}\left[\frac{4}{\sqrt{5}}\right]-\frac{\mu_{0} i}{3 \pi a}\left[\frac{4}{\sqrt{13}}\right]\left[\frac{4}{\sqrt{13}}\right]$

$B=\frac{4 \mu_{0} i}{2 \pi a}\left[\frac{1}{\sqrt{5}}-\frac{1}{3 \sqrt{13}}\right]$

Answer 2

$\huge{\boxed{\mathcal\pink{\fcolorbox{red}{yellow}{Answer}}}}$

Above answer is correct

hope it help