Question

Figure shows a square loop ABCD with edge-length a. The resistance of the wire ABC is r and that of ADC is 2r. Find the magnetic field B at the centre of the loop assuming uniform wires.

Answer 1

Let the currents in wires ABC and ADC be i₁ and i₂, respectively.
The resistances in wires ABC and ADC are r and 2r, respectively.

$$\begin{gathered} \therefore \frac{i_1}{i_2}=\frac{2}{1} \\ \Rightarrow i_1-2i_2=0...\left(1\right) \end{gathered}$$
And,
$i_1+i_2=i...\left(2\right)$
Using (1) and (2), we get
$i_1=\frac{2i}{3}$ and $i_2=\frac{i}{3}$
The angles made by points A and D with point O are ${\theta }_1=45°\mathrm{and}{\theta }_2=45°$, respectively.
Separation of the point from the wire, d = a/2
Now,
The magnetic field due to current in wire AD is given by
$$\begin{gathered} B=\frac{{\mathrm{\mu }}_0{i}_2}{4\mathrm{\pi }d}(\sin {\theta }_1+\sin {\theta }_2) \\ \Rightarrow B=\frac{{\mathrm{\mu }}_0{\displaystyle \frac{i}{3}}}{4\mathrm{\pi }{\displaystyle \frac{a}{2}}}(\sin 45+\sin 45) \end{gathered}$$
The magnetic field at centre due to wire ADC is given by
$$\begin{gathered} B\text{'}=2B=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{i}{3}\frac{a}{a^2}\times 4\times \sqrt{2} \\ =\frac{\sqrt{2}{\mathrm{\mu }}_{0}i}{3\mathrm{\pi }a} \end{gathered}$$
(Perpendicular to the plane in outward direction)

The magnetic field at centre due to wire ABC is given by
$$\begin{gathered} B\text{'}\text{'}=\frac{\mathrm{\mu }}{4\mathrm{\pi }}\frac{2i}{3}\frac{a}{a^2}\times 4\times \sqrt{2} \\ =\frac{2\sqrt{2}{\mathrm{\mu }}_{0}i}{3\mathrm{\pi }a} \end{gathered}$$
(Perpendicular to the plane in inward direction)
$\therefore B_{\mathrm{net}}=B\text{'}\text{'}-B\text{'}=\frac{\sqrt{2}{\mathrm{\mu }}_{0}i}{3\mathrm{\pi }a}$
(Perpendicular to the plane in inward direction)