Question

Consider the situation described in the previous problem. Suppose the current i enters the loop at the points A and leaves it at the point B. Find the magnetic field at the centre of the loop.

Answer 1

The loop ABCD can be considered as a circuit with two resistances in parallel, one along branch AB and other along branch ADC.

As, the sides of the loop are identical, their resistances are also same.

Let the resistance of each side be r.

The resistance of branch AB = r

The resistance of branch ADC = 3r

The current in the branches are calculated as:

$i_{\mathrm{AB}}=i\times \frac{3r}{3r+r}=\frac{3i}{4}$

$i_{\mathrm{ADC}}=i\times \frac{r}{3r+r}=\frac{i}{4}$
As current follow the least resistive path so

Current in branch AB = $\frac{3i}{4}$
Current in branch ADC = $\frac{i}{4}$
At the centre of the loop:

Magnetic field due to wire AD, DC and CB will be into the plane of paper according to right hand thumb rule.
Magnetic field due to wire AB will be out of the plane of paper according to right hand thumb rule.
Net magnetic field at the centre = B_AD +B_DC +B_CB − B_AB which will be out of the plane of paper.
As, perpendicular distance of the centre from every wire will be equal to $\frac{a}{\sqrt{2}}$ and angle made by corner points of each side at the centre is $45°$.
$$\begin{gathered} B_{AD}=\frac{{\mu }_0\left({\displaystyle \frac{i}{4}}\right)}{4\mathrm{\pi }\left({\displaystyle \frac{\mathrm{a}}{\sqrt{2}}}\right)}(\sin 45°+\sin 45°) \\ =\frac{{\mu }_0{\displaystyle i}}{8\mathrm{\pi a}} \\ {B}_{AD}=B_{DC}=B_{CB} \\ {B}_{AD}+B_{DC}+B_{CB}=B^{,}=\frac{3{\mu }_0{\displaystyle i}}{8\mathrm{\pi a}} \\ {B}_{AB}=\frac{{\mu }_0\left({\displaystyle \frac{3i}{4}}\right)}{4\mathrm{\pi }\left({\displaystyle \frac{\mathrm{a}}{\sqrt{2}}}\right)}(\sin 45°+\sin 45°) \\ =\frac{3{\mu }_0{\displaystyle i}}{8\mathrm{\pi a}} \\ {B}_{net}=B^{,}-B_{AB} \\ =0 \end{gathered}$$