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. 

\[\therefore \frac{i_1}{i_2} = \frac{2}{1}\]
\[ \Rightarrow i_1 - 2 i_2 = 0 . . . \left( 1 \right)\]

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 \[\theta_1  = 45^\circ \text{ and }   \theta_2  = 45^\circ \] , respectively.

 

Separation of the point from the wire, d = a/2
Now,
The magnetic field due to current in wire AD is given by

\[B = \frac{\mu_0 i_2}{4\pi d}(\sin \theta_1 + \sin \theta_2 )\]
\[ \Rightarrow B = \frac{\mu_0 \frac{i}{3}}{4\pi\frac{a}{2}}(\sin45 + \sin45)\]

The magnetic field at centre due to wire ADC is given by

\[B'   =   2B   =   \frac{\mu_0}{4\pi}\frac{i}{3}\frac{a}{a^2} \times 4 \times \sqrt{2}\] 

\[         = \frac{\sqrt{2} \mu_0 i}{3\pi a}\] 

(Perpendicular to the plane in outward direction)

The magnetic field at centre due to wire ABC is given by

\[B'' = \frac{\mu}{4\pi}\frac{2i}{3}\frac{a}{a^2} \times 4 \times \sqrt{2}\]
\[ = \frac{2\sqrt{2} \mu_0 i}{3\pi a}\]

(Perpendicular to the plane in inward direction)

\[\therefore B_{net} = B'' - B' = \frac{\sqrt{2} \mu_0 i}{3\pi a}\]

(Perpendicular to the plane in inward direction)