Figure shows a current carrying square loop ABCD of edge length 'a' lying in a plane. If the resistance of the ABC part is r and that of ADC part is 2r, then the magnitude of the resultant magnetic field at centre of the square loop is

Resistance of $$ABC = r$$

Resistance of $$ADC = 2r$$

$$i_1 = I \times \frac{2r}{r + 2r} = \frac{2}{3}I$$

$$i_2 = I \times \frac{r}{r + 2r} = \frac{1}{3}I$$

$$B_0 = \frac{\mu_0 i}{4\pi \left(\frac{a}{2}\right)}(\sin 45^\circ + \sin 45^\circ) = \frac{\mu_0 i}{2\pi a}\left(\frac{2}{\sqrt{2}}\right) = \frac{\sqrt{2}\mu_0 i}{2\pi a}$$

$$B_{ABC} = 2 \times \left(\frac{\sqrt{2}\mu_0 i_1}{2\pi a}\right) = \frac{\sqrt{2}\mu_0 \left(\frac{2}{3}I\right)}{\pi a} = \frac{2\sqrt{2}\mu_0 I}{3\pi a} \quad (\odot)$$

$$B_{ADC} = 2 \times \left(\frac{\sqrt{2}\mu_0 i_2}{2\pi a}\right) = \frac{\sqrt{2}\mu_0 \left(\frac{1}{3}I\right)}{\pi a} = \frac{\sqrt{2}\mu_0 I}{3\pi a} \quad (\otimes)$$

$$B_{\text{net}} = B_{ABC} - B_{ADC} = \frac{2\sqrt{2}\mu_0 I}{3\pi a} - \frac{\sqrt{2}\mu_0 I}{3\pi a} = \frac{\sqrt{2}\mu_0 I}{3\pi a}$$