CGPET 2021 Solved Paper

Given, $i = i_1 + i_2$ .......(i) ∵ The length of part (ABC) is equal to length of part (ABD). Here, resistance of part (ABC) is half that of part (ABD), i.e. $i_1 = 2 i_2$ ........(ii) From eqs. (i) and (ii), we get $i_1 = \frac{2i}{3}, i_2 = \frac{i}{3}$ Magnetic field due to AB and BC wire should be equal, i.e. $B_{AB} = B_{BC}$ Magnetic field due to straight wire, $B = \frac{\mu_0 i}{4 \pi r} (\sin \alpha + \sin \beta)$ For wire AB, From figure, $\alpha = \beta = 45^{\circ}, \text{ and } \text{MO} = \frac{l}{2}$ ∴ $B_{AB} = \frac{\mu_0}{4 \pi} \cdot \frac{i_1}{\frac{l}{2}} (\sin 45^{\circ} + \sin 45^{\circ})$ ⇒ $B_{AB} = \frac{\mu_0}{4 \pi} \cdot \frac{2\sqrt{2} I_1}{l} = B_{BC}$ Also, the magnetic field due to AD and DC wire should be equal, i.e. $B_{AD} = B_{DC}$ For wire AD, $B_{AD} = \frac{\mu_0}{4 \pi} \cdot \frac{2\sqrt{2} i_2}{l} = B_{DC}$ Net magnetic field at centre O is $B_{\text{net}} = B_{AB} + B_{BC} - B_{AD} - B_{DC}$ Since, $(B_{AB}+B_{BC}) > (B_{AD}+B_{DC})$ $B_{\text{net}} = \frac{\mu_0}{4 \pi} \cdot \frac{2\sqrt{2} \times \left(\frac{2}{3} i\right) \times 2}{l} - \frac{\mu_0}{4 \pi} \cdot \frac{2\sqrt{2} \cdot \left(\frac{i}{3}\right) \times 2}{l}$ ⇒ $B_{\text{net}} = \frac{\sqrt{2} \mu_0 i}{3 \pi a}$ The direction of $B_{\text{net}}$ will be inward to the plane of paper.