KCET 2020 Physics Solved Paper

The magnetic field due to a long straight current carrying wire is given by $B=\frac{\mu_{0} I}{2\pi r}$ or, $B\propto\frac{1}{r}$ At the point exactly mid-way between the conductors, the net magnetic field is zero. Using right hand thumb rule, we find that the magnetic field due to left wire will be in $\hat{j}$ direction while due to the right wire is in $(-\hat{j})$ direction. Magnetic field at a distance x from the left wire, lying between the wires. $B=\frac{\mu_{0} I}{2\pi x}\hat{j}+\frac{\mu_{0} I}{2\pi(2d-x)}(-\hat{j})$ or, $B=\frac{\mu_{0} I}{2\pi}\left(\frac{1}{x}-\frac{1}{2d-x}\right)$ At $x=d, B=0$ For $x < d, B$ is along $\hat{j}$ For $x>d, B$ is along $(-\hat{j})$ On the left side of the left conductor, magnetic fields due to the currents will add up and the net magnetic field will be along$(-\hat{j})$ direction. To the right side of second conductor, the total magnetic field will be along $\hat{j}$ direction.