JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Magnetic Effects of Current

Each wire is bent at a right angle, so at $O$ one arm of each lies along the line through $O$ and the other is a semi-infinite straight segment at perpendicular distance $4\text{cm}$. A semi-infinite wire produces a field $\frac{{\mu }_{0}i}{4\pi d}$ at a point level with its end, and the two contributing arms add, giving $B=2\cdot \frac{{\mu }_{0}i}{4\pi d}$. Putting $B=10^{-4}\text{T}$, $d=4\times 10^{-2}\text{m}$ and ${\mu }_0=4\pi \times 10^{-7}$, solving $10^{-4}=2\cdot \frac{(4\pi \times 10^{-7})i}{4\pi (4\times 10^{-2})}$ gives $i=20\text{A}$. Applying the right-hand rule to the currents shown, both contributions point into the page, so the field at $O$ is that of a $20\text{A}$ current, directed into the page.