JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Magnetic Effects of Current

The point $O$ lies on the common axis a distance $r$ from each loop, and each loop has radius $r$, so the axial field of a loop carrying current $i$ is $B=\frac{{\mu }_{0}i{r}^2}{2{\left(r^2+r^2\right)}^{3/2}}=\frac{{\mu }_{0}i}{4\sqrt{2}r}$. Loop $P$ with current $I$ gives $B_P=\frac{{\mu }_{0}I}{4\sqrt{2}r}$, and loop $Q$ with current $4I$ gives $B_Q=\frac{4{\mu }_{0}I}{4\sqrt{2}r}=\frac{{\mu }_{0}I}{\sqrt{2}r}$. Both currents appear clockwise from $O$, so the field of $P$ points towards $P$ while that of $Q$ points towards $Q$; the two are opposite and the stronger loop $Q$ prevails. The resultant is $B_Q-B_P=\frac{{\mu }_{0}I}{\sqrt{2}r}-\frac{{\mu }_{0}I}{4\sqrt{2}r}=\frac{3{\mu }_{0}I}{4\sqrt{2}r}$, directed towards $Q$.