MHTCET 19 Apr 2026 Shift 2 Paper

Concept:Magnetic field on the axis of a circular coil depends on the axial distance $x$, while the field at the centre is the maximum value when $x = 0$.Explanation:For a circular coil of radius $R$ carrying current $I$, the magnetic field at an axial point $P$ at distance $x$ from the centre is:$$B_{1} = \frac{\mu_{0} I R^{2}}{2\left(R^{2} + x^{2}\right)^{3/2}}$$At the centre of the coil, $x = 0$, so the magnetic field is:$$B_{2} = \frac{\mu_{0} I}{2R}$$Given that $B_{2} = 8B_{1}$, we can write:$$\frac{B_{2}}{B_{1}} = 8$$Substitute the expressions for $B_{1}$ and $B_{2}$:$$\frac{\frac{\mu_{0} I}{2R}}{\frac{\mu_{0} I R^{2}}{2\left(R^{2} + x^{2}\right)^{3/2}}} = \frac{\left(R^{2} + x^{2}\right)^{3/2}}{R^{3}} = 8$$Raising both sides to the power $\frac{2}{3}$, we get:$$\frac{R^{2} + x^{2}}{R^{2}} = 8^{2/3} = 4$$Therefore:$$R^{2} + x^{2} = 4R^{2}$$$$x^{2} = 3R^{2}$$$$x = \sqrt{3}R$$Answer:The value of $x$ is $\sqrt{3}R$, which corresponds to Option B.