MHTCET 15 Apr 2026 Shift 2 Paper

Concept:For a circular coil, the magnetic field at the centre is given by $B = \frac{\mu_0 N I}{2r}$, where $N$ is the number of turns and $r$ is the radius.Explanation:Let the total length of the wire be $L$.When the wire forms one circular turn of radius $R$, we have:$$L = 2\pi R$$So the magnetic field for one turn is:$$B_1 = \frac{\mu_0 I}{2R}$$When the same wire is bent into a coil of $n$ turns with smaller radius $r$, the total length still equals $L$:$$L = n(2\pi r)$$Since $L = 2\pi R$, we get:$$2\pi R = 2\pi n r$$$$\Rightarrow R = nr$$$$\Rightarrow r = \frac{R}{n}$$For the $n$-turn coil, the magnetic field is:$$B_2 = \frac{\mu_0 n I}{2r}$$Substitute $r = \frac{R}{n}$:$$B_2 = \frac{\mu_0 n I}{2\left(\frac{R}{n}\right)}$$$$B_2 = \frac{\mu_0 n^2 I}{2R}$$Therefore, the ratio is:$$\frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{2R}}{\frac{\mu_0 n^2 I}{2R}} = \frac{1}{n^2}$$$$B_1 : B_2 = 1 : n^2$$Answer:Option C: $1 : n^{2}$