MHTCET 23 Apr 2025 Shift 1 Paper

Concept:The magnetic moment of a coil is the product of the current flowing through it and its area, while the magnetic field at the centre relates the current to the coil's radius.Explanation:For a single-turn circular coil of radius $r$ carrying current $I$, the magnetic field at its centre is given by:$$B = \frac{\mu_0 I}{2r}$$Rearrange this equation to express the current $I$ in terms of $B$ and $r$:$$I = \frac{2Br}{\mu_0}$$The magnetic moment $M$ of the coil is defined as:$$M = I \times A$$Substitute the expression for $I$ into this equation:$$M = \frac{2Br}{\mu_0} \times A$$The area of the circular coil is $A = \pi r^2$, therefore the radius can be written as:$$r = \sqrt{\frac{A}{\pi}}$$Replace $r$ in the magnetic moment equation:$$M = \frac{2B}{\mu_0} \times \sqrt{\frac{A}{\pi}} \times A$$Simplify by combining the terms involving $A$:$$M = \frac{2B A^{3/2}}{\mu_0 \sqrt{\pi}}$$This result matches Option B.Answer:Option B: $\frac{2 B A^{3/2}}{\mu_0 \sqrt{\pi}}$