Q 12-04-151JEE MainJEE Main 2022 (25 Jul, Shift 2)Medium
The electric current in a circular coil of $2$ turns produces a magnetic induction $B_1$ at its centre. The coil is unwound and is rewound into a circular coil of $5$ turns and the same current produces a magnetic induction $B_2$ at its centre. The ratio of $\dfrac{B_2}{B_1}$ is:
Answer: (B) $\dfrac{25}{4}$
Same wire length: $2(2\pi r_1) = 5(2\pi r_2)\Rightarrow r_2 = \dfrac25r_1$. With $B = \dfrac{\mu_0NI}{2r}$:
$$\frac{B_2}{B_1} = \frac{N_2}{N_1}\cdot\frac{r_1}{r_2} = \frac52\times\frac52 = \frac{25}{4}$$
Solution by Sreeraj P, M.Sc Physics