A wire is first bent in the form of a circular coil of 5 turns and the same wire is then bent in the form of…
A wire is first bent in the form of a circular coil of 5 turns and the same wire is then bent in the form of another circular coil of 10 turns. If same current is passed in both the coils, then the ratio of the magnetic fields at their centres is
$1: 8$
$1: 1$
$1: 2$
$1: 4$
Solution
Magnetic field at the centre of a circular coil $\mathrm{B}=\frac{\mu_0 \mathrm{ni}}{2 \mathrm{r}}$ where $\mathrm{n}=$ no. of turns and $\mathrm{r}=$ radius of circular coil.
'B' when the wire is bent in the form of a circular coil of 5 turns ' $\mathrm{B}$ ' when the wire is bent in the form of a circular coil of 10 turns
$=\frac{\mu_0 \mathrm{ni} / 2 \mathrm{r}}{\mu_{0^n} \mathrm{n}^{\prime} i / 2 \mathrm{r}^{\prime}}=\frac{\mathrm{n}}{\mathrm{n}^{\prime}} \times \frac{\mathrm{r}^{\prime}}{\mathrm{r}}=\frac{5}{10} \times \frac{\mathrm{R} / 10}{\mathrm{R} / 5}=\frac{25}{100}=\frac{1}{4}$