A wire of certain length carries a steady current. It is first bent to form a circular coil of one turn. The…
A wire of certain length carries a steady current. It is first bent to form a circular coil of one turn. The same wire is then bent to form a circular coil of three turns. The ratio of magnetic inductions at the centre of the coil in the two cases is
1:9
1:3
3:1
9:1
Solution
Magnetic induction at the centre of the coil with $\mathrm{N}$ turns is given by
$B_{\mathrm{N}}=\left(\frac{\mu_0 I}{2 r}\right) \mathrm{N}$
Where $r$ is the radius of each coil is, $I$ is the current flowing through.
$\therefore$ for single loop coil $2 \pi r=L$
$B_1=\frac{\mu_0 I(1)}{2\left(\frac{L}{2 \pi}\right)}=\frac{\mu_0 I \pi}{L}$
& For the three-loop coil $(2 \pi r) 3=L$
$B_3=\frac{\mu_0 I(3)}{2\left(\frac{L}{6 \pi}\right)}=\frac{9 \mu_0 I \pi}{L}$
Or $\frac{B_1}{B_2}=\frac{1}{9}$