Question

The coil having one turn is made into coil of two turns. If the same current is passed in both the cases, then the ratio of the magnetic induction at their centers will be

• A. $3:1$
• B. $1:4$
• C. $2:1$
• D. $1:2$

Answer 1

The correct option is B $1:4$

Magnetic field at the center of the coil,
$B=\frac{{\mu }_0}{2\pi }\frac{NI}{a}$
Let $l$ be the length of the wire, then
$B_1=\frac{{\mu }_0}{2\pi }\frac{1\times I}{\frac{l}{2\pi }}=\frac{{\mu }_{0}I}{l}{B}_2=\frac{{\mu }_0}{2\pi }\frac{2\times I}{\frac{l}{4\pi }}=\frac{4{\mu }_{0}I}{l}$
Therefore, $\frac{B_1}{B_2}=\frac{1}{4}$
$B_1:B_2$ = 1 : 4