Question

A coil of one turn is made of a wire of certain length and then from the same length a coil of two turns is made. If the same current is passed in both the cases, then the ratio of the magnetic inductions at their centres will be

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

Answer 1

Answer: (B)

1 : 4

Answer 2

Magnetic field at the centre of the coil, $B= \frac{{\mu}_0}{2 \pi } \frac{ NI}{ a}$
Let l be the length of the wire, then
$B_1 = \frac{ {\mu}_0}{ 2 \pi }\cdot \frac{1 \times I }{ l /2 \pi } =\frac{ {\mu}_0 I}{ l }$
and $B_2 = \frac{ {\mu_0}}{ 2 \pi } . \frac{2 \times I}{ l \times 4 \pi } = \frac{ 4 {\mu}_0 I}{l }$
Therefore, $\frac{B_1}{B_2} = \frac{1}{4}$ or $B_1: B_2 = 1:4$