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
AIPMT 1998, Medium
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Magnetic field at the centre of the coil, $B=\frac{\mu_0}{2 \pi} \frac{N I}{a}$
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 $B_2=\frac{\mu_0}{2 \pi} \cdot \frac{2 \times I}{l / 4 \pi}=\frac{4 \mu_0 I}{l}$
Therefore, $\frac{B_1}{B_2}=\frac{1}{4}$
or, $B_1: B_2=1: 4$
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