A coil of one turn is made of a wire of certa | vedclass

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Question

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}

Vedclass · Accepted Answer

$1 : 4$

Vedclass · Answer

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

$B_1=\frac{\mu_0}{2 \pi} \cdot \frac{1 	imes I}{l / 2 \pi}=\frac{\mu_0 I}{l}$

and $B_2=\frac{\mu_0}{2 \pi} \cdot \frac{2 	imes I}{l / 4 \pi}=\frac{4 \mu_0 I}{l}$

Therefore, $\frac{B_1}{B_2}=\frac{1}{4}$

or, $B_1: B_2=1: 4$

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

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