Question

A square frame of side L carries a current i The magnetic field at its centre is B. The same current is passed through a circular coil having the same perimeter as the square. The field at the centre of the circular coil is B'. What is the ratio of B/B' ?

Answer 1

Let's consider the square loop ABCD with side length = l.

We will use the formula
$B =$μ₀$\frac{i}{4\pi d} (sin x_{1} + sin x_{2})$

where $x_1$ and $x_2$ are the angles subtended from the given point to the ends of the straight current carrying conductor.

Diagonals bisect each other at 90° for a square.
By symmetry, in the given scenario, $x_1$ and $x_2$ are each 45°.
For each side, the magnetic field at the centre = μ₀$\frac{i}{4 \pi \frac{l}{2}}(sin 45+sin 45)$
= μ₀$\frac{i}{\pi l}$ = $B_{1}$ (say)

Now, each side of the loop carrying current i contributes to the magnetic field, in the same direction. There are four sides contributing the same.
So the resultant magnetic field

$B = 4B_1$
⇒$B =$ 4μ₀ $\frac{i}{\pi l}$

Perimeter of the given loop = 4l
If this same loop is made into a circle, perimeter will be same.
Let the radius of the circle be r.
Hence, 2πr = 4l.
⇒r = $\frac{2l}{\pi}$

For a circular loop carrying current i and of radius r,
The magnetic field at the centre (B) = μ₀$\frac{i}{2r}$

Therefore, B' = μ₀$\frac{i}{2.\frac{2l}{\pi}}$ = μ₀$\frac{ \pi i}{4l}$

Ratio asked for = B/B'.

$\frac{B}{B'} = \frac{16}{ \pi ^{2} }$

And that's the answer.

[ANSWERED]

Answer 2

ans I hope ans is correct