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A square frame of side $1 \mathrm{~m}$ carrying a current $I$ produces a magnetic field $B$ at its centre. The same current is passed through a circular coil having the same perimeter as the square. The magnetic field at the centre of the circular coil is $B^{\prime}$. The ratio of $\frac{B}{B^{\prime}}$ is
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Verified Answer
The correct answer is:
$\frac{8 \sqrt{2}}{\pi^2}$
Given,
Side of square frame, $a=1 \mathrm{~m}$
Electric current $=I$
For square,
Let $r$ be the distance of centre $O$ from all sides, $r=\frac{a}{2}$
Since, all sides of square produces equal field in the same direction.
Using, $B=4\left[\frac{\mu_0 I}{4 \pi r}\left(\sin 45^{\circ}+\sin 45^{\circ}\right)\right]$
$\Rightarrow \quad B=\frac{4 \mu_0 I}{4 \pi\left(\frac{a}{2}\right)}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right) \Rightarrow B=\frac{2 \mu_0 I}{\pi a} \frac{2}{\sqrt{2}}$
$B=\frac{2 \sqrt{2} \mu_0 I}{\pi a}$ ...(i)
For circle,
Let $R$ be the radius of circular coil.
Then, $2 \pi R=4 a$
$\Rightarrow \quad R=\frac{4 a}{2 \pi}=\frac{2 a}{\pi}$
Using the expression for magnetic field at centre of circular coil,
$B^{\prime}=\frac{\mu_0 I}{2 R}$
$\Rightarrow \quad B^{\prime}=\frac{\mu_0 I}{2\left(\frac{2 a}{\pi}\right)}=\frac{\mu_0 \pi I}{4 a}$ ...(ii)
From Eqs. (i) and (ii), we get
$\frac{B}{B^{\prime}}=\frac{\frac{2 \sqrt{2} \mu_0 I}{\pi a}}{\frac{\mu_0 \pi I}{4 a}}=\frac{8 \sqrt{2}}{\pi^2}$
Side of square frame, $a=1 \mathrm{~m}$
Electric current $=I$
For square,
Let $r$ be the distance of centre $O$ from all sides, $r=\frac{a}{2}$
Since, all sides of square produces equal field in the same direction.
Using, $B=4\left[\frac{\mu_0 I}{4 \pi r}\left(\sin 45^{\circ}+\sin 45^{\circ}\right)\right]$
$\Rightarrow \quad B=\frac{4 \mu_0 I}{4 \pi\left(\frac{a}{2}\right)}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right) \Rightarrow B=\frac{2 \mu_0 I}{\pi a} \frac{2}{\sqrt{2}}$
$B=\frac{2 \sqrt{2} \mu_0 I}{\pi a}$ ...(i)
For circle,
Let $R$ be the radius of circular coil.
Then, $2 \pi R=4 a$
$\Rightarrow \quad R=\frac{4 a}{2 \pi}=\frac{2 a}{\pi}$
Using the expression for magnetic field at centre of circular coil,
$B^{\prime}=\frac{\mu_0 I}{2 R}$
$\Rightarrow \quad B^{\prime}=\frac{\mu_0 I}{2\left(\frac{2 a}{\pi}\right)}=\frac{\mu_0 \pi I}{4 a}$ ...(ii)
From Eqs. (i) and (ii), we get
$\frac{B}{B^{\prime}}=\frac{\frac{2 \sqrt{2} \mu_0 I}{\pi a}}{\frac{\mu_0 \pi I}{4 a}}=\frac{8 \sqrt{2}}{\pi^2}$
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