Answer

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**Hint:**The magnetic field is the vector field caused due to the influence of the moving electric charges, electric current and also from magnetized materials. The formula of the magnetic field can be used to solve this problem.

**Formula used:**The magnetic field for the current carrying coil in square shape is given by, $B = \dfrac{{{\mu _o}I}}{{4\pi L}}\left[ {\sin {\theta _1} + \sin {\theta _2}} \right]$ where I is the current in the coil L is the length of the side of the square also ${\theta _1}$ and ${\theta _2}$ be the angle of the length from the centre to the ends of the square of the length L and the formula for the magnetic field for the current carrying coil in the circular shape is given by$B = \dfrac{{{\mu _o}I}}{{2r}}$.

**Step by step solution:**

It is given that the case one is the coil made into the shape of the square of each side with length L.

The magnetic field for the coil carrying current in the coil in the shape of square is given by,

$B = \dfrac{{{\mu _o}I}}{{4\pi L}}\left[ {\sin {\theta _1} + \sin {\theta _2}} \right]$ Where I is the current in the coil L is the length of the side of the square also ${\theta _1}$ and ${\theta _2}$ be the angle of the length from the centre to the ends of the square of the length L.

The angle made by the ends of any length L of the square to the centre is ${\theta _1} = {\theta _2} = 45^\circ $ as the diagonal of the square makes an angle of $45^\circ $ with the sides of the square.

The magnetic field for the current carrying coil in the shape of square is equal to,

$ \Rightarrow B = \dfrac{{{\mu _o}I}}{{4\pi L}}\left[ {\sin {\theta _1} + \sin {\theta _2}} \right]$

There are four sides in a square. The total magnetic field of the coil will be equal to.

$ \Rightarrow {B_1} = \dfrac{{{\mu _o}I}}{{4\pi L}}\left[ {\sin {\theta _1} + \sin {\theta _2}} \right] \times 4$

Replace the value of ${\theta _1} = {\theta _2} = 45^\circ $ and $L = \dfrac{L}{4}$ in the above relation.

$ \Rightarrow {B_1} = \dfrac{{{\mu _o}I}}{{4\pi L}}\left[ {\sin {\theta _1} + \sin {\theta _2}} \right] \times 4$

$ \Rightarrow {B_1} = \dfrac{{{\mu _o}I}}{{4\pi \left( {\dfrac{L}{4}} \right)}}\left[ {\sin 45^\circ + \sin 45^\circ } \right] \times 4$

$\Rightarrow {B_1} = \dfrac{{{\mu _o}I}}{{4\pi \left( {\dfrac{L}{4}} \right)}}\left[ {\dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }}} \right] \times 4$

$ \Rightarrow {B_1} = \dfrac{{{\mu _o}I}}{{\pi L}}\left( {\sqrt 2 } \right) \times 4$

$ \Rightarrow {B_1} = \dfrac{{4\sqrt 2 {\mu _o}I}}{{\pi L}}$………eq. (1)

The magnetic field for the current coil in circular coil is given by,

$B = \dfrac{{{\mu _o}I}}{{2r}}$ Where I is the current in the coil and r is the radius of the circle.

The perimeter of a circle with radius r is given by,

$ \Rightarrow p = 2\pi r$

$ \Rightarrow L = 2\pi r$

$ \Rightarrow r = \dfrac{L}{{2\pi }}$

Replace the value of the r in the relation of the magnetic field of the current.

$ \Rightarrow {B_2} = \dfrac{{{\mu _o}I}}{{2r}}$

$ \Rightarrow {B_2} = \dfrac{{{\mu _o}I}}{{2\left( {\dfrac{L}{{2\pi }}} \right)}}$

$ \Rightarrow {B_2} = \dfrac{{{\mu _o}I\pi }}{L}$………eq. (2)

Divide the equation (1) with equation (2) we get.

$ \Rightarrow \dfrac{{{B_1}}}{{{B_2}}} = \dfrac{{\left( {\dfrac{{4\sqrt 2 {\mu _o}I}}{{\pi L}}} \right)}}{{\left( {\dfrac{{{\mu _o}I\pi }}{L}} \right)}}$

$ \Rightarrow \dfrac{{{B_1}}}{{{B_2}}} = \dfrac{{4\sqrt 2 }}{{{\pi ^2}}}$

The ratio of the magnetic field of the current carrying coil in the shape of square to the magnetic field in the current carrying coil in circular shape is equal to$\dfrac{{{B_1}}}{{{B_2}}} = \dfrac{{4\sqrt 2 }}{{{\pi ^2}}}$.

**The correct answer for this problem is option A.**

**Note:**The formula for magnetic field in a current carrying coil for different shapes should be remembered. The total length of the circular coil is the perimeter of the circular coil with the radius of the circle.

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