Question

A square frame of side I m carries a current I, it produces a magnetic field B1 at its centre. Same current is passed through a circular coil having the same perimeter as the square. Magnetic field at the centre of the circular coil is B2. Then, the ratio\[\dfrac{{{B_1}}}{{{B_2}}}\].is
\[
A. \dfrac{2\sqrt{2}}{3\pi } \\
B. \dfrac{8\sqrt{2}}{{{\pi }^{2}}} \\
C. \dfrac{5\sqrt{3}}{\sqrt{7}\pi } \\
D. 1 \\
\]

Answer 1

Hint:We know that the perimeter of the square is equal to the circumference of the circle; hence we can obtain the relationship between the side length of the square and the radius of the circle. Then, we find the magnetic fields for the frame and the coil in terms of side length of the square and take ratios of the magnetic fields.

Complete step by step answer:
Here, the square has a side length of l meters. Hence if $P$ is the perimeter of the square, then the perimeter $P$ would be $P=4l$. The current passing through both the square frame as well as the circular coil is denoted by \[I\] . The perimeter of the square is equal to the circumference of the circle. Let the radius of the circle is r. The magnetic field of the square is B1 and the magnetic field of the circle is denoted as B2. The circumference of the circle is denoted as C. Thus,
\[
P = C \\
\Rightarrow P = 2\pi r \\
\Rightarrow\dfrac{{4l}}{{2\pi }} = r \\
\Rightarrow\dfrac{{2l}}{\pi } = r \\
\]
Now, the magnetic field of the circular coil is given by:
\[{B_2} = \dfrac{{{\mu _0}I}}{{2r}} = \dfrac{{{\mu _0}I\pi }}{{4l}}\]
Here, \[{\mu _0}\] is the magnetic permeability of the material.
The magnetic field at the center of a square is given by the following formula:
\[
{B_1} = 4\dfrac{{{\mu _0}I}}{{2\pi l}}(\sin 45^\circ + \sin 45^\circ ) \\
\dfrac{{{B_1}}}{{{B_2}}}{B_1} = \dfrac{{4{\mu _0}I}}{{2\pi l}}(\dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }}) \\
\Rightarrow{B_1} = \dfrac{{2{\mu _0}I\sqrt 2 }}{{\pi l}} \\
 \]
The ratio of the magnetic field of the square frame and the magnetic field of the circular coil is given by:
\[\dfrac{{{B_1}}}{{{B_2}}} = \dfrac{{\dfrac{{2{\mu _0}I\sqrt 2 }}{{\pi l}}}}{{\dfrac{{{\mu _0}I\pi }}{{4l}}}}\\
\therefore\dfrac{{{B_1}}}{{{B_2}}} = \dfrac{{8\sqrt 2 }}{{{\pi ^2}}}\]

Thus, option B is the correct answer.

Note: Here, instead of taking the side length of the term as the subject, the radius of the circle can be taken as the subject and we can obtain the same solution. Here, for the magnetic field of a square, we took one side length as a finite wire and found its magnetic field, and for a square we considered four such wires and found its magnetic field.