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# A thin circular wire carrying a current $I$ has a magnetic moment $M$. The shape of the wire is changed to a square and it carries the same current. It will have a magnetic moment.A) $M$B) $\dfrac{4}{{{\pi ^2}}}M$C) $\dfrac{4}{\pi }M$D) $\dfrac{\pi }{4}M$

Last updated date: 11th Aug 2024
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Hint: Magnetic moment of a current carrying wire can be calculated as the product of the current flowing through the wire and area of the loop. Since the same wire is changed from circular shape to square shape, the circumference of the circle can be equated to the perimeter of the square. Using this, we can calculate new areas and new magnetic moments. By comparing this new magnetic moment with the initial magnetic moment, we can reach the final answer.

Complete step by step solution:
It is given that the shape of a wire is changed from circular shape to square shape. When the wire is in circular shape the current flowing is given as $I$ and the magnetic moment is given as $M$. We need to find the magnetic moment when the shape of the wire is changed to square. We know that magnetic moment is given as,
$M = NIA$
Where $N$ is the number of turns, $I$ is the current, $A$ is the area.
Let us write the equation for the magnetic moment when the wire is circular.
$M = NIA$
$\Rightarrow M = IA$
Since number of turns is 1
We know that area of circle is,
$A = \pi {r^2}$
$\therefore M = I\pi {r^2}$................................(1)
Now let us write the magnetic moment when the shape is square.
$M' = NIA'$
$\Rightarrow M' = IA'$
Area of the square is given as .
$A' = a \times a$
When a is the length of side,
$\Rightarrow M' = I{a^2}$
Now let us calculate the value of a.
Since the same circular wire is changed to square we can equate the circumference of the circle with the perimeter of the square.
We know the circumference of a circle is given as $2\pi r$, where r is the radius and perimeter of square is four times the length of one side, $4a$ .
$\Rightarrow 2\pi r = 4a$
$\Rightarrow a = \dfrac{{\pi r}}{2}$
$\therefore$ Magnetic moment of square can be written as
$\Rightarrow M' = I{\left( {\dfrac{{\pi r}}{2}} \right)^2}$
$\Rightarrow M' = \dfrac{{I{\pi ^2}{r^2}}}{4}$...........................(2)
Now we know that from equation 1
$M = I\pi {r^2}$
On substituting this in equation 2, we get
$\Rightarrow M' = \dfrac{{\left( {I\pi {r^2}} \right) \times \pi }}{4}$
$\Rightarrow M' = \dfrac{\pi }{4}M$
This is the new magnetic moment.

So, the correct answer is option (D).

Note: Remember that the magnetic moment depends directly on the area. So, even when current flowing remains the same the magnetic moment will decrease when area is decreased. For a given length of wire the area will be maximum when it is bent in the form of a circle. Hence the new magnetic moment of the square will be less than that of circle shape.