A square loop is carrying a steady current $I$ and the magnitude of its magnetic dipole moment is $m$. If this square loop is changed to a circular loop and it carries the same current, the magnitude of the magnetic dipole moment of the circular loop will be:
$\left( a \right)$$\dfrac{{3m}}{\pi }$
$\left( b \right)$$\dfrac{{4m}}{\pi }$
$\left( c \right)$$\dfrac{{2m}}{\pi }$
$\left( d \right)$$\dfrac{m}{\pi }$
Answer
302.7k+ views
Hint So here in this question we have a steady current and their magnetic dipole moment is also given here. It is saying that if the loop of squares gets changed to a circular loop then it will carry the same current then we have to find the magnitude. So here we will use the concept which is of magnetic dipole and through this, we will get the relation.
Formula:
Magnetic dipole,
$m = nIA$ ;
Where,
$m$ , will be the magnetic dipole.
$n$ , will be equal to the turn in the loop and $I$ will be current and $A$ will be the area of current-conducting conductor
Complete Step by Step Solution Here we will first see the formula and accordingly we will put the given values and will get the result.
So from the formula, know the magnetic dipole is equal to the
$ \Rightarrow m = nIA$
Or we can write it as
$ \Rightarrow m = 1 \times I \times {a^2}$
Now we know from the question,
$ \Rightarrow 4a = 2\pi r$
Now we will solve the above equation for the value of $r$ , we get
$ \Rightarrow r = \dfrac{{2a}}{\pi }$
And as we already have seen the formula for the loop which is being circular, so
$ \Rightarrow m' = 1 \times I \times \pi {r^2}$
On substituting the values in the above equation, we will get
$ \Rightarrow 1 \times I \times \pi \times {\left( {\dfrac{{2a}}{\pi }} \right)^2}$
So from here, after solving the above equation, we will get the value for the magnitude of the magnetic dipole.
$ \Rightarrow m' = \dfrac{{4m}}{\pi }$
Therefore, $\dfrac{{4m}}{\pi }$ is the required magnitude, and hence the correct option is B .
Note A magnetic dipole is akin to an electric dipole, it exists in reality and has physical representation and meaning. The magnetic moment is just a consequence of mathematical building for physical proof. So we can say that when the field is produced because the magnetic field is proportional to the magnetic moment.
Formula:
Magnetic dipole,
$m = nIA$ ;
Where,
$m$ , will be the magnetic dipole.
$n$ , will be equal to the turn in the loop and $I$ will be current and $A$ will be the area of current-conducting conductor
Complete Step by Step Solution Here we will first see the formula and accordingly we will put the given values and will get the result.
So from the formula, know the magnetic dipole is equal to the
$ \Rightarrow m = nIA$
Or we can write it as
$ \Rightarrow m = 1 \times I \times {a^2}$
Now we know from the question,
$ \Rightarrow 4a = 2\pi r$
Now we will solve the above equation for the value of $r$ , we get
$ \Rightarrow r = \dfrac{{2a}}{\pi }$
And as we already have seen the formula for the loop which is being circular, so
$ \Rightarrow m' = 1 \times I \times \pi {r^2}$
On substituting the values in the above equation, we will get
$ \Rightarrow 1 \times I \times \pi \times {\left( {\dfrac{{2a}}{\pi }} \right)^2}$
So from here, after solving the above equation, we will get the value for the magnitude of the magnetic dipole.
$ \Rightarrow m' = \dfrac{{4m}}{\pi }$
Therefore, $\dfrac{{4m}}{\pi }$ is the required magnitude, and hence the correct option is B .
Note A magnetic dipole is akin to an electric dipole, it exists in reality and has physical representation and meaning. The magnetic moment is just a consequence of mathematical building for physical proof. So we can say that when the field is produced because the magnetic field is proportional to the magnetic moment.
Recently Updated Pages
If the magnetizing field on a ferromagnetic material class 12 physics JEE_Main

A point charge is placed at the corner of a cube The class 12 physics JEE_Main

What changes occur if the monochromatic light used class 12 physics JEE_Main

The unit of specific conductance is A Ohm B Ohmmetre class 12 physics JEE_Main

Which lens is used in magnifying glass A Concave lens class 12 physics JEE_Main

Sir C V Raman won the Nobel Prize in which year A 1928 class 12 physics JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

How to Convert a Galvanometer into an Ammeter or Voltmeter

Electron Gain Enthalpy and Electron Affinity Explained

Other Pages
CBSE Class 12 Physics Question Paper 2026: Download SET-wise PDF with Answer Key & Analysis

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

Hybridisation in Chemistry – Concept, Types & Applications

What Are Current and Potential Difference in Electricity?

What Are Elastic Collisions in One Dimension?

Understanding Uniform Acceleration in Physics

