
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
217.2k+ 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
Wheatstone Bridge Explained: Working, Formula & Uses

Young’s Double Slit Experiment Derivation Explained

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Electricity and Magnetism Explained: Key Concepts & Applications

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Understanding Atomic Structure for Beginners

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Understanding Electromagnetic Waves and Their Importance

Understanding the Electric Field of a Uniformly Charged Ring

