The magnetic field due to a current carrying circular loop of radius $3m$ at a point on the axis at a distance of $4m$ from the centre is $54\mu T$ .What will be its value at the centre of the loop?
A. $250\mu T$
B. $150\mu T$
C. $125\mu T$
D. $75\mu T$
Answer
249.6k+ views
Hint: Relate the given value of magnetic field at the centre of the loop with the formulae for both the cases.
The magnetic field on the axis of a current carrying loop is given by ${B_z} = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{2\pi {R^2}I}}{{{{\left( {{z^2} + {R^2}} \right)}^{3/2}}}}$ where $z$ is the distance of that point on the axis from centre of the loop, $R$ is the radius of the loop and $I$ is the current flowing in the loop. For the magnetic field at the centre of the loop, put $z = 0$ in the above expression.
Complete step by step answer:
In the question we are supposed to deal with the magnetic field due to the same circular loop at two different points on the axis and at the centre. Magnetic field on the axis is given ${B_z} = 54\mu T$ .
So we can relate the given value of the magnetic field at the centre of the loop with the formulae for both the cases.
We know that the magnetic field on the axis of a current carrying loop is given by ${B_z} = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{2\pi {R^2}I}}{{{{\left( {{z^2} + {R^2}} \right)}^{3/2}}}}$ where $z$ is the distance of that point on the axis from centre of the loop, $R$ is the radius of the loop and $I$ is the current flowing in the loop.
So, let us suppose that current $I$ is flowing in the loop.
Now, for magnetic field at the centre of the loop, we can put $z = 0$ in the above expression which gives
$\implies$ ${B_c} = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{2\pi I}}{R}$
On dividing both the equation we have
$\implies$ $\dfrac{{{B_c}}}{{{B_z}}} = \dfrac{{{{\left( {{z^2} + {R^2}} \right)}^{3/2}}}}{{{R^3}}}$
Therefore, substituting the values given in the question we have
$\implies$ $\dfrac{{{B_c}}}{{{B_z}}} = \dfrac{{{{\left( {{4^2} + {3^2}} \right)}^{3/2}}}}{{{3^3}}} = \dfrac{{125}}{{27}}$
So, we get the magnetic field at the centre as
$\implies$ ${B_c} = \dfrac{{125}}{{27}} \times {B_z} = \dfrac{{125}}{{27}} \times 54 = 250\mu T$
Hence, option A is correct.
Note: The two formulae which are used here can be derived using Biot-Savart’s Law. The direction of the magnetic field due to the loop can be determined using the right hand thumb rule which states that when we roll our fingers according to the direction of current flowing then the direction of thumb gives us the direction of the magnetic field due to the loop.
The magnetic field on the axis of a current carrying loop is given by ${B_z} = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{2\pi {R^2}I}}{{{{\left( {{z^2} + {R^2}} \right)}^{3/2}}}}$ where $z$ is the distance of that point on the axis from centre of the loop, $R$ is the radius of the loop and $I$ is the current flowing in the loop. For the magnetic field at the centre of the loop, put $z = 0$ in the above expression.
Complete step by step answer:
In the question we are supposed to deal with the magnetic field due to the same circular loop at two different points on the axis and at the centre. Magnetic field on the axis is given ${B_z} = 54\mu T$ .
So we can relate the given value of the magnetic field at the centre of the loop with the formulae for both the cases.
We know that the magnetic field on the axis of a current carrying loop is given by ${B_z} = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{2\pi {R^2}I}}{{{{\left( {{z^2} + {R^2}} \right)}^{3/2}}}}$ where $z$ is the distance of that point on the axis from centre of the loop, $R$ is the radius of the loop and $I$ is the current flowing in the loop.
So, let us suppose that current $I$ is flowing in the loop.
Now, for magnetic field at the centre of the loop, we can put $z = 0$ in the above expression which gives
$\implies$ ${B_c} = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{2\pi I}}{R}$
On dividing both the equation we have
$\implies$ $\dfrac{{{B_c}}}{{{B_z}}} = \dfrac{{{{\left( {{z^2} + {R^2}} \right)}^{3/2}}}}{{{R^3}}}$
Therefore, substituting the values given in the question we have
$\implies$ $\dfrac{{{B_c}}}{{{B_z}}} = \dfrac{{{{\left( {{4^2} + {3^2}} \right)}^{3/2}}}}{{{3^3}}} = \dfrac{{125}}{{27}}$
So, we get the magnetic field at the centre as
$\implies$ ${B_c} = \dfrac{{125}}{{27}} \times {B_z} = \dfrac{{125}}{{27}} \times 54 = 250\mu T$
Hence, option A is correct.
Note: The two formulae which are used here can be derived using Biot-Savart’s Law. The direction of the magnetic field due to the loop can be determined using the right hand thumb rule which states that when we roll our fingers according to the direction of current flowing then the direction of thumb gives us the direction of the magnetic field due to the loop.
Recently Updated Pages
JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Isoelectronic Definition in Chemistry: Meaning, Examples & Trends

Ionisation Energy and Ionisation Potential Explained

Iodoform Reactions - Important Concepts and Tips for JEE

Introduction to Dimensions: Understanding the Basics

Instantaneous Velocity Explained: Formula, Examples & Graphs

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

Hybridisation in Chemistry – Concept, Types & Applications

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

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

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

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Understanding the Angle of Deviation in a Prism

Understanding Centrifugal Force in Physics

