
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
216.3k+ 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
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

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

Understanding Average and RMS Value in Electrical Circuits

