![SearchIcon](https://vmkt.vedantu.com/vmkt/PROD/png/bdcdbbd8-08a7-4688-98e6-4aa54e5e0800-1733305962725-4102606384256179.png)
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
124.8k+ 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
Young's Double Slit Experiment Step by Step Derivation
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Circuit Switching and Packet Switching
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Mass and Weight
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
JEE Main Participating Colleges 2024 - A Complete List of Top Colleges
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
JEE Main Maths Paper Pattern 2025 – Marking, Sections & Tips
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Sign up for JEE Main 2025 Live Classes - Vedantu
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The formula of the kinetic mass of a photon is Where class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Learn About Angle Of Deviation In Prism: JEE Main Physics 2025
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
JEE Main Login 2045: Step-by-Step Instructions and Details
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Ideal and Non-Ideal Solutions Raoult's Law - JEE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
JEE Mains 2025 Correction Window Date (Out) – Check Procedure and Fees Here!
![arrow-right](/cdn/images/seo-templates/arrow-right.png)