
The field due to a magnet at a distance R from the centre of the magnet is proportional to
(A) ${R^2}$
(B) ${R^3}$
(C) $\dfrac{1}{{{R^2}}}$
(D) $\dfrac{1}{{{R^3}}}$
Answer
162.6k+ views
Hint:
In order to solve this question, we will first write down the formula of the magnetic field due to a bar magnet at some point on its axis and then we will determine the relationship between the magnetic field and distance and will check for the correct option.
Complete step by step solution:
As we know that, the magnetic field due to a bar magnet having magnetic moment M at a distance of R from its centre on its axis is calculated using the formula $B = \dfrac{{{\mu _o}}}{{4\pi }}\dfrac{{2M}}{{{R^3}}}$ where, ${\mu _o}$ is known as the relative permeability of free space so, we see that magnetic field due to a bar magnetic B on its axis at a point of distance R is related as $B \propto \dfrac{1}{{{R^3}}}$ , also let us check magnetic field due to bar magnet at a distance of R on the equatorial point and it’s given by $B = \dfrac{{{\mu _o}}}{{4\pi }}\dfrac{M}{{{R^3}}}$ here also we see that, $B \propto \dfrac{1}{{{R^3}}}$ so we see that magnetic field is always inversely proportional to cube of distance R irrespective of point we choose either the point on axis or the point on equatorial line.
Hence, the correct answer is option (D) $\dfrac{1}{{{R^3}}}$
Therefore, the correct option is D.
Note:
It should also be noted that the magnetic field due to a bar magnet on its axis point is always twice the value of the magnetic field on its equatorial point while keeping the distance the same from the center of the bar magnet in both cases.
In order to solve this question, we will first write down the formula of the magnetic field due to a bar magnet at some point on its axis and then we will determine the relationship between the magnetic field and distance and will check for the correct option.
Complete step by step solution:
As we know that, the magnetic field due to a bar magnet having magnetic moment M at a distance of R from its centre on its axis is calculated using the formula $B = \dfrac{{{\mu _o}}}{{4\pi }}\dfrac{{2M}}{{{R^3}}}$ where, ${\mu _o}$ is known as the relative permeability of free space so, we see that magnetic field due to a bar magnetic B on its axis at a point of distance R is related as $B \propto \dfrac{1}{{{R^3}}}$ , also let us check magnetic field due to bar magnet at a distance of R on the equatorial point and it’s given by $B = \dfrac{{{\mu _o}}}{{4\pi }}\dfrac{M}{{{R^3}}}$ here also we see that, $B \propto \dfrac{1}{{{R^3}}}$ so we see that magnetic field is always inversely proportional to cube of distance R irrespective of point we choose either the point on axis or the point on equatorial line.
Hence, the correct answer is option (D) $\dfrac{1}{{{R^3}}}$
Therefore, the correct option is D.
Note:
It should also be noted that the magnetic field due to a bar magnet on its axis point is always twice the value of the magnetic field on its equatorial point while keeping the distance the same from the center of the bar magnet in both cases.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

Young's Double Slit Experiment Step by Step Derivation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Wheatstone Bridge for JEE Main Physics 2025

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

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

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Uniform Acceleration

Degree of Dissociation and Its Formula With Solved Example for JEE
