The earth's magnetic field at the equator is approximately 0.4G. Estimate the earth's dipole moment.
A) $1.05 \times {10^{23}}A{m^2}$
B) $1.05 \times {10^{ - 23}}A{m^2}$
C) $5.01 \times {10^{23}}A{m^2}$
D) $5.01 \times {10^{ - 23}}A{m^2}$
Answer
298.5k+ views
Hint: To solve this question, we have to imagine that a bar magnet is kept at the equator of the earth. After that, we just have to find the relation between the dipole moment of that bar magnet and the magnetic field of that bar magnet. We can easily substitute the values to find the dipole moment of that magnet which will be the required dipole moment of the earth.
Formulae used:
${B_{equatorial}} = \dfrac{{{\mu _0}}}{{4\pi }} \times \dfrac{M}{{{d^3}}}$
Here ${\mu _0}$ is the permeability of free space, $M$ is the dipole moment of the magnet, $d$ is the distance of the point from the equator of the magnet, and ${B_{equatorial}}$ is the magnetic field of the magnetic field at the equator of the magnet.
Complete step by step solution:
Let us consider that a bar magnet is placed along the equator of the earth.

We know that,
$ \Rightarrow {B_{equatorial}} = \dfrac{{{\mu _0}}}{{4\pi }} \times \dfrac{M}{{{d^3}}}$
Here ${\mu _0}$ is the permeability of free space, $M$ is the dipole moment of the magnet, $d$ is the distance of the point from the equator of the magnet, and ${B_{equatorial}}$ is the magnetic field of the magnetic field at the equator of the magnet.
Let this be equation 1.
From the above diagram, we can say that
$ \Rightarrow d = R = 6400Km = 6400 \times {10^3}m$
Also, the value of the magnetic field is given as
$ \Rightarrow B = 0.4G = 0.4 \times {10^{ - 4}}T$
So equation 1 becomes,
$ \Rightarrow 0.4 \times {10^{ - 4}} = \dfrac{{{\mu _0}}}{{4\pi }} \times \dfrac{M}{{{{\left( {6400 \times {{10}^3}} \right)}^3}}}$
$ \Rightarrow M = \dfrac{{0.4 \times {{10}^{ - 4}} \times {{\left( {6400 \times {{10}^3}} \right)}^3}}}{{{{10}^{ - 7}}}} = 1.05 \times {10^{23}}A{m^2}$
$ \Rightarrow M = 1.05 \times {10^{23}}A{m^2}$
Earth’s dipole moment at the equator will be $1.05 \times {10^{23}}A{m^2}$.
From this, we can conclude that option (A) is the correct answer.
Note: The values of magnetic fields at the equator and the axis of the bar magnet are different. So we have to be very careful while solving questions related to the magnetic field or dipole moment of any magnet to avoid incorrect answers.
Formulae used:
${B_{equatorial}} = \dfrac{{{\mu _0}}}{{4\pi }} \times \dfrac{M}{{{d^3}}}$
Here ${\mu _0}$ is the permeability of free space, $M$ is the dipole moment of the magnet, $d$ is the distance of the point from the equator of the magnet, and ${B_{equatorial}}$ is the magnetic field of the magnetic field at the equator of the magnet.
Complete step by step solution:
Let us consider that a bar magnet is placed along the equator of the earth.

We know that,
$ \Rightarrow {B_{equatorial}} = \dfrac{{{\mu _0}}}{{4\pi }} \times \dfrac{M}{{{d^3}}}$
Here ${\mu _0}$ is the permeability of free space, $M$ is the dipole moment of the magnet, $d$ is the distance of the point from the equator of the magnet, and ${B_{equatorial}}$ is the magnetic field of the magnetic field at the equator of the magnet.
Let this be equation 1.
From the above diagram, we can say that
$ \Rightarrow d = R = 6400Km = 6400 \times {10^3}m$
Also, the value of the magnetic field is given as
$ \Rightarrow B = 0.4G = 0.4 \times {10^{ - 4}}T$
So equation 1 becomes,
$ \Rightarrow 0.4 \times {10^{ - 4}} = \dfrac{{{\mu _0}}}{{4\pi }} \times \dfrac{M}{{{{\left( {6400 \times {{10}^3}} \right)}^3}}}$
$ \Rightarrow M = \dfrac{{0.4 \times {{10}^{ - 4}} \times {{\left( {6400 \times {{10}^3}} \right)}^3}}}{{{{10}^{ - 7}}}} = 1.05 \times {10^{23}}A{m^2}$
$ \Rightarrow M = 1.05 \times {10^{23}}A{m^2}$
Earth’s dipole moment at the equator will be $1.05 \times {10^{23}}A{m^2}$.
From this, we can conclude that option (A) is the correct answer.
Note: The values of magnetic fields at the equator and the axis of the bar magnet are different. So we have to be very careful while solving questions related to the magnetic field or dipole moment of any magnet to avoid incorrect answers.
Recently Updated Pages
If the magnetizing field on a ferromagnetic material class 12 physics JEE_Main

A point charge is placed at the corner of a cube The class 12 physics JEE_Main

What changes occur if the monochromatic light used class 12 physics JEE_Main

The unit of specific conductance is A Ohm B Ohmmetre class 12 physics JEE_Main

Which lens is used in magnifying glass A Concave lens class 12 physics JEE_Main

Sir C V Raman won the Nobel Prize in which year A 1928 class 12 physics JEE_Main

Trending doubts
Electron Gain Enthalpy and Electron Affinity Explained

What Are Current and Potential Difference in Electricity?

Understanding Uniform Acceleration in Physics

Understanding Geostationary and Geosynchronous Satellites

Isoelectronic Species: Definition, Examples & Importance

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Other Pages
JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

JEE Advanced 2026 Notification Out with Exam Date, Registration (Extended), Syllabus and More

JEE Advanced Weightage Chapter Wise 2026 for Physics, Chemistry, and Mathematics

Units and Measurements Mock Test 2025-26: Free Practice for Students

Free Radical Substitution and Its Stepwise Mechanism

Essential Derivations for CBSE Class 12 Physics: Stepwise & PDF Solutions

