Question:
The equitorial magnetic field of earth on its surface at equator is $0.4\text{ G}$. Then its dipole moment is (The radius of the earth $R=6400\text{ km}$):
The equitorial magnetic field of earth on its surface at equator is $0.4\text{ G}$. Then its dipole moment is (The radius of the earth $R=6400\text{ km}$):
Show Hint
Remember that $1\text{ Gauss} = 10^{-4}\text{ Tesla}$.
Be careful with the power of ten when cubing the Earth's radius: $(6.4 \times 10^6)^3 \approx 262 \times 10^{18}$.
Be careful with the power of ten when cubing the Earth's radius: $(6.4 \times 10^6)^3 \approx 262 \times 10^{18}$.
Updated On: Jul 22, 2026
- $1.05 \times 10^{23}\text{ Am}^2$
- $2.05 \times 10^{23}\text{ Am}^2$
- $1.05 \times 10^{21}\text{ Am}^2$
- $2.05 \times 10^{21}\text{ Am}^2$
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
We need to calculate the magnetic dipole moment of the Earth given its equatorial magnetic field strength on the surface and the Earth's radius.
Step 2: Key Formula and Approach:
The magnetic field at a point on the equatorial line of a magnetic dipole is:
\[ B_e = \frac{\mu_0}{4\pi} \frac{M}{R^3} \] where:
- $M$ is the magnetic dipole moment.
- $R$ is the distance from the center (Earth's radius).
- $\frac{\mu_0}{4\pi} = 10^{-7}\text{ T m A}^{-1}$.
Step 3: Detailed Explanation:
• Identify and convert units:
Equatorial magnetic field $B_e = 0.4\text{ G} = 0.4 \times 10^{-4}\text{ T}$
Radius of Earth $R = 6400\text{ km} = 6.4 \times 10^6\text{ m}$
• Rearrange the formula for $M$:
\[ M = \frac{B_e R^3}{\mu_0/4\pi} \]
• Substitute the values:
\[ M = \frac{\left(0.4 \times 10^{-4}\right) \times \left(6.4 \times 10^6\right)^3}{10^{-7}} \] \[ M = \frac{0.4 \times 10^{-4} \times 262.144 \times 10^{18}}{10^{-7}} \] \[ M = 0.4 \times 262.144 \times 10^{21} \] \[ M = 104.8576 \times 10^{21}\text{ Am}^2 = 1.048 \times 10^{23}\text{ Am}^2 \] Rounding to two decimal places:
\[ M \approx 1.05 \times 10^{23}\text{ Am}^2 \]
Step 4: Final Answer:
The magnetic dipole moment of the Earth is approximately $1.05 \times 10^{23}\text{ Am}^2$, which corresponds to Option (A).
We need to calculate the magnetic dipole moment of the Earth given its equatorial magnetic field strength on the surface and the Earth's radius.
Step 2: Key Formula and Approach:
The magnetic field at a point on the equatorial line of a magnetic dipole is:
\[ B_e = \frac{\mu_0}{4\pi} \frac{M}{R^3} \] where:
- $M$ is the magnetic dipole moment.
- $R$ is the distance from the center (Earth's radius).
- $\frac{\mu_0}{4\pi} = 10^{-7}\text{ T m A}^{-1}$.
Step 3: Detailed Explanation:
• Identify and convert units:
Equatorial magnetic field $B_e = 0.4\text{ G} = 0.4 \times 10^{-4}\text{ T}$
Radius of Earth $R = 6400\text{ km} = 6.4 \times 10^6\text{ m}$
• Rearrange the formula for $M$:
\[ M = \frac{B_e R^3}{\mu_0/4\pi} \]
• Substitute the values:
\[ M = \frac{\left(0.4 \times 10^{-4}\right) \times \left(6.4 \times 10^6\right)^3}{10^{-7}} \] \[ M = \frac{0.4 \times 10^{-4} \times 262.144 \times 10^{18}}{10^{-7}} \] \[ M = 0.4 \times 262.144 \times 10^{21} \] \[ M = 104.8576 \times 10^{21}\text{ Am}^2 = 1.048 \times 10^{23}\text{ Am}^2 \] Rounding to two decimal places:
\[ M \approx 1.05 \times 10^{23}\text{ Am}^2 \]
Step 4: Final Answer:
The magnetic dipole moment of the Earth is approximately $1.05 \times 10^{23}\text{ Am}^2$, which corresponds to Option (A).
Was this answer helpful?
0
0
Top AP EAPCET Physics Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The radii of the two columns in a U - tube are r1 and r2. When a liquid of density p (angle of contact is $ 0{}^\circ $ ) is filled in it, the level difference of the liquid in the two arms is h. The surface tension of the liquid is: (g = acceleration due to gravity)
- A rigid metallic sphere is spinning around its own axis in the absence of external torque. If the temperature is raised, its volume increases by $9 \%$. The change in its angular speed is
View More Questions
Top AP EAPCET The Earth’s Magnetism Questions
- If the radius of the Earth is \( 6.4 \times 10^6 \, m \) and its magnetic dipole moment is \( 1.3 \times 10^{23} \, Am^2 \), then the magnetic field at the equator is:
- The magnetic susceptibility is negative for:
- A short bar magnet placed in a horizontal plane has its axis aligned along north-south direction. Null points are found on the axis of the magnet at \(20\ \text{cm}\) from the center of magnet. The earth's magnetic field at the place is \(B\) and angle of dip is \(0^\circ\). If the total magnetic field on the normal bisector of the magnet \(20\ \text{cm}\) from the center of the magnet is \(0.6\ \text{G}\), then the magnitude of \(B\) is:
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions