Home
Class 12
PHYSICS
The earth's magnetic field at the equato...

The earth's magnetic field at the equator is approximately 0.4 G. Estimate the earth's dipole moment.

Text Solution

Verified by Experts

The equatorial magnetic field is , `B_E=(mu_0m)/(4pir^3)`
We are given that `B_B~0.4 G = 4 xx10^(-5)T.` For r, we take the earth `6.4xx10^6` m .
Hence , `m=(4xx10^(-5)xx(6.4xx10^6)^3)/(mu_0//4pi)=4xx10^2xx(6.4xx10^6)^3" "((mu_0)/(4pi)=10^(-7))`
This is close to the value `8xx10^(22) Am^(-2)` as obtained in geomagnetic experiments .
Promotional Banner

Topper's Solved these Questions

  • MAGNETISM AND MATTER

    NEW JOYTHI PUBLICATION|Exercise SOLUTIONS TO EXERCISES FROM NCERT TEXT|21 Videos

  • MAGNETISM AND MATTER

    NEW JOYTHI PUBLICATION|Exercise PRACTICE PROBLEMS FOR SELF ASSESSMENT|12 Videos

  • ELECTROSTATIC POTENTIAL AND CAPACITANCE

    NEW JOYTHI PUBLICATION|Exercise Competitive Exam Corner|22 Videos

  • MOVING CHARGES AND MAGNETISM

    NEW JOYTHI PUBLICATION|Exercise COMPETITIVE EXAM CORNER|28 Videos

Similar Questions

Explore conceptually related problems

Define magnetic dipole moment.

A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22^@ with the horizontal . The horizontal component of the earth's magnetic field at the place is known to be 0.35 G . Determine the magnitude of the earth's magnetic field at the place.

Discuss Earth's magnetic field in detail.

A long straight horizontal cable carries a current of 2.5 A in the direction 10^@ north of east . The magnetic meridian of the place happens to be 10^@ west of the geographic meridian The earth's magnetic field at the location is 0.33 G , ad the angle of dip is zero. Locate the line of neutral points (ignore the thickness of the cable ) (At neutral points , magnetic field due to a current - carrying cable is equal and opposite to the horizontal component of earth's magnetic field ).

A magnetic needle is free to rotate in a vertical plans parallel to the magnetic meridian has its north tip down at 60^(@) with horizontal . The horizontal component of the earth's magnetic field at the place is known to be 0.4 xx 10^(-4) T determine the magnitude of earth's magnetic field at the place.

The frequency of oscillation of the magnet in an oscillation magnetometer in the earth's magnetic field is 40 oscillations per minute. A short bar magnet is placed to the north of the magnetometer, at a separation of 20cm from the oscillating magnet, with its north pole pointing towards north (figure). The frequency of oscillation is found to increase to 60 oscillations per minute. Calculate the magnetic moment of this short bar magnet. Horizonatl component of the earth's magnetic field is 24 muT .