The earth’s magnetic field at the equator is approximately $0.4 \;G$. Estimate the earth’s dipole moment.
the equatorial magnetic field is,
$B_{E}=\frac{\mu_{0} m}{4 \pi r^{3}}$
We are given that $B_{E} \sim 0.4 G =4 \times 10^{-5} T .$
For $r,$ we take the radius of the earth $6.4 \times 10^{6} m .$ Hence,
$m=\frac{4 \times 10^{-5} \times\left(6.4 \times 10^{6}\right)^{3}}{\mu_{0} / 4 \pi}$$=4 \times 10^{2} \times\left(6.4 \times 10^{6}\right)^{3} \;\;\left(\mu_{0} / 4 \pi=10^{-7}\right)$
$=1.05 \times 10^{23} Am ^{2}$
This is close to the value $8 \times 10^{22}\; A m ^{2}$ quoted in geomagnetic texts.
A straight magnetic strip has a magnetic moment of $44 \mathrm{Am}^2$. If the strip is bent in a semicircular shape, its magnetic moment will be .. . . . . . .$\mathrm{Am}^2$ $\left(\right.$ Given $\left.\pi=\frac{22}{7}\right)$
Two identical short bar magnets, each having magnetic moment $M$ are placed a distance of $2d$ apart with axes perpendicular to each other in a horizontal plane. The magnetic induction at a point midway between them is
A sheet is placed on a horizontal surface in front of a strong magnetic pole. A force is needed to:
$A$. hold the sheet there if it is magnetic.
$B$. hold the sheet there if it is non-magnetic.
$C$. move the sheet away from the pole with uniform velocity if it is conducting.
$D$. move the sheet away from the pole with uniform velocity if it is both, non-conducting and non-polar.
Choose the correct statement($s$) from the options given below:
A closely wound solenoid of $800$ turns and area of cross section $2.5 \times 10^{-4} \;m ^{2}$ carries a current of $3.0\; A .$ Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?
The ultimate individual unit of magnetism in any magnet is called