Question

The earth’s magnetic field lines resemble that of a dipole at the center of the earth. If the magnetic moment of this dipole is close to \[8 \times {10^{22}}A{m^2}\], the value of earth’s magnetic field near the equator is closed to (radius of the earth $ = 6.4 \times {10^6}m$)
\[(A)0.6{\text{ Gauss}}\]
$(B)1.2{\text{ Gauss}}$
$(C)1.8{\text{ Gauss}}$
$(D)0.32{\text{ Gauss}}$

Answer 1

Hint: The magnitude of the earth’s magnetic flux at its surface ranges from 0.5 to 6.5 microteslas. Under the solution, we find out the value of the earth’s magnetic field near the equator.

Complete step by step answer:
The magnetic moment of dipole $ = 8 \times {10^{22}}A{m^2}$and
The radius of the earth $ = 6.4 \times {10^6}m$
The magnetic field strength at the earth’s magnetic equator $0.0000305$ tesla, or $0.305 \times {10^{ - 4}}T$.
 Maps of the surface magnetic fields near the poles where the magnetic fields lines congregate, at roughly twice the strength of the field at the equator.
Now, we can find the earth’s magnetic field near the equator
$B = \dfrac{{{\mu _0}}}{{4\pi }} \times \dfrac{{2m}}{{{d^3}}}$
B is the earth's magnetic field.
${\text{when }}{\mu _0}{\text{ value is 4}}\pi \times {\text{1}}{{\text{0}}^{22}}$
\[m\]Is the mass and $d$is the radius of the earth.
$ \Rightarrow \dfrac{{4\pi \times {{10}^{ - 7}}}}{{4\pi }} \times \dfrac{{2 \times 8 \times {{10}^{22}}}}{{{{(6.4 \times {{10}^6})}^3}}}$
Canceling the same term we get,
$ \Rightarrow {10^{ - 7}} \times \dfrac{{2 \times 8 \times {{10}^{22}}}}{{{{(6.4 \times {{10}^6})}^3}}}$
$ \Rightarrow 0.6{\text{ Gauss}}$

Hence, the correct answer is option (A).

Additional information:
Magnetic field due to a hypothetical dipole. The moment of a magnet $(\mu )$ may be a vector quantity wont to measure the tendency of an object to interact with an external magnetic flux.
In NMR, the thing of interest is usually a molecule, atom, nucleus, or elementary particle.
SI units the dipole moment,$\mu $ for Earth is $7.95 \times 1022A/{m^2}$ (amperes per square meter).
Since $\mu = IA$ (current times area), a loop the size of the liquid core ( $RC = 3.48 \times {10^6}m$) would require an equivalent current of nearly $2 \times {10^6}A$.

Note:
• A magnetic flux may be a vector field that describes the magnetic influence on moving electric charges, electric currents, and magnetized materials.
• A charge that occupies the magnetic flux experiences a force perpendicular to its velocity and the magnetic flux.
• The unit for a magnetic moment in the International System of Units (SI) base units is $A$. ${m^2}$, where A is ampere (SI base unit of current) and m is the meter (SI base unit of distance).