Explain the magnetic fields of the Earth. and Give the value of magnified of earth.

Vedclass pdf generator app on play store
Vedclass iOS app on app store

To describe the magnetic field of the earth at a point on its surface, we need to specify three quantities.

$(i)$ The declination $D$

$(ii)$ The angle of dip or the inclination $I$

$(iii)$ The horizontal component of Earth's field $\mathrm{H}_{\mathrm{E}}$

$(iv)$ The vertical component of earth's field $Z_{E}$

All these are known as the elements of the earth's magnetic field.

From figure,

$\mathrm{H}_{\mathrm{E}}=\mathrm{B}_{\mathrm{E}} \cos \mathrm{I}...(1)$

$\mathrm{Z}_{\mathrm{E}} =\mathrm{B}_{\mathrm{E}} \sin \mathrm{I}....(2)$

$\therefore \quad \tan \mathrm{I}=\frac{\mathrm{Z}_{\mathrm{E}}}{\mathrm{H}_{\mathrm{E}}}...(3)$

$\text { and } \mathrm{B}_{\mathrm{E}}=\sqrt{\mathrm{H}_{\mathrm{E}}^{2}+\mathrm{Z}_{\mathrm{E}}^{2}}...(4)$

Of the order of $10^{-5} \mathrm{~T}$

Similar Questions

Let $V $ and $H $ be the vertical and horizontal components of earth's magnetic field at any point on earth. Near the north pole

The angle of dip is the angle

A fighter plane of length $20\, m$, wing span (distance from tip of one wing to the tip of the other wing) of $15\,m$ and height $5\,m$ is lying towards east over Delhi. Its speed is $240\, ms^{-1}$ . The earth's magnetic field over Delhi is $5 \times 10^{-5}\,T$ with the declination angle $ \sim {0^o}$ and dip of $\theta$ such that $\sin \,\theta  = \frac{2}{3}$. If the voltage developed is $V_B$ between the lower and upper side of the plane and $V_W$ between the tips of the wings then $V_B$ and $V_W$ are close to

  • [JEE MAIN 2016]

If the dip circle is set at $45^o$ to the magnetic meridian, then the apparent dip is $30^o$. The true dip. of the place is

At a place, the horizontal and vertical intensities of earth's magnetic field is $0.30$ Gauss and $0.173$  Gauss respectively. The angle of dip at this place is.....$^o$