(B) The angle of dip (or inclination) $\theta$ is given by the relation $\tan \theta = \frac{B_V}{B_H}$,where $B_V$ is the vertical component and $B_H

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(B) The angle of dip (or inclination) $\theta$ is given by the relation $\tan \theta = \frac{B_V}{B_H}$,where $B_V$ is the vertical component and $B_H$ is the horizontal component of the Earth's magnetic field.Given that $B_V = \sqrt{3} B_H$.Substituting this into the formula,we get $\tan \theta = \frac{\sqrt{3} B_H}{B_H} = \sqrt{3}$.Since $\tan 60^{\circ} = \sqrt{3}$,the angle of dip is $\theta = 60^{\circ}$.Therefore,the magnetic needle will incline $60^{\circ}$ below the horizontal.

Class 12 Physics Magnetism and Matter Earth Magnetism

Medium

At a certain place,the vertical component of the Earth's magnetic field is $\sqrt{3}$ times the horizontal component of the Earth's magnetic field. If a magnetic needle is suspended freely in the air,then it will incline:

A
$30^{\circ}$ below horizontal
B
$60^{\circ}$ below horizontal
C
$30^{\circ}$ above horizontal
D
$45^{\circ}$ above horizontal

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Class 12 Questions

Class 12

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Magnetism and Matter

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Earth Magnetism

The horizontal component of the earth's magnetic field at any place is $0.36 \times 10^{-4} \; Wb/m^2$. If the angle of dip at that place is $60^{\circ}$,then the value of the vertical component of the earth's magnetic field will be ........ $\times 10^{-4} \; Wb/m^2$.

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Earth's magnetic field always has a horizontal component except at,or the horizontal component of Earth's magnetic field remains zero at:

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In the magnetic meridian of a certain place,the horizontal component of the earth's magnetic field is $0.26 \ G$ and the dip angle is $60^{\circ}$. What is the magnetic field of the earth at this location (in $G$)?

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The Earth's magnetic field at the equator is approximately $0.4 \, G$. Calculate the Earth's magnetic dipole moment (nearly). (Radius of Earth $R = 6.4 \times 10^6 \, m$)

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The vertical component of the earth's magnetic field is zero at a place where the angle of dip is.....$^o$

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At a certain place,the vertical component of the Earth's magnetic field is $\sqrt{3}$ times the horizontal component of the Earth's magnetic field. If a magnetic needle is suspended freely in the air,then it will incline:

Similar Questions

The horizontal component of the earth's magnetic field at any place is $0.36 \times 10^{-4} \; Wb/m^2$. If the angle of dip at that place is $60^{\circ}$,then the value of the vertical component of the earth's magnetic field will be ........ $\times 10^{-4} \; Wb/m^2$.

Earth's magnetic field always has a horizontal component except at,or the horizontal component of Earth's magnetic field remains zero at:

In the magnetic meridian of a certain place,the horizontal component of the earth's magnetic field is $0.26 \ G$ and the dip angle is $60^{\circ}$. What is the magnetic field of the earth at this location (in $G$)?

The Earth's magnetic field at the equator is approximately $0.4 \, G$. Calculate the Earth's magnetic dipole moment (nearly). (Radius of Earth $R = 6.4 \times 10^6 \, m$)

The vertical component of the earth's magnetic field is zero at a place where the angle of dip is.....$^o$

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