Earth Magnetism Questions in English

(A) When a magnet is suspended freely,it aligns itself with the Earth's magnetic field. The vertical plane passing through the magnetic axis of the freely suspended magnet is known as the magnetic meridian. Therefore,the magnet comes to rest in the magnetic meridian.

(D) The phenomenon of magnetic shielding occurs when a material with high magnetic permeability (like steel) is used to enclose a region.
When a hollow steel box is placed in an external magnetic field,the magnetic field lines prefer to pass through the steel walls due to their high permeability rather than through the hollow cavity inside.
Consequently,the magnetic field inside the cavity becomes effectively zero.
Therefore,the correct option is $D$.

(B) The horizontal component of the Earth's magnetic field is given by the formula $B_H = B \cos \theta$,where $B$ is the total magnetic field and $\theta$ is the angle of dip (inclination).
At the magnetic poles,the magnetic field lines are perpendicular to the Earth's surface,meaning the angle of dip $\theta = 90^o$.
Substituting this into the formula: $B_H = B \cos(90^o) = B \times 0 = 0$.
Therefore,the horizontal component of the Earth's magnetic field is zero at the magnetic poles.

(A) The magnetic meridian is the vertical plane passing through the magnetic North and South poles of the Earth. The Earth's magnetic field vector $\vec{B}$ lies entirely within this plane.
When a dip needle is placed in a plane perpendicular to the magnetic meridian,the horizontal component of the Earth's magnetic field $(B_H)$ is perpendicular to this plane. Therefore,the effective horizontal component of the magnetic field acting on the needle in this plane is zero.
Only the vertical component of the Earth's magnetic field $(B_V)$ acts on the needle in this plane. Since the magnetic force on the needle is directed along the vertical component,the needle will align itself vertically.
Thus,the dip needle will remain vertical.

(C) The angle of dip (or magnetic inclination) is the angle that the total magnetic field of the Earth makes with the horizontal direction.
At the magnetic equator,the magnetic field lines are parallel to the Earth's surface,so the angle of dip is $0^{\circ}$.
At the magnetic poles,the magnetic field lines are perpendicular to the Earth's surface (pointing vertically downwards or upwards),so the angle of dip is $90^{\circ}$.
Therefore,the correct option is $(C)$.

(C) The horizontal and vertical components of the Earth's magnetic field are $B_H$ and $B_V$ respectively.
We know that the horizontal component is given by $B_H = B \cos \delta$,where $B$ is the total magnetic field and $\delta$ is the angle of dip.
The vertical component is given by $B_V = B \sin \delta$.
Squaring and adding both equations:
$B_H^2 + B_V^2 = (B \cos \delta)^2 + (B \sin \delta)^2$
$B_H^2 + B_V^2 = B^2 (\cos^2 \delta + \sin^2 \delta)$
Since $\cos^2 \delta + \sin^2 \delta = 1$,we get:
$B^2 = B_H^2 + B_V^2$
$B = \sqrt{B_H^2 + B_V^2}$

(D) At the magnetic equator,the angle of dip (inclination) is $0^{\circ}$. The vertical component of the Earth's magnetic field is given by $V = B \sin \delta$,where $B$ is the total magnetic field and $\delta$ is the angle of dip. Since $\delta = 0^{\circ}$ at the magnetic equator,$V = B \sin(0^{\circ}) = 0$. Therefore,the vertical component is zero at the magnetic equator.

(B) The angle between the magnetic meridian and the geographical meridian at a given place is defined as the angle of declination. It represents the deviation of the magnetic compass needle from the true geographical north.

(A) The Earth's magnetic field is approximately uniform over a small region on its surface. The horizontal component of the Earth's magnetic field $(B_H)$ represents the magnetic field lines running parallel to the Earth's surface in a specific direction. In a localized region,these magnetic field lines are represented as parallel straight lines.

(C) The angle of dip $\phi$ is related to the vertical component $B_V$ and the horizontal component $B_H$ of the Earth's magnetic field by the formula: $\tan \phi = \frac{B_V}{B_H}$.
Given that the horizontal and vertical components are equal,we have $B_V = B_H$.
Substituting this into the formula,we get $\tan \phi = \frac{B_H}{B_H} = 1$.
Since $\tan \phi = 1$,the angle of dip is $\phi = \tan^{-1}(1) = 45^o$.

(D) The horizontal component of the earth's magnetic field is given by $B_H = B \cos \phi$,and the vertical component is given by $B_V = B \sin \phi$,where $\phi$ is the angle of dip.
Dividing the two equations,we get $\frac{B_V}{B_H} = \tan \phi$.
Therefore,the vertical component is $B_V = B_H \tan \phi$.
Given $B_H = 0.36 \times 10^{-4} \ Wb/m^2$ and $\phi = 60^o$.
Substituting the values: $B_V = (0.36 \times 10^{-4}) \times \tan(60^o)$.
Since $\tan(60^o) = \sqrt{3} \approx 1.732$,
$B_V = 0.36 \times 1.732 \times 10^{-4} \approx 0.623 \times 10^{-4} \ Wb/m^2$.
Thus,the correct option is $D$.

(D) The vertical component of the earth's magnetic field is given by the formula: $V = I \sin \phi$,where $I$ is the total intensity and $\phi$ is the angle of dip.
Given: $V = 6 \times 10^{-5} \text{ Tesla}$ and $\phi = 40.6^\circ$.
Rearranging the formula to solve for $I$: $I = \frac{V}{\sin \phi}$.
Substituting the values: $I = \frac{6 \times 10^{-5}}{\sin 40.6^\circ}$.
Using $\sin 40.6^\circ \approx 0.65$,we get: $I = \frac{6 \times 10^{-5}}{0.65} \approx 9.23 \times 10^{-5} \text{ Tesla}$.
Thus,the total intensity is approximately $9.2 \times 10^{-5} \text{ Tesla}$.

(C) The angle of dip (or magnetic inclination) is defined as the angle that the total magnetic field vector of the earth makes with the horizontal direction in the magnetic meridian.
Let $B$ be the total magnetic field of the earth.
The horizontal component is given by $B_{H} = B \cos \theta$.
The vertical component is given by $B_{V} = B \sin \theta$.
Here,$\theta$ represents the angle of dip. Therefore,the angle of dip is the angle between the earth's magnetic field direction and the horizontal direction.

(C) The horizontal component of the Earth's magnetic field $(B_H)$ is related to the total magnetic field $(B)$ and the angle of dip $(\phi)$ by the formula: $B_H = B \cos \phi$.
Given: $B_H = 0.50 \text{ Oersted}$ and $\phi = 30^{\circ}$.
Substituting the values: $0.50 = B \cos(30^{\circ})$.
Since $\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$,we have $0.50 = B \times \frac{\sqrt{3}}{2}$.
Therefore,$B = \frac{0.50 \times 2}{\sqrt{3}} = \frac{1}{\sqrt{3}} \text{ Oersted}$.

(A) At the magnetic equator,the vertical component of the Earth's magnetic field,$B_{V}$,is equal to $0$.
The angle of dip $\phi$ is defined by the relation $\tan \phi = \frac{B_{V}}{B_{H}}$,where $B_{H}$ is the horizontal component of the Earth's magnetic field.
Since $B_{V} = 0$,we have $\tan \phi = \frac{0}{B_{H}} = 0$.
Therefore,$\phi = 0^{\circ}$.

(D) The earth's magnetic field is a vector quantity that can be resolved into horizontal $(B_H)$ and vertical $(B_V)$ components.
At the magnetic poles,the magnetic field lines are perpendicular to the earth's surface. Therefore,the field is entirely vertical,and the horizontal component is zero.
At the magnetic equator,the magnetic field lines are parallel to the earth's surface. Therefore,the field is entirely horizontal,and the vertical component is zero.
The line on the earth's surface joining all points where the magnetic field is purely horizontal (i.e.,the angle of dip is $0^{\circ}$) is defined as the magnetic equator.
Thus,the correct option is $(D)$.

(D) The Earth's magnetic axis is tilted at an angle of approximately $11.3^\circ$ (often rounded to $11^\circ$ or $12^\circ$ in various textbooks) with respect to the Earth's geographical axis (the axis of rotation). Since $11.3^\circ$ is not explicitly listed in the options provided $(0, 17, 23)$,the correct choice is $D$ (None of these).

(D) The lines joining the places of the same horizontal intensity are known as Isodynamic lines.
Isogonic lines join places of the same magnetic declination.
Isoclinic lines join places of the same magnetic inclination (dip).
Aclinic lines are lines where the magnetic dip is zero (magnetic equator).

(B) The magnetic field of the Earth can be modeled as a magnetic dipole.
At the magnetic equator,the horizontal component of the Earth's magnetic field is $B_e = \frac{\mu_0}{4\pi} \frac{M}{R^3}$.
At the magnetic poles,the vertical component of the Earth's magnetic field is $B_p = \frac{\mu_0}{4\pi} \frac{2M}{R^3}$.
Therefore,the ratio of the intensity at the equator to the poles is $\frac{B_e}{B_p} = \frac{\frac{\mu_0 M}{4\pi R^3}}{\frac{2\mu_0 M}{4\pi R^3}} = \frac{1}{2}$.
Thus,the ratio is $1:2$.

(D) The magnetic dip (or inclination) is the angle that the Earth's magnetic field makes with the horizontal at any point on the surface.
At the magnetic equator,the magnetic field lines are parallel to the Earth's surface,meaning the angle of dip is $0^{\circ}$.
$A$ line connecting all such points on the Earth's surface where the magnetic dip is zero is known as the magnetic equator or the Aclinic line.
Therefore,the correct option is $D$.

(A) The angle of dip $\phi$ is given by the relation $\tan \phi = \frac{B_V}{B_H}$,where $B_V$ is the vertical component and $B_H$ is the horizontal component of the Earth's magnetic field.
Given: $B_H = 0.30 \ G$ and $B_V = 0.173 \ G$.
Substituting the values: $\tan \phi = \frac{0.173}{0.30} = \frac{1.73}{3.0}$.
Since $\sqrt{3} \approx 1.732$,we have $\tan \phi = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$.
Therefore,$\phi = \tan^{-1}(\frac{1}{\sqrt{3}}) = 30^o$.

(D) The horizontal component of the Earth's magnetic field $(B_H)$ is related to the total magnetic field $(B)$ and the angle of dip $(\phi)$ by the formula:
$B_H = B \cos \phi$
Given:
Total intensity $B = 0.64 \text{ units}$
Angle of dip $\phi = 60^{\circ}$
Substituting the values:
$B_H = 0.64 \times \cos 60^{\circ}$
Since $\cos 60^{\circ} = 0.5$
$B_H = 0.64 \times 0.5 = 0.32 \text{ units}$
Therefore,the correct option is $D$.

(B) magnetic compass needle is designed to rotate in a horizontal plane to indicate the direction of the horizontal component of the Earth's magnetic field $(B_H)$.
Near the magnetic poles,the Earth's magnetic field lines are almost perpendicular to the surface of the Earth,meaning the field is almost entirely vertical.
Consequently,the horizontal component $(B_H)$ becomes negligible or zero.
Since the compass needle cannot align itself horizontally in the absence of a horizontal magnetic field,it becomes useless for navigation in these regions.

(D) The angle of dip at a place on the Earth represents the direction of the Earth's total magnetic field $\vec{B}$ relative to the horizontal direction in the magnetic meridian.
By definition,the angle of dip (or magnetic inclination) is the angle made by the Earth's total magnetic field vector with the horizontal plane.
Experimentally,if a magnetic needle is pivoted such that it is free to rotate in a vertical plane coinciding with the magnetic meridian,it will align itself along the direction of the Earth's total magnetic field.
The angle between the horizontal and the final orientation of this needle provides the angle of dip at that specific location.
Therefore,the correct option is $D$.

(A) The earth's magnetic field $B$ can be resolved into two components: the horizontal component $H$ and the vertical component $V$.
At the magnetic poles, the magnetic field lines are perpendicular to the earth's surface.
Therefore, the angle of dip $\delta$ is $90^{\circ}$.
The horizontal component is given by $H = B \cos(\delta)$.
Substituting $\delta = 90^{\circ}$, we get $H = B \cos(90^{\circ}) = B \times 0 = 0$.
Thus, at the magnetic north pole, the horizontal component is $0$ and the angle of dip is $90^{\circ}$.

(A) Given: Horizontal component of Earth's magnetic field,$B_H = 0.3 \; Oersted$.
Total intensity of Earth's magnetic field,$I = 0.6 \; Oersted$.
The relationship between the horizontal component,total intensity,and the angle of dip $(\phi)$ is given by: $B_H = I \cos \phi$.
Rearranging for $\cos \phi$: $\cos \phi = \frac{B_H}{I}$.
Substituting the values: $\cos \phi = \frac{0.3}{0.6} = \frac{1}{2}$.
Since $\cos \phi = \frac{1}{2}$,the angle of dip is $\phi = 60^o$.

(D) The relation between the apparent dip $(\alpha_{1})$ and the true dip $(\alpha)$ when the dip circle is at an angle $\alpha_{2}$ with the magnetic meridian is given by:
$\cot \alpha_{1} = \cot \alpha \cdot \cos \alpha_{2}$
Here, $\alpha_{2} = 90^{\circ}$ because the dip circle is at a right angle to the magnetic meridian.
Substituting the value:
$\cot \alpha_{1} = \cot \alpha \cdot \cos 90^{\circ}$
Since $\cos 90^{\circ} = 0$, we get:
$\cot \alpha_{1} = 0$
This implies $\alpha_{1} = 90^{\circ}$.
Therefore, the apparent dip is $90^{\circ}$.

(A) The Earth's magnetic field is often modeled as a magnetic dipole located at the center of the Earth. However,in terms of the physical representation of the field lines and the magnetic moment,it is equivalent to a large bar magnet placed at the center of the Earth with a length equal to the diameter of the Earth. Therefore,option $A$ is the most accurate description of the equivalent magnet model.

(C) The total magnetic field $B$ is related to its horizontal component $B_H$ and vertical component $B_V$ by the equation $B^2 = B_V^2 + B_H^2$.
Given $B = 0.5 \ G$ and $B_H = 0.3 \ G$,we can find $B_V$:
$B_V = \sqrt{B^2 - B_H^2} = \sqrt{(0.5)^2 - (0.3)^2} = \sqrt{0.25 - 0.09} = \sqrt{0.16} = 0.4 \ G$.
The angle of dip $\phi$ is given by the relation $\tan \phi = \frac{B_V}{B_H}$.
Substituting the values,$\tan \phi = \frac{0.4}{0.3} = \frac{4}{3}$.
Therefore,$\phi = \tan^{-1} \left( \frac{4}{3} \right)$.

(A) The horizontal component of the earth's magnetic field is given by $B_H = B \cos \phi$,where $B$ is the total intensity and $\phi$ is the angle of dip.
Given: $B_H = 1.8 \times 10^{-5} \, Wb/m^2$ and $\phi = 30^{\circ}$.
We need to find the total intensity $B = \frac{B_H}{\cos \phi}$.
Substituting the values: $B = \frac{1.8 \times 10^{-5}}{\cos 30^{\circ}}$.
Since $\cos 30^{\circ} = \frac{\sqrt{3}}{2} \approx 0.866$,we have $B = \frac{1.8 \times 10^{-5}}{0.866} \approx 2.08 \times 10^{-5} \, Wb/m^2$.

(C) Lines that connect places on the Earth's surface having the same angle of dip (or magnetic inclination) are known as Isoclinic lines.
Isogonic lines connect places with the same magnetic declination,while Isobaric lines connect places with the same atmospheric pressure.

(A) The vertical component $(B_v)$ of the earth's magnetic field is given by the formula $B_v = B \sin(\delta)$,where $B$ is the total magnetic field and $\delta$ is the angle of dip.
For the vertical component to be zero $(B_v = 0)$,we must have $\sin(\delta) = 0$.
This implies that the angle of dip $\delta = 0^o$.
This condition occurs at the magnetic equator of the earth.

(D) The total magnetic field intensity $B$ of the Earth is given by the vector sum of its horizontal component ${B_0}$ and vertical component ${V_0}$.
Mathematically,$B = \sqrt{B_0^2 + V_0^2}$.
Given that the magnitudes are equal,we have ${B_0} = {V_0}$.
Substituting this into the formula,we get $B = \sqrt{B_0^2 + B_0^2}$.
$B = \sqrt{2B_0^2}$.
Therefore,$B = \sqrt{2}{B_0}$.

(A) At the Earth's magnetic poles,the magnetic field lines are perpendicular to the Earth's surface.
Since a compass needle aligns itself with the local magnetic field,it will point vertically downward at the North Magnetic Pole and vertically upward at the South Magnetic Pole.
Therefore,the needle will show a vertical direction.

(A) The magnetic poles of the Earth are inverted relative to its geographic poles.
By convention,the north pole of a compass needle points towards the geographic north pole. Since opposite poles attract,the magnetic pole located near the geographic north pole must be a magnetic south pole.
Conversely,the magnetic pole located near the geographic south pole must be a magnetic north pole.
Therefore,the north pole of the Earth's internal magnet is situated near the geographic south pole.
Hence,the correct option is $A$.

(A) The Earth's magnetic field is primarily generated by the dynamo effect,which involves the motion of molten iron and nickel in the Earth's outer core. These convection currents,combined with the Earth's rotation,create electric currents that generate the magnetic field. Therefore,the magnetic field of the Earth is due to the motion and distribution of material in and outside the Earth.

(C) The earth's magnetic field lines are perpendicular to the surface at the poles and parallel to the surface at the equator. Charged particles moving towards the equator experience a magnetic force $F = q(v \times B)$ that deflects them. To overcome this magnetic influence and reach the equatorial region,the charged particles must possess higher kinetic energy compared to those reaching the polar regions.

(C) When a magnetic system is placed in a uniform magnetic field $B_H$,it comes to equilibrium when its net magnetic moment $M_{net}$ aligns with the direction of the magnetic field.
Let the magnetic moment $2M$ be along the $x$-axis and $M$ be along the $y$-axis.
The net magnetic moment vector is $\vec{M}_{net} = 2M \hat{i} + M \hat{j}$.
In equilibrium,the net magnetic moment aligns with the external magnetic field $B_H$.
The angle $\theta$ that the magnet with moment $2M$ makes with the magnetic meridian (direction of $B_H$) is given by the ratio of the perpendicular component to the parallel component of the net magnetic moment.
$\tan \theta = \frac{M}{2M} = \frac{1}{2}$.
Therefore,$\theta = \tan^{-1}\left(\frac{1}{2}\right)$.

(A) The angle of dip (or magnetic inclination) is defined as the angle that the total magnetic field of the Earth makes with the surface of the Earth.
At the magnetic poles,the Earth's magnetic field lines are perpendicular to the surface of the Earth.
Therefore,the angle of dip at the poles is $90^{\circ}$.

(B) The horizontal component of the Earth's magnetic field is given by $B_H = B \cos \delta$,where $B$ is the total intensity and $\delta$ is the angle of dip.
Given $B_H = B_0$ and $\delta = 45^o$.
Substituting the values,we get $B_0 = B \cos 45^o$.
Since $\cos 45^o = \frac{1}{\sqrt{2}}$,we have $B_0 = B \times \frac{1}{\sqrt{2}}$.
Therefore,the total intensity $B = \sqrt{2} B_0$.

(C) The angle of dip (or magnetic inclination) $\delta$ is defined by the relation $\tan \delta = \frac{V}{H}$,where $V$ is the vertical component of the Earth's magnetic field and $H$ is the horizontal component.
At the magnetic equator,the Earth's magnetic field lines are parallel to the surface of the Earth.
Therefore,the vertical component $V$ of the Earth's magnetic field is zero at the magnetic equator.
Substituting $V = 0$ into the formula,we get $\tan \delta = \frac{0}{H} = 0$,which implies $\delta = 0^\circ$.
Thus,the correct option is $C$.

(A) In the study of Earth's magnetism,the total intensity of the Earth's magnetic field $(I)$ can be resolved into two rectangular components:
$1$. The horizontal component $(H)$ acting along the Earth's surface.
$2$. The vertical component $(V)$ acting perpendicular to the Earth's surface.
According to the Pythagorean theorem for the vector triangle formed by these components,the relationship is given by $I^2 = V^2 + H^2$,where $I$ is the resultant magnetic field intensity.

(A) Null points are the points where the magnetic field due to the bar magnet is equal and opposite to the horizontal component of the Earth's magnetic field $(B_H)$.
When the north pole of a bar magnet points towards the geographic North,the magnetic field lines of the magnet on its equatorial line are directed from the magnetic North pole to the magnetic South pole (i.e.,from North to South).
Since the Earth's horizontal magnetic field $(B_H)$ is directed from geographic South to geographic North,the magnetic field of the magnet $(B)$ on the equatorial line opposes $B_H$.
Thus,the null points $N_1$ and $N_2$ are located on the equatorial line of the magnet when the north pole of the magnet points towards the geographic North.

(C) The magnetic meridian at any point on the Earth's surface is defined as the vertical plane passing through the magnetic North and South poles of the Earth at that location.
It represents the plane in which a freely suspended magnetic needle aligns itself under the influence of the Earth's magnetic field.
Therefore,the magnetic meridian is a vertical plane.

(B) The horizontal component of the Earth's magnetic field $(B_H)$ is related to the total magnetic field $(B)$ and the angle of dip $(\phi)$ by the formula: $B_H = B \cos \phi$.
Given that $B_H = H$ and $\phi = 30^o$,we can substitute these values into the equation:
$H = B \cos 30^o$
Since $\cos 30^o = \frac{\sqrt{3}}{2}$,we have:
$H = B \times \frac{\sqrt{3}}{2}$
Solving for the total magnetic field $B$:
$B = \frac{2H}{\sqrt{3}}$
Therefore,the intensity of the total magnetic field is $\frac{2H}{\sqrt{3}}$.

(C) Given:
Horizontal component of Earth's magnetic field,$B_H = 0.22 \ G$
Total magnetic field,$B = 0.4 \ G$
Let $\phi$ be the angle of dip.
The horizontal component is given by $B_H = B \cos \phi$.
Therefore,$\cos \phi = \frac{B_H}{B} = \frac{0.22}{0.4} = 0.55$.
The vertical component is $B_V = B \sin \phi = \sqrt{B^2 - B_H^2} = \sqrt{(0.4)^2 - (0.22)^2} = \sqrt{0.16 - 0.0484} = \sqrt{0.1116} \approx 0.334 \ G$.
The angle of dip $\phi$ is given by $\tan \phi = \frac{B_V}{B_H} = \frac{\sqrt{B^2 - B_H^2}}{B_H} = \frac{\sqrt{(0.4)^2 - (0.22)^2}}{0.22} = \frac{\sqrt{0.1116}}{0.22} \approx \frac{0.334}{0.22} \approx 1.518$.
Thus,$\phi = \tan^{-1}(1.518)$.

(D) The angle of dip (or magnetic inclination) is the angle that the total magnetic field of the Earth makes with the surface of the Earth.
At the magnetic poles,the magnetic field lines are perpendicular to the surface,so the angle of dip is $90^{\circ}$.
At the magnetic equator,the magnetic field lines are parallel to the surface of the Earth,meaning the angle of dip is $0^{\circ}$.
When the angle of dip is $0^{\circ}$,the Earth's magnetic field is purely horizontal.
Therefore,the correct answer is the magnetic equator.

(C) Isogonic lines (or isogonal lines) are lines drawn on a map connecting points on the Earth's surface that have the same magnetic declination.
Magnetic declination is the angle between magnetic north and true (geographic) north at a particular location.
Therefore,all points along an isogonic line share the same value of magnetic declination.
Thus,the correct option is $C$.

(A) The Earth's magnetic field $B_E$ can be resolved into two components: the horizontal component $H = B_E \cos \delta$ and the vertical component $V = B_E \sin \delta$,where $\delta$ is the angle of dip.
Near the North Pole,the magnetic field lines are almost perpendicular to the Earth's surface.
Therefore,the angle of dip $\delta$ is nearly $90^{\circ}$.
Since $\cos 90^{\circ} = 0$,the horizontal component $H$ becomes very small $(H \approx 0)$.
Since $\sin 90^{\circ} = 1$,the vertical component $V$ becomes nearly equal to the total magnetic field $B_E$ $(V \approx B_E)$.
Thus,near the North Pole,$V >> H$.

(A) The magnetic field lines of the Earth are perpendicular to the surface at the magnetic poles.
Since a compass needle aligns itself along the direction of the magnetic field lines,it will point vertically downwards or upwards at the poles.
Therefore,the compass needle will be in a vertical position at the magnetic poles of the Earth.