Earth Magnetism Questions in English

(N/A) The dynamo effect is a physical process that explains how a celestial body,such as the Earth,generates its magnetic field.
It occurs in the Earth's outer core,which consists of molten iron and nickel.
As the Earth rotates,the molten metal in the outer core undergoes convective motion due to heat from the inner core.
This moving,electrically conducting fluid interacts with the Earth's existing magnetic field,inducing electric currents within the fluid.
These currents,in turn,generate their own magnetic fields,which reinforce and sustain the Earth's overall magnetic field.
This self-sustaining mechanism is known as the dynamo effect.

(B) The Earth's magnetic axis is tilted relative to its geographic axis (axis of rotation).
This angle of inclination between the magnetic axis and the geographic axis is approximately $11.3^{\circ}$.

(N/A) $(i)$ The magnetic north pole is located at a latitude of $79.74^{\circ} N$ and a longitude of $71.8^{\circ} W$ in northern Canada.
$(ii)$ The magnetic south pole is located at $79.74^{\circ} S$ and $108.22^{\circ} E$ in Antarctica.

(A) The Earth acts as a giant bar magnet. The magnetic pole of the Earth located near the geographic north pole is known as the $ \text{north magnetic pole} $ (or magnetic north pole).
Conversely, the magnetic pole of the Earth located near the geographic south pole is known as the $ \text{south magnetic pole} $ (or magnetic south pole).
Therefore, the correct sequence is: $(i)$ $ \text{north magnetic pole} $, $(ii)$ $ \text{south magnetic pole} $.

(N/A) $1$. Geographic Meridian: The geographic meridian at any place is a vertical plane passing through the geographic north and south poles of the Earth. It is the plane that contains the geographic axis of rotation of the Earth.
$2$. Magnetic Meridian: The magnetic meridian at any place is a vertical plane passing through the magnetic north and south poles of the Earth. It is the plane that contains the magnetic axis of the Earth,which is the line joining the Earth's magnetic poles.

(N/A) Magnetic declination is the angle between the magnetic meridian and the geographic meridian at a particular location on the Earth's surface.
It represents the difference between magnetic north (the direction a compass needle points) and true geographic north (the direction toward the North Pole).
This angle varies depending on the observer's position on the Earth and changes slowly over time due to the movement of the Earth's magnetic poles.

(N/A) $(i)$ The declination is greater at higher latitudes because the magnetic meridians converge towards the magnetic poles.
$(ii)$ The declination in India is small,as India is located at a lower latitude relative to the magnetic poles.

(N/A) $1$. Angle of Dip (Magnetic Inclination): The angle of dip is the angle that the total magnetic field of the Earth makes with the surface of the Earth (horizontal direction) at a given location. It is $0^{\circ}$ at the magnetic equator and $90^{\circ}$ at the magnetic poles.
$2$. Angle of Declination: The angle of declination is the angle between the geographic meridian (true north-south line) and the magnetic meridian (the plane containing the magnetic axis of the Earth) at a specific point on the Earth's surface.

(N/A) The earth's magnetic field at any point on the surface is completely specified by three physical quantities, which are known as the elements of the earth's magnetic field:
$1$. Magnetic Declination $(\theta)$: The angle between the geographic meridian and the magnetic meridian at a place.
$2$. Magnetic Inclination or Angle of Dip $(\delta)$: The angle made by the total magnetic field vector of the earth with the horizontal direction in the magnetic meridian.
$3$. Horizontal Component of the Earth's Magnetic Field $(B_H)$: The component of the earth's total magnetic field vector along the horizontal direction in the magnetic meridian.

(A) Given that,$B_{V} = \frac{\mu_{0}}{4\pi} \frac{2m \cos \theta}{r^{3}}$ and $B_{H} = \frac{\mu_{0}}{4\pi} \frac{m \sin \theta}{r^{3}}$.
The magnitude of the magnetic field $B$ is given by $B = \sqrt{B_{V}^{2} + B_{H}^{2}}$.
Substituting the expressions:
$B^{2} = \left(\frac{\mu_{0}}{4\pi} \frac{m}{r^{3}}\right)^{2} (4 \cos^{2} \theta + \sin^{2} \theta)$
Using $\sin^{2} \theta = 1 - \cos^{2} \theta$:
$B^{2} = \left(\frac{\mu_{0}}{4\pi} \frac{m}{r^{3}}\right)^{2} (3 \cos^{2} \theta + 1)$
$B = \frac{\mu_{0}}{4\pi} \frac{m}{r^{3}} \sqrt{3 \cos^{2} \theta + 1}$.
For $B$ to be minimum,the term $(3 \cos^{2} \theta + 1)$ must be minimum. This occurs when $\cos \theta = 0$,which implies $\theta = 90^{\circ}$.
Since $\theta = 90^{\circ} - \text{latitude}$,$\theta = 90^{\circ}$ corresponds to a latitude of $0^{\circ}$,which is the magnetic equator. Thus,the locus of points where $|\vec{B}|$ is minimum is the magnetic equator.

(A) The dip angle $\delta$ is defined by the relation $\tan \delta = \frac{B_v}{B_H}$.
For the dip angle to be zero,we must have $\tan \delta = 0$,which implies $B_v = 0$.
Substituting the given expression for $B_v$:
$B_v = \frac{\mu_0}{4\pi} \frac{2m \cos \theta}{r^3} = 0$
Since $\frac{\mu_0}{4\pi}$,$m$,and $r^3$ are non-zero,we must have $\cos \theta = 0$.
This implies $\theta = 90^\circ$.
Given that $\theta = 90^\circ - \text{latitude}$,we have $90^\circ = 90^\circ - \text{latitude}$,which means $\text{latitude} = 0^\circ$.
Therefore,the locus of points where the dip angle is zero is the magnetic equator.

(A) The dip angle $\delta$ is defined by the relation $\tan \delta = \frac{B_v}{B_H}$.
Given the expressions for the vertical component $B_v$ and the horizontal component $B_H$:
$B_v = \frac{\mu_0}{4\pi} \frac{2m \cos \theta}{r^3}$
$B_H = \frac{\mu_0}{4\pi} \frac{m \sin \theta}{r^3}$
Substituting these into the expression for $\tan \delta$:
$\tan \delta = \frac{\frac{\mu_0}{4\pi} \frac{2m \cos \theta}{r^3}}{\frac{\mu_0}{4\pi} \frac{m \sin \theta}{r^3}} = \frac{2 \cos \theta}{\sin \theta} = 2 \cot \theta$
We are looking for the loci of points where the dip angle $\delta = \pm 45^{\circ}$.
Since $\tan(\pm 45^{\circ}) = \pm 1$,we have:
$\pm 1 = 2 \cot \theta$
$\cot \theta = \pm 0.5$
$\tan \theta = \pm 2$
Since $\theta = 90^{\circ} - \lambda$ (where $\lambda$ is the latitude),the locus of points is defined by the condition $\tan(90^{\circ} - \lambda) = \pm 2$,which simplifies to $\cot \lambda = \pm 2$ or $\tan \lambda = \pm 0.5$.

(N/A) $1$. At point $P$: Since point $P$ lies in the plane $S$ formed by the dipole axis and the Earth's axis,the magnetic meridian coincides with the geographical meridian. Therefore,the declination at $P$ is $0^{\circ}$. Since $P$ lies on the magnetic equator,the angle of dip at $P$ is $0^{\circ}$.
$2$. At point $Q$: Point $Q$ is the intersection of the geographical and magnetic equators. The magnetic axis is tilted at an angle of $11.3^{\circ}$ with respect to the geographical axis. Consequently,the magnetic meridian at $Q$ makes an angle of $11.3^{\circ}$ with the geographical meridian. Therefore,the declination at $Q$ is $11.3^{\circ}$. Since $Q$ lies on the magnetic equator,the angle of dip at $Q$ is $0^{\circ}$.

(D) The horizontal component of the earth's magnetic field is given by $B_H = 0.36 \times 10^{-4} \; Wb/m^2$.
The angle of dip is $\delta = 60^{\circ}$.
The vertical component $B_V$ is related to the horizontal component $B_H$ by the formula:
$B_V = B_H \tan \delta$
Substituting the given values:
$B_V = (0.36 \times 10^{-4}) \times \tan 60^{\circ}$
Since $\tan 60^{\circ} = \sqrt{3} \approx 1.732$:
$B_V = 0.36 \times 10^{-4} \times 1.732$
$B_V \approx 0.6235 \times 10^{-4} \; Wb/m^2$.
Rounding to the provided option,the value is $0.622 \times 10^{-4} \; Wb/m^2$.

(B) The relationship between the true dip $(\delta)$ and the apparent dip $(\delta^{\prime})$ at an angle $\theta$ to the magnetic meridian is given by the formula: $\tan \delta^{\prime} = \frac{\tan \delta}{\cos \theta}$.
Rearranging for the true dip: $\tan \delta = \tan \delta^{\prime} \cos \theta$.
Given: $\delta^{\prime} = 45^{\circ}$ and $\theta = 30^{\circ}$.
Substituting the values: $\tan \delta = \tan 45^{\circ} \cos 30^{\circ}$.
Since $\tan 45^{\circ} = 1$ and $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$,we get: $\tan \delta = 1 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$.
Therefore,the true dip is $\delta = \tan^{-1} \left( \frac{\sqrt{3}}{2} \right)$.

(B) The true dip $(\phi)$ is defined in the magnetic meridian,where the horizontal component of the Earth's magnetic field is $B_H$. The relation is given by $\tan \phi = \frac{B_V}{B_H}$,where $B_V$ is the vertical component.
When the dip circle is rotated by an angle $\alpha$ from the magnetic meridian,the horizontal component becomes $B_H \cos \alpha$,while the vertical component $B_V$ remains unchanged.
The apparent dip $(\phi^{\prime})$ is then given by $\tan \phi^{\prime} = \frac{B_V}{B_H \cos \alpha} = \frac{\tan \phi}{\cos \alpha}$.
Since $\cos \alpha \leq 1$,it follows that $\tan \phi^{\prime} \geq \tan \phi$. Because the tangent function is increasing in the range $[0, \pi/2]$,we have $\phi^{\prime} \geq \phi$.
Therefore,the true dip $(\phi)$ is always less than or equal to the apparent dip $(\phi^{\prime})$.

(A) The horizontal component of the Earth's magnetic field is given by the formula: $B_H = B \cos \delta$,where $B$ is the total magnetic field and $\delta$ is the angle of dip.
Given: $B_H = 0.5 \ G$ and $\delta = 30^{\circ}$.
Substituting the values into the formula: $0.5 = B \cos 30^{\circ}$.
Since $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$,we have $0.5 = B \times \frac{\sqrt{3}}{2}$.
Solving for $B$: $B = \frac{0.5 \times 2}{\sqrt{3}} = \frac{1}{\sqrt{3}} \ G$.

(A) The apparent angle of dip $\theta^{\prime}$ in a plane making an angle $\alpha$ with the magnetic meridian is given by the relation: $\tan \theta^{\prime} = \frac{\tan \theta}{\cos \alpha}$,where $\theta$ is the actual angle of dip.
Given,$\theta^{\prime} = 60^{\circ}$ and $\alpha = 45^{\circ}$.
Substituting these values into the formula:
$\tan 60^{\circ} = \frac{\tan \theta}{\cos 45^{\circ}}$
$\sqrt{3} = \frac{\tan \theta}{1/\sqrt{2}}$
$\tan \theta = \sqrt{3} \times \frac{1}{\sqrt{2}} = \sqrt{\frac{3}{2}}$
Therefore,the actual angle of dip is $\theta = \tan ^{-1}\left(\sqrt{\frac{3}{2}}\right)$.

(D) The vertical component of the earth's magnetic field $(B_V)$ is related to the resultant magnetic field $(B)$ and the angle of dip $(\delta)$ by the formula:
$B_V = B \sin \delta$
Given that $B_V = 6 \times 10^{-5} \text{ T}$ and $\delta = 37^{\circ}$.
Since $\tan 37^{\circ} = \frac{3}{4}$,we can form a right-angled triangle with opposite side $3$ and adjacent side $4$. The hypotenuse is $\sqrt{3^2 + 4^2} = 5$.
Therefore,$\sin 37^{\circ} = \frac{3}{5}$.
Substituting the values into the formula:
$6 \times 10^{-5} = B \times \frac{3}{5}$
$B = \frac{6 \times 10^{-5} \times 5}{3}$
$B = 2 \times 10^{-5} \times 5$
$B = 10 \times 10^{-5} \text{ T} = 10^{-4} \text{ T}$

(B) The total change in orientation is $180^{\circ}$.
Given that the duration of the flip is $10^5 \, \text{years}$.
The average change in orientation per year in degrees is $\frac{180^{\circ}}{10^5} = 1.8 \times 10^{-3} \, ^{\circ}/\text{year}$.
We know that $1^{\circ} = 60 \, \text{minutes}$ and $1 \, \text{minute} = 60 \, \text{seconds}$, so $1^{\circ} = 3600 \, \text{seconds}$.
Therefore, the average change in orientation per year in seconds is $1.8 \times 10^{-3} \times 3600 \, \text{s/year} = 6.48 \, \text{s/year}$.
Comparing this value with the given options, it is closest to $5 \, \text{s}$.

(B) The magnetic field due to a long straight wire at a distance $r$ is given by $B_{\text{wire}} = \frac{\mu_0 I}{2 \pi r}$.
Given $I = 5.0 \, A$ and $r = 10.0 \, cm = 0.1 \, m$.
$B_{\text{wire}} = \frac{4 \pi \times 10^{-7} \times 5.0}{2 \pi \times 0.1} = 1.0 \times 10^{-5} \, T$.
The compass needle aligns with the resultant magnetic field. The deflection is $60^{\circ}$ north of east. From the geometry of the vector addition,$\tan 60^{\circ} = \frac{B_{\text{wire}}}{B_H}$,where $B_H$ is the horizontal component of the earth's magnetic field.
Therefore,$B_H = \frac{B_{\text{wire}}}{\tan 60^{\circ}} = \frac{1.0 \times 10^{-5}}{\sqrt{3}} \approx 0.577 \times 10^{-5} \, T \approx 0.6 \times 10^{-5} \, T$.
Thus,option $(B)$ is the correct answer.

(A) The correct answer is $A$.
According to the dynamo theory,the magnetic field of a planet is generated by the motion of molten,electrically conducting material in its core.
The strength of this magnetic field is directly related to the rotational speed of the planet.
Planets that rotate faster tend to have more vigorous convection and motion in their cores,which results in a stronger magnetic field.
Therefore,planets producing a larger magnetic field have a larger rotational speed.

(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.

(B) Let $\phi$ be the true angle of dip,and $\alpha_1$ and $\alpha_2$ be the apparent angles of dip in two mutually perpendicular planes.
The relationship between the true angle of dip and the apparent angles of dip is given by the formula:
$\cot^2 \phi = \cot^2 \alpha_1 + \cot^2 \alpha_2$
Given:
$\alpha_1 = 45^{\circ}$
$\alpha_2 = 30^{\circ}$
Substituting the values:
$\cot^2 \phi = \cot^2 45^{\circ} + \cot^2 30^{\circ}$
Since $\cot 45^{\circ} = 1$ and $\cot 30^{\circ} = \sqrt{3}$:
$\cot^2 \phi = (1)^2 + (\sqrt{3})^2$
$\cot^2 \phi = 1 + 3 = 4$
Taking the square root on both sides:
$\cot \phi = 2$
Therefore,the true angle of dip is:
$\phi = \cot^{-1}(2)$

(D) The horizontal component of the earth's magnetic field is given by $B_H = 0.35 \times 10^{-4} \,T$.
The angle of dip is $\delta = 60^{\circ}$.
The relationship between the vertical component $(B_V)$,horizontal component $(B_H)$,and the angle of dip $(\delta)$ is given by the formula: $B_V = B_H \tan \delta$.
Substituting the given values into the formula:
$B_V = (0.35 \times 10^{-4}) \times \tan(60^{\circ})$.
Since $\tan(60^{\circ}) = \sqrt{3} \approx 1.732$,we have:
$B_V = 0.35 \times 1.732 \times 10^{-4} \,T$.
$B_V \approx 0.6062 \times 10^{-4} \,T$.
Rounding to two decimal places,we get $B_V \approx 0.61 \times 10^{-4} \,T$.
Therefore,the correct option is $D$.

(B) Magnetic declination is defined as the angle between the geographic meridian and the magnetic meridian at a given location.
Given that the declination is $15^{\circ} \, E$,it means the magnetic north is $15^{\circ}$ to the east of the geographic north.
In the provided figure,the compass needle points along the central arrow. Since the magnetic north is $15^{\circ}$ east of the geographic north,the geographic north must be $15^{\circ}$ to the west of the magnetic north (the direction of the compass needle).
Looking at the diagram,direction $2$ is $15^{\circ}$ to the left (west) of the central arrow (magnetic north).
Therefore,direction $2$ represents the geographic north.

(D) The frequency of oscillation $f$ of a magnetic needle in a magnetic field is given by $f \propto \sqrt{B_H}$,where $B_H$ is the horizontal component of the Earth's magnetic field.
Given $B_H = B \cos \theta$,where $B$ is the total magnetic field and $\theta$ is the angle of dip.
Thus,$f \propto \sqrt{B \cos \theta}$.
Let $f_1 = 20 \text{ oscillations/min}$ at $\theta_1 = 30^{\circ}$ and $f_2 = 30 \text{ oscillations/min}$ at $\theta_2 = 60^{\circ}$.
$\frac{f_1}{f_2} = \sqrt{\frac{B_1 \cos \theta_1}{B_2 \cos \theta_2}}$
$\frac{20}{30} = \sqrt{\frac{B_1 \cos 30^{\circ}}{B_2 \cos 60^{\circ}}}$
$\frac{2}{3} = \sqrt{\frac{B_1 (\sqrt{3}/2)}{B_2 (1/2)}} = \sqrt{\frac{B_1 \sqrt{3}}{B_2}}$
Squaring both sides: $\frac{4}{9} = \frac{B_1 \sqrt{3}}{B_2} \Rightarrow \frac{B_1}{B_2} = \frac{4}{9\sqrt{3}} = \frac{4}{\sqrt{81 \times 3}} = \frac{4}{\sqrt{243}}$.
Comparing this with $\frac{4}{\sqrt{x}}$,we get $x = 243$.

(A) The horizontal component of the Earth's magnetic field is $B_H = 3.5 \times 10^{-5} \,T$.
The current in the conductor is $i = \sqrt{2} \,A$.
The conductor is placed from South-East to North-West,which makes an angle of $\theta = 45^\circ$ with the horizontal magnetic field (North-South direction).
The force per unit length on a current-carrying conductor is given by $\frac{F}{\ell} = i B_H \sin \theta$.
Substituting the values: $\frac{F}{\ell} = \sqrt{2} \times (3.5 \times 10^{-5}) \times \sin(45^\circ)$.
Since $\sin(45^\circ) = \frac{1}{\sqrt{2}}$,we get $\frac{F}{\ell} = \sqrt{2} \times 3.5 \times 10^{-5} \times \frac{1}{\sqrt{2}} = 3.5 \times 10^{-5} \,N/m$.
Converting to the required format: $3.5 \times 10^{-5} = 35 \times 10^{-6} \,N/m$.
Therefore,the value is $35$.

(D) The horizontal component of the Earth's magnetic field is given by the formula $H = B_e \cos \delta$,where $B_e$ is the total magnetic field of the Earth and $\delta$ is the angle of dip.
At place $A$,the angle of dip $\delta_A = 30^{\circ}$. Therefore,the horizontal component $H_A = B_e \cos 30^{\circ}$.
At place $B$,the angle of dip $\delta_B = 45^{\circ}$. Therefore,the horizontal component $H_B = B_e \cos 45^{\circ}$.
The ratio of the horizontal component at $A$ to that at $B$ is given by:
$\frac{H_A}{H_B} = \frac{B_e \cos 30^{\circ}}{B_e \cos 45^{\circ}} = \frac{\cos 30^{\circ}}{\cos 45^{\circ}}$
Substituting the values: $\frac{H_A}{H_B} = \frac{\sqrt{3}/2}{1/\sqrt{2}} = \frac{\sqrt{3}}{2} \times \sqrt{2} = \frac{\sqrt{3}}{\sqrt{2}}$.
Thus,the ratio is $\sqrt{3}: \sqrt{2}$.

(B) The angle of dip $(I)$ is defined by the relationship between the vertical component $(Z_E)$ and the horizontal component $(H_E)$ of the earth's magnetic field as:
$\tan I = \frac{Z_E}{H_E}$
Given that the horizontal component is equal to the vertical component,we have:
$Z_E = H_E$
Substituting this into the formula:
$\tan I = \frac{Z_E}{Z_E} = 1$
Therefore,the angle of dip is:
$I = \tan^{-1}(1) = 45^{\circ}$

(D) The magnetic declination at a specific location is the angle between the geographic meridian and the magnetic meridian. According to standard geographical and magnetic data for India,the magnetic declination at Delhi is approximately $0^{\circ} 41^{\prime} E$.

(A) In the context of the Earth's magnetic field,$B_E$ represents the total magnetic field intensity,$H_E$ represents the horizontal component,and $Z_E$ represents the vertical component.
According to the vector resolution of the Earth's magnetic field,the total magnetic field $B_E$ is the vector sum of its horizontal and vertical components.
Since these components are perpendicular to each other,the magnitude is given by the Pythagorean theorem: $B_E = \sqrt{Z_E^2 + H_E^2}$.

(B) The angle of dip $(I)$ is defined by the relationship between the vertical component $(Z_E)$ and the horizontal component $(H_E)$ of the Earth's magnetic field as:
$\tan I = \frac{Z_E}{H_E}$
Given that the vertical component is $\sqrt{3}$ times the horizontal component:
$Z_E = \sqrt{3} H_E$
Substituting this into the formula:
$\tan I = \frac{\sqrt{3} H_E}{H_E} = \sqrt{3}$
Since $\tan 60^{\circ} = \sqrt{3}$,we have:
$I = 60^{\circ}$

(B) The magnetic field $B$ at the magnetic equator of the Earth is given by the formula:
$B = \frac{\mu_0}{4\pi} \frac{m}{R^3}$
where $m$ is the magnetic dipole moment and $R$ is the radius of the Earth.
Given values:
$B = 0.5 \times 10^{-4} \text{ T}$
$R = 6400 \text{ km} = 6.4 \times 10^6 \text{ m}$
$\frac{\mu_0}{4\pi} = 10^{-7} \text{ T m A}^{-1}$
Substituting these values into the formula:
$0.5 \times 10^{-4} = 10^{-7} \times \frac{m}{(6.4 \times 10^6)^3}$
$m = \frac{0.5 \times 10^{-4} \times (6.4 \times 10^6)^3}{10^{-7}}$
$m = 0.5 \times 10^3 \times (6.4)^3 \times 10^{18}$
$m = 0.5 \times 262.144 \times 10^{21}$
$m = 131.072 \times 10^{21} = 1.31 \times 10^{23} \text{ Am}^2$
Thus,the magnetic dipole moment of the Earth is $1.31 \times 10^{23} \text{ Am}^2$.

(B) The angle of dip $(I)$ is defined by the ratio of the vertical component $(Z_{E})$ to the horizontal component $(H_{E})$ of the Earth's magnetic field as follows:
$\tan I = \frac{Z_{E}}{H_{E}}$
Given that the angle of dip $I = 60^{\circ}$,we substitute this value into the equation:
$\tan 60^{\circ} = \frac{Z_{E}}{H_{E}}$
Since $\tan 60^{\circ} = \sqrt{3}$,we have:
$\sqrt{3} = \frac{Z_{E}}{H_{E}}$
Therefore,the relationship is:
$Z_{E} = \sqrt{3} H_{E}$

(D) The angle between the Geographic Meridian and the Magnetic Meridian at any point on the Earth's surface is known as the Magnetic Declination or simply Declination.
It represents the deviation of the magnetic compass needle from the true geographic north.

(D) Given:
Horizontal component of Earth's magnetic field,$B_H = 3 \times 10^{-5} \,T$
Angle of dip,$\delta = 45^{\circ}$
We know that the horizontal component $B_H$ is related to the resultant magnetic field $B$ by the formula:
$B_H = B \cos \delta$
Therefore,the resultant magnetic field $B$ is given by:
$B = \frac{B_H}{\cos \delta}$
Substituting the given values:
$B = \frac{3 \times 10^{-5}}{\cos 45^{\circ}}$
Since $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$,we have:
$B = \frac{3 \times 10^{-5}}{1 / \sqrt{2}}$
$B = 3 \sqrt{2} \times 10^{-5} \,T$

(D) The intensity of the cosmic ray stream of charged particles is greater at the poles than at the equator because the Earth's magnetic field is strongest at the poles.
Charged particles (cosmic rays) are deflected by the Earth's magnetic field due to the Lorentz force,$F = q(v \times B)$.
This deflection is more pronounced near the equator,where the magnetic field lines are horizontal,causing many charged particles to be diverted away.
At the poles,the magnetic field lines are vertical,allowing more charged particles to reach the surface.
This variation in the intensity of cosmic rays provides direct evidence that the Earth possesses a magnetic field.

(B) At the magnetic poles,the angle of dip (inclination),$\delta = 90^{\circ}$.
The horizontal component of the Earth's magnetic field $(B_{H})$ is given by the formula:
$B_{H} = B \cos \delta$
where $B$ is the net magnetic field of the Earth.
Substituting the value of $\delta$ at the poles:
$B_{H} = B \cos 90^{\circ}$
Since $\cos 90^{\circ} = 0$,we get:
$B_{H} = 0$
Therefore,at the magnetic poles,the horizontal component of the Earth's magnetic field is zero.

(D) The strength of the Earth's magnetic field is not uniform across the globe.
It varies from place to place on the Earth's surface.
This variation occurs because the strength of a magnetic field is directly related to the density of magnetic field lines at a particular location.
Since the magnetic field lines of the Earth are not distributed with uniform density,the magnetic field strength changes depending on the geographic location.

(C) Given: Horizontal component of Earth's magnetic field $(B_H)$ = $3.0 \ G$. Angle of dip $(\delta)$ = $30^{\circ}$.
We know that the horizontal component of the Earth's magnetic field is given by the relation: $B_H = B \cos \delta$,where $B$ is the total magnetic field of the Earth.
Substituting the given values:
$3.0 = B \cos 30^{\circ}$
$3.0 = B \times \frac{\sqrt{3}}{2}$
$B = \frac{3.0 \times 2}{\sqrt{3}}$
$B = \frac{6.0}{1.732} \approx 3.464 \ G$
Rounding to one decimal place,we get $B \approx 3.5 \ G$.
Therefore,the magnetic field of the Earth at that location is $3.5 \ G$.

(B) Given:
Vertical component of Earth's magnetic field,$B_v = 0.45 \ G$
Angle of dip,$\delta = 60^{\circ}$
We know that the vertical component of the Earth's magnetic field is given by the formula:
$B_v = B \sin \delta$
Where $B$ is the total magnetic field of the Earth.
Rearranging the formula to solve for $B$:
$B = \frac{B_v}{\sin \delta}$
Substituting the given values:
$B = \frac{0.45}{\sin 60^{\circ}}$
Since $\sin 60^{\circ} = \frac{\sqrt{3}}{2} \approx 0.866$:
$B = \frac{0.45}{0.866} \approx 0.5196 \ G$
Rounding to two decimal places,we get:
$B \approx 0.52 \ G$

(D) Given: Horizontal component of Earth's magnetic field $B_{H} = 3 \times 10^{-5} \ T$.
Magnetic declination $\phi = 30^{\circ}$.
Magnetic moment of the compass needle $M = 18 \ Am^2$.
The torque $\tau$ experienced by a magnetic dipole in a magnetic field is given by the formula $\tau = M B_{H} \sin \phi$.
Substituting the given values:
$\tau = 18 \times (3 \times 10^{-5}) \times \sin 30^{\circ}$
Since $\sin 30^{\circ} = 0.5$,
$\tau = 18 \times 3 \times 10^{-5} \times 0.5$
$\tau = 54 \times 10^{-5} \times 0.5$
$\tau = 27 \times 10^{-5} \ Nm$.

(A) The angle of dip $\delta$ is the angle that the total magnetic field of the earth makes with the horizontal direction. Here,$\delta = 30^{\circ}$.
Given,the horizontal component of the earth's magnetic field is $B_H = 0.3 \ G$.
The relationship between the total magnetic field $B$,the horizontal component $B_H$,and the angle of dip $\delta$ is given by $B_H = B \cos \delta$.
Substituting the given values: $0.3 = B \cos 30^{\circ}$.
Since $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$,we have $0.3 = B \times \frac{\sqrt{3}}{2}$.
Solving for $B$: $B = \frac{0.3 \times 2}{\sqrt{3}} = \frac{0.6}{\sqrt{3}} = \frac{0.6 \times \sqrt{3}}{3} = 0.2 \sqrt{3} \ G$.
Alternatively,$B = \frac{0.6}{\sqrt{3}} = \frac{6}{10 \sqrt{3}} = \frac{3}{5 \sqrt{3}} = \frac{\sqrt{3}}{5} \ G$.

(C) Let $\phi$ be the true dip angle and $\theta$ be the magnetic declination. When the dip circle is in the geographic meridian,the angle between the plane and the magnetic meridian is $\theta$. The apparent dip $\delta_1$ is given by $\tan \delta_1 = \frac{\tan \phi}{\cos \theta}$.
When the plane is perpendicular to the geographic meridian,the angle between the plane and the magnetic meridian is $(90^\circ - \theta)$. The apparent dip $\delta_2$ is given by $\tan \delta_2 = \frac{\tan \phi}{\cos(90^\circ - \theta)} = \frac{\tan \phi}{\sin \theta}$.
From these two equations,we have $\tan \phi = \tan \delta_1 \cos \theta$ and $\tan \phi = \tan \delta_2 \sin \theta$.
Equating the two expressions for $\tan \phi$: $\tan \delta_1 \cos \theta = \tan \delta_2 \sin \theta$.
Rearranging gives $\frac{\sin \theta}{\cos \theta} = \frac{\tan \delta_1}{\tan \delta_2}$,which implies $\tan \theta = \frac{\tan \delta_1}{\tan \delta_2}$.
Therefore,$\theta = \tan ^{-1}\left(\frac{\tan \delta_1}{\tan \delta_2}\right)$.

(A) Let $B_e$ be the net magnetic field of the Earth at both locations. The horizontal component of the Earth's magnetic field $H$ is given by $H = B_e \cos \delta$,where $\delta$ is the angle of dip.
For the two places,the ratio of the horizontal components $H_1$ and $H_2$ is:
$\frac{H_1}{H_2} = \frac{B_e \cos \delta_1}{B_e \cos \delta_2} = \frac{\cos \delta_1}{\cos \delta_2}$
Given $\delta_1 = 30^{\circ}$ and $\delta_2 = 45^{\circ}$,we have:
$\frac{H_1}{H_2} = \frac{\cos 30^{\circ}}{\cos 45^{\circ}} = \frac{\sqrt{3}/2}{1/\sqrt{2}} = \frac{\sqrt{3}}{2} \times \sqrt{2} = \frac{\sqrt{3}}{\sqrt{2}}$
Therefore,the ratio $H_1: H_2 = \sqrt{3}: \sqrt{2}$.

(C) Let $\delta$ be the true angle of dip and $\theta$ be the magnetic declination.
When the dip circle is in the magnetic meridian,the apparent dip is $\delta$.
However,the problem states the plane is set in the geographic meridian. Let $\theta$ be the angle between the geographic meridian and the magnetic meridian.
The apparent dip $\delta_1$ in a plane making an angle $\theta$ with the magnetic meridian is given by $\tan \delta_1 = \frac{\tan \delta}{\cos \theta}$.
The apparent dip $\delta_2$ in a plane perpendicular to the first (making an angle $90^\circ - \theta$ with the magnetic meridian) is given by $\tan \delta_2 = \frac{\tan \delta}{\cos(90^\circ - \theta)} = \frac{\tan \delta}{\sin \theta}$.
From these two equations,we have $\tan \delta = \tan \delta_1 \cos \theta$ and $\tan \delta = \tan \delta_2 \sin \theta$.
Equating them: $\tan \delta_1 \cos \theta = \tan \delta_2 \sin \theta$.
Therefore,$\frac{\sin \theta}{\cos \theta} = \frac{\tan \delta_1}{\tan \delta_2}$,which implies $\tan \theta = \frac{\tan \delta_1}{\tan \delta_2}$.
Thus,$\theta = \operatorname{Tan}^{-1}\left(\frac{\tan \delta_1}{\tan \delta_2}\right)$.

(B) The relationship between the vertical component $(B_V)$,horizontal component $(B_H)$,and the angle of dip $(\delta)$ is given by the formula: $\tan \delta = \frac{B_V}{B_H}$.
Given: $B_V = 0.3464 \ G$ and $\delta = 30^{\circ}$.
Rearranging the formula to solve for $B_H$: $B_H = \frac{B_V}{\tan \delta}$.
Substituting the values: $B_H = \frac{0.3464}{\tan 30^{\circ}}$.
Since $\tan 30^{\circ} = \frac{1}{\sqrt{3}} \approx 0.577$,we have $B_H = 0.3464 \times \sqrt{3}$.
Using $\sqrt{3} \approx 1.732$,we get $B_H = 0.3464 \times 1.732 \approx 0.6 \ G$.

(C) The angle of dip $\theta$ is related to the vertical component $B_V$ and the horizontal component $B_H$ of the Earth's magnetic field by the formula: $\tan \theta = \frac{B_V}{B_H}$.
Given that the horizontal component $B_H$ is $\frac{1}{\sqrt{3}}$ times the vertical component $B_V$,we have $B_H = \frac{1}{\sqrt{3}} B_V$.
Substituting this into the formula: $\tan \theta = \frac{B_V}{\frac{1}{\sqrt{3}} B_V} = \sqrt{3}$.
Since $\tan 60^{\circ} = \sqrt{3}$,the angle of dip is $\theta = 60^{\circ}$.