Q: If a magnet is suspended at angle \${30}^{o}\$ to the magnetic meridian, the dip needle makes an angle of \${45}^{o}\$ with the horizontal. The real dip is :

Angle of dip, \$\delta ={ 45 }^{ o }\$\$\therefore \quad \tan { \delta ' } =\cfrac { \tan { \delta } }{ \cos { \theta } } \$\$=\cfrac { \tan { 45 } }{ \cos { { 30 }^{ o } } } =\cfrac { 1 }{ \sqrt { 3 } /2 } =\cfrac { 2 }{ \sqrt { 3 } } \$\$\therefore\$ Real dip \$ \quad \delta '=\tan ^{ -1 }{ \left( 2/\sqrt { 3 } \right) } \$

Q: At a certain place, horizontal component is \$\sqrt 3\$ times the vertical component. The angle of dip at this place is:

\$\tan \delta = \dfrac{B_V}{B_H} \$\$= \dfrac{B_V}{\sqrt{3}B_V}=\dfrac{1}{ \sqrt{3}}=\tan 30^{\circ}\$\$ \Rightarrow \delta =30^{\circ}=\dfrac{\pi}{6}\$

Q: In the magnetic meridian of a certain place, the horizontal component of the earth's magnetic field is \$0.26\ G\$ and the angle of dip is \$60^o\$. The magnetic field of the earth at this location is

The earth's magnetic field is \$B_e\$ and its horizontal and vertical components are \$H_e\$ and \$H_v\$\$\cos \theta =\dfrac{H_e}{B_e}\$\$ \therefore \cos 60^{\circ} = \dfrac{0.26 \times 10^{-4}}{B_e}\:T\$\$ \Rightarrow B_e = \dfrac{ 0.26 \times 10^{-4}}{ \dfrac{1}{2}}=0.52 \times 10^{-4}\:T=0.52\:G\$

Q: If \$\theta_1\$ and \$\theta_2\$ be the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip \$\theta\$ is given by:

Note that \$\tan \theta = \dfrac{V}{H}\$ ...(1)where \$\theta\$ is true value of dip . Here we assumed \$\theta = \propto\$Now, angle of dips, \$\theta_1\$ and \$\theta_2\$ have different formula\$\tan\theta_1=\dfrac{V}{H\times cos \propto}\$ ...(2)\$\tan \theta_2=\dfrac{V}{H\times sin \propto}\$ ...(3)From eqn (1), (2), (3), we can find value of H/Vwhich \$\cot \theta\$, which when calculate comes option B\$\cot^2\theta_1+\cot^2\theta_2=\dfrac{H_2}{V^2}(\cos^2\propto +\sin^2\propto)=\cot^2\theta\$\$\cot^2\theta=\cot^2\theta_1+\cot^2\theta_2\$

Q: The earth's magnetic field at the equator is approximately \$0.4 G\$. The earth's dipole moment is :\$(R_e = 6.4 \times 10^6 m)\$

The equatorial magnetic field is: \$B=\dfrac{\mu_0}{4\pi} \dfrac{m}{r^3} \$Given,\$B = 0.4\:G = 0.4 \times 10^{-4}\: T\$Radius of earth \$r=6.4 \times 10^{6}\: m\$\$ \therefore m = \dfrac{B}{\dfrac{\mu_0}{4\pi}}r^3=\dfrac{0.4 \times 10^{-4}}{10^{-7}}(6.4 \times 10^{6})^3=104.9 \times 10^{21}=1.05 \times 10^{23}\:A\: m^2\$

Q: The vertical component of earth's magnetic field at a place is \$\sqrt{3}\$ times the horizontal component the value of angle of dip at this place is?

As, \$B_V=\sqrt{3}B_H\$Also, \$\tan\delta =\dfrac{B_V}{B_H}=\dfrac{\sqrt{3}B_H}{B_H}=\sqrt{3}\$ or \$\delta =\tan^{-1}(\sqrt{3})=60^o\$\$\therefore\$ Angle of dip, \$\delta =60^o\$.

Q: The earth's magnetic field at a certain place has a horizontal component of \$0.3 G\$ and total strength \$0.5 G\$. Find angle of dip in \$\tan^{-1}\$.

Given that,Total strength of field \$B = 0.5 G\$,Horizontal component of the field \$H = 0.3 G\$,Total strength of field is given by\$B = \sqrt {(H^2+V^2)}\$where, V is the vertical component of the field.\$B^2 = H^2+V^2\$\$V^2 = B^2-H^2 = (0.5)^2-(0.3)^2 = 0.4 G\$Now, angle of dip is\$\tan \delta = \dfrac{V}{H} = \dfrac{0.4}{0.3} = \dfrac{4}{3}\$\$\delta = \tan^{-1} \dfrac{4}{3}\$

Q: At a certain place, the angle of dip \${30}^{o}\$ and the horizontal component of earth's magnetic field is \$0.50\$ oerested.The earth's total magnetic field (in oerested) is

\$B=\cfrac { H }{ \cos { \theta } } \$\$=\cfrac { 0.50 }{ \cos { { 30 }^{ o } } } \$\$=\cfrac { 0.50\times 2 }{ \sqrt { 3 } } \$\$=1/\sqrt { 3 } \$

Question 1

If a magnet is suspended at angle ${30}^{o}$ to the magnetic meridian, the dip needle makes an angle of ${45}^{o}$ with the horizontal. The real dip is :

Accepted Answer

Angle of dip, $\delta ={ 45 }^{ o }$$\therefore \quad \tan { \delta ' } =\cfrac { \tan { \delta  }  }{ \cos { \theta  }  } $$=\cfrac { \tan { 45 }  }{ \cos { { 30 }^{ o } }  } =\cfrac { 1 }{ \sqrt { 3 } /2 } =\cfrac { 2 }{ \sqrt { 3 }  } $$\therefore$ Real dip $ \quad \delta '=\tan ^{ -1 }{ \left( 2/\sqrt { 3 }  \right)  } $

Question 2

At a certain place, horizontal component is $\sqrt 3$ times the vertical component. The angle of dip at this place is:

Accepted Answer

$\tan \delta = \dfrac{B_V}{B_H} $$= \dfrac{B_V}{\sqrt{3}B_V}=\dfrac{1}{ \sqrt{3}}=\tan 30^{\circ}$$ \Rightarrow \delta =30^{\circ}=\dfrac{\pi}{6}$

Question 3

In the magnetic meridian of a certain place, the horizontal component of the earth's magnetic field is $0.26\  G$ and the angle of dip is $60^o$. The magnetic field of the earth at this location is

Accepted Answer

The earth's magnetic field is $B_e$ and its horizontal and vertical components are $H_e$ and $H_v$$\cos \theta =\dfrac{H_e}{B_e}$$ \therefore \cos 60^{\circ} = \dfrac{0.26 \times 10^{-4}}{B_e}\:T$$ \Rightarrow B_e = \dfrac{ 0.26 \times 10^{-4}}{ \dfrac{1}{2}}=0.52 \times 10^{-4}\:T=0.52\:G$

Question 4

If $\theta_1$ and $\theta_2$ be the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip $\theta$ is given by:

Accepted Answer

Note that $\tan \theta = \dfrac{V}{H}$     ...(1)where $\theta$ is true value of dip . Here we assumed $\theta = \propto$Now, angle of dips, $\theta_1$ and $\theta_2$ have different formula$\tan\theta_1=\dfrac{V}{H\times cos \propto}$     ...(2)$\tan \theta_2=\dfrac{V}{H\times sin \propto}$     ...(3)From eqn (1), (2), (3), we can find value of H/Vwhich $\cot \theta$, which when calculate comes option B$\cot^2\theta_1+\cot^2\theta_2=\dfrac{H_2}{V^2}(\cos^2\propto +\sin^2\propto)=\cot^2\theta$$\cot^2\theta=\cot^2\theta_1+\cot^2\theta_2$

Question 5

The earth's magnetic field at the equator is approximately $0.4 G$. The earth's dipole moment is :$(R_e = 6.4 \times 10^6 m)$

Accepted Answer

The equatorial magnetic field is: $B=\dfrac{\mu_0}{4\pi} \dfrac{m}{r^3} $Given,$B = 0.4\:G = 0.4 \times 10^{-4}\: T$Radius of earth $r=6.4 \times 10^{6}\: m$$ \therefore m = \dfrac{B}{\dfrac{\mu_0}{4\pi}}r^3=\dfrac{0.4 \times 10^{-4}}{10^{-7}}(6.4 \times 10^{6})^3=104.9 \times 10^{21}=1.05 \times 10^{23}\:A\: m^2$

Question 6

If the angles of dip at two places are $30^{\circ}$ and $45^{\circ}$ respectively, then the ratio of horizontal components of earth's magnetic field at the two places will be

Accepted Answer

The horizontal component of earth's magnetic field is given by , $H=R\cos{\delta}$where, $\delta$ is the angle of dip so ${ H }_{ 1 }=R\cos { { 30 }^{ o } } $and ${ H }_{ 2 }=R\cos { { 45 }^{ o } } $$\therefore \cfrac { { H }_{ 1 } }{ { H }_{ 2 } } =\cfrac { R\cos { { 30 }^{ o } }  }{ R\cos { { 45 }^{ o } }  } =\cfrac { \sqrt { 3 } /2 }{ 1/\sqrt { 2 }  } =\cfrac { \sqrt { 3 }  }{ \sqrt { 2 }  } $

Question 7

The vertical component of earth's magnetic field at a place is $\sqrt{3}$ times the horizontal component the value of angle of dip at this place is?

Accepted Answer

As, $B_V=\sqrt{3}B_H$Also, $\tan\delta =\dfrac{B_V}{B_H}=\dfrac{\sqrt{3}B_H}{B_H}=\sqrt{3}$ or $\delta =\tan^{-1}(\sqrt{3})=60^o$$\therefore$ Angle of dip, $\delta =60^o$.

Question 8

The earth's magnetic field at a certain place has a horizontal component of $0.3 G$ and total strength $0.5 G$. Find angle of dip in $\tan^{-1}$.

Accepted Answer

Given that,Total strength of field $B = 0.5 G$,Horizontal component of the field $H = 0.3 G$,Total strength of field is given by$B = \sqrt {(H^2+V^2)}$where, V is the vertical component of the field.$B^2 = H^2+V^2$$V^2 = B^2-H^2 = (0.5)^2-(0.3)^2  = 0.4 G$Now, angle of dip is$\tan \delta = \dfrac{V}{H} =  \dfrac{0.4}{0.3} = \dfrac{4}{3}$$\delta = \tan^{-1} \dfrac{4}{3}$

Question 9

A compass needle whose magnetic moment is $60 A m^{2}$ pointing geographical north at a certain place where the horizontal component of earth's magnetic field is $40 \times 1 0^{- 6} W b m^{- 2}$ experiences a torque of $1.2 \times 1 0^{- 3} N m .$ The declination of the place is

Accepted Answer

In stable equilibrium, a compass needle points along the magnetic north and experiences no torque.When it is turned through declination $\alpha$, it points along geographic north and experiences torque, $ T = mB sin \alpha$$\therefore sin \alpha = \dfrac{T}{mB} = \dfrac{1.2 \times 10^{-3}}{60 \times 40 \times 10^{-6}} = \dfrac{1}{2}$  or $ \alpha = 30^0$

Question 10

At a certain place, the angle of dip ${30}^{o}$ and the horizontal component of earth's magnetic field is $0.50$ oerested.The earth's total magnetic field (in oerested) is

Accepted Answer

$B=\cfrac { H }{ \cos { \theta  }  } $$=\cfrac { 0.50 }{ \cos { { 30 }^{ o } }  } $$=\cfrac { 0.50\times 2 }{ \sqrt { 3 }  } $$=1/\sqrt { 3 } $

Basic Introduction to Earths magnetism

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REVISE WITH CONCEPTS

Basic Knowledge of Magnetic Field of Earth

Earth's Magnetism

Earth's Magnet Declination

Terrestrial Magnetism And Terms Related To Geomagnetism

QUICK SUMMARY WITH STORIES

Navigation Using Magnets

Earth as a Magnet

Dip Circle

Declination and Dip