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 } $

Declination and Dip

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REVISE WITH CONCEPTS

Earth's Magnetism

Earth's Magnet Declination

Angle of Declination And Angle of Dip

QUICK SUMMARY WITH STORIES

Earth as a Magnet

Dip Circle