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Magnetism and Matter — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsMagnetism and Matter5 Marks
Question
Define magnetic elements of earth's magnetic field. Establish the relation between them.
On suspending a magnet at an angle of 30° with magnetic meridian it makes an angle of 45° with horizontal. Calculate the correct value of angle of dip.
✓

Answer

Magnetic Elements of Earth's Magnetic Field : (i) Angle Declination, (ii) Angle of dip, (iii) Horizontal component of Earth's Magnetic field.
Horizontal Component of earth's magnetic field : The knowledge of declination and inclination at a place on the earth's surface completely specifies the direction of the earth's magnetic field. It is along the axis of the needle suspended freely through its centre of gravity lying in magnetic meridian from south pole towards north pole and is denoted by $\overrightarrow{ B _e}$. It can be resolved into two mutually perpendicular components. The component of earth's magnetic field in the horizontal direction in the magnetic meridian is called horizontal component of earth's magnetic field which is directed towards the magnetic south. It is denoted by $\overrightarrow{ B _{ H }}$.
In adjoining figure all the three elements of earth's magnetic field are shown. Starting from the geographical meridian OLMR the magnetic meridian OPOR is drawn at an angle equal to angle of declination. In the magnetic meridian horizontal direction is drawn-specifying magnetic north. The intensity of magnetic field $\overrightarrow{ B _e}$ is at an angle equal to angle of dip $\theta$ from this direction.
Image
Hence the horizontal component $B _{ H }$ may be expressed as follows :
$B _{ H }= B _{ e } \cdot \cos \theta$
Relation between the elements: From the knowledge of $\alpha$ and $\theta$ the direction of earth's magnetic field $\overrightarrow{ B _e}$ can be specified and from the knowledge of BH and $\theta$ the magnitude of the earth's magnetic field $\left|\overrightarrow{ B }_e\right|=$ $B _e$, can be calculated by the use of above equation.
Hence from the knowledge of all the three elements $\alpha,$ Q and H both the magnitude and direction of the earth's magnetic field can be determined i.e. complete information about earth's magnetic field at any place can be obtained due to which these three quantities are called fundamental quantities of earth's magnetic field or elements of the earth's magnetic field.
Magnitude and direction of $\overrightarrow{ B _e}$ can also be calculated as follows :
From the figure given above :
Horizontal component of earth's magnetic field :
$B_{H}=B_e \cos \theta$ ... (1)
and Vertical component of earth's magnetic field :
$B _{ V }= B _e \sin \theta$ ...(2)
On squaring equations (1) and (2) and then adding them, we get
$B _{ H }^2+ B _{ V }^2= B _e^2 \cos ^2 \theta+ B _e^2 \cos ^2 \theta$
$= B _e^2\left(\cos ^2 \theta+\sin ^2 \theta\right)$
But $\cos ^2 \theta+\sin ^2 \theta=1$
Hence $B _{ H }^2+ B _{ V }^2= B _e^2$
$\Rightarrow \quad B _e=\sqrt{ B _{ H }^2+ B _{ V }^2}$ ...(3)
On dividing equation (2) by (1), we get
$\frac{ B _{ V }}{ B _{ H }}=\tan \theta$ ...(4)
Solution of Numerical Question : If $B _{ H }$ and $B _{ V }$ be the be the horizontal and vertical components earth's magnetic field in the magnetic meridian then the true value of angle of dip $\theta$ is given by :
$\tan \theta=\frac{B_V}{B_H}$ ...(1)
Image
But the adjoining figure lies in a vertical plane at 30° from the magnetic meridian, the horizontal component is given by: BH' = BH cos 30° while the vertical con component is still BV. Hence the apparent value of the angle of dip $\theta^{\prime}$ in this plane is given by :
$\tan \theta^{\prime}=\frac{ B _{ V }}{ B _{ H }{ }^{\prime}}$, where $\theta^{\prime}$ is given to be equal to $45^{\circ}$
$\therefore \quad \tan 45^{\circ}=\frac{ B _{ V }}{ B _{ H } \cos 30^{\circ}}$ ...(2)
On dividing equation (1) by equation (2), we get
$\frac{\tan \theta}{\tan 45^{\circ}}=\cos 30^{\circ}$
$\Rightarrow \quad \tan \theta=\tan 45^{\circ} \times \cos 30^{\circ}$
or $\tan \theta=1 \times 0.8660=0.8660$
$\Rightarrow \quad \theta=\tan ^{-1}(0.8660)$

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