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Magnetism and Matter · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 5 questions.

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Question 15 Marks
Write the properties of diamagnetic, paramagnetic and ferromagnetic materials.
Answer
Properties of Diamagnetic Materials :
(i) When a bar of diamagnetic material is suspended freely between two magnetic poles of a magnet then it turns to become perpendicular to the magnetic field and the poles produced on the two sides of the bar are similar to the nearer magnetic poles.
(ii) When a diamagnetic material is placed in a strong uniform external magnetising field, the field lines prefer to pass through the surrounding air than to pass through the material itself i.e. field lines get repelled or expelled. Consequently, the magnetic induction 'B' inside material becomes less than magnetising field H. Thus permeability for diamagnetic substance is less but never negative.
(iii) In a non-uniform field a diamagnetic substance moves from stronger to the weaker parts of the field. For instance, if a diamagnetic liquid is placed in a watch glass resting on the pole pieces not more than 2 mm apart, the liquid accumulates on the sides where the field is weaker and creates depression in the middle. The reverse is the effect when the poles are far apart because now the field is stronger near the poles.
(iv) If a diamagnetic solution is poured into a U-tube and one arm of this U-tube is placed between the poles of a strong magnet, the level of the solution in that arm is depressed.
(v) A diamagnetic gas when allowed to ascend in between the poles of a magnet spreads across the field.
(vi) The susceptibility of a diamagnetic substance is independent of temperature and has a negative value e.g. for lead it is $-1.7 \times 10^5$.
Properties of Paramagnetic Materials :
(i) When a rod of a paramagnetic materials is suspen-ded freely between two magnetic poles of a magnet then it rotates until its axis becomes parallel to the magnetising field. The poles produced at the ends of the rod are opposite to the nearer magnetic poles of the magnet.
(ii) When a rod of paramagnetic substance is placed in a strong external magnetising field, the field lines prefer to pass through it than through the surrounding air, i.e. these field lines get slightly more concentrated inside the material. Consequently, the magnetic induction B inside the paramagnetic substance becomes slightly greater than the magnetising field H. Thus permeability for paramagnetic substance is greater than 1, but increase is very small, about I part in $10^5$. For Al and platinum the value of $\chi$ is 1.76 x $10^{-6}$ and 2.88 × $10^{6}$ respectively.
(iii) In a non-uniform field they experience an attractive force towards the stronger part of the field. For example, if a paramagnetic liquid be placed in a watch glass resting on the pole pieces not more than 2 mm apart, the liquid accumulates in the middle where the field is strongest. If the pole-pieces are far apart say, 2 cm or so, the field is strongest near the poles, then the liquid moves away from the centre producing a depression in the middle.
(iv) If the solution of paramagnetic substance is poured in a U-tube and one-arm of the U-tube is placed between two strong poles the level of the solution in that arm rises.
(v) When paramagnetic gases (like traces of ammonia and hydrogen chloride) are allowed to ascend between the poles of a magnet they spread along the field.
(vi) The susceptibility $\chi$ for paramagnetic susbtances is positive and mall e.g. for A l it is $1.76 \times 10^{-6}$ and for platnium it is 2.88 × 10-6.
(vii) $\chi$ and $\mu$ for paramagnetic substances do not change with variation in magnetising field H. $\chi$ varies inversely with the absolute temperature. At some high temperature becomes negative and substance become diamagnetic.
(viii) A paramagnetic substance acts like a diamagnetic substance if it is surrounded by a medium which is more paramagnetic than itself.
Properties of Ferromagnetic Materials
(i) The dipole moment of atoms is permanent and exists in domains.
(ii) The alignment of atomic dipoles and external magnetic fields are in the same direction.
(iii) There is a significant strength of magnetic dipole moment.
(iv) The magnetisation intensity varies linearly with the magnetising field. Moreover, the magnetisation intensity is quite high and positive.
(v) Magnetic susceptibility is relatively high and positive.
(vi) The magnetic flux density is also high and positive. The magnetic field lines inside ferromagnetic materials are dense.
(vii) The relative permeability is also relatively high. It varies linearly with the magnetic field. The magnetic field inside the material is substantially more vital than outside the material. They tend to pull in many force lines from the material.
(viii) The field aggressively attracts ferromagnetic materials. They tend to adhere to the poles where the field is more potent in a non-uniform field.
(ix) Because the field is more vital at poles, if the ferromagnetic powder is placed in a watch glass between two properly faraway poles, powder accumulates on the sides, and the centre is depressed.
(x) At high temperatures, a ferromagnetic substance loses its ferromagnetic characteristics.
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Question 25 Marks
Explain end-on position and broad side-on position of a bar magnet.
Answer
(a) End-on-position of a bar magnet : Let NS be a bar magnet of length $2l$ and 'm' be the pole strength each of its magnetic poles. Suppose P is a point along the axis of the bar magnet at a distance r from the middle point O of the magnet at which the magnetic field intensity due to the bar magnet is to be determined.
This position is called a end on position or longitudinal position. The distance of point P from north pole N is equal to $(r-l)$ and from south pole S is equal to $(r+l)$ [Fig.(a)].
Image
The magnitude of the magnetic field intensity vec $\overrightarrow{ B _1}$ due to north pole N is given by:
$B_1=\frac{\mu_0}{4 \pi}\left[\frac{m}{(r-l)^2}\right]$ ...(1)
Direction of $\overrightarrow{ B _1}$ is away from the magnet along the axis.
The magnitude of the magnetic field intensity $\overrightarrow{ B _2}$ due to south pole S is given by :
$B_2=\frac{\mu_0}{4 \pi}\left[\frac{m}{(r+l)^2}\right]$ ...(2)
Direction of $\overrightarrow{ B _2}$ is towards the magnet along the axis.
From equations (1) and (2) it is obvious that $B _1> B _2$ and $\overrightarrow{ B _1}$ and $\overrightarrow{ B _2}$ both are oppositely directed. Hence the magnitude of the resultant magnetic field $\overrightarrow{ B }$ at P is given by:
$B = B _1- B _2$
$=\frac{\mu_0}{4 \pi}\left[\frac{m}{(r-l)^2}\right]-\frac{\mu_0}{4 \pi}\left[\frac{m}{(r+l)^2}\right]$
$=\frac{\mu_0 m}{4 \pi}\left[\frac{1}{(r-l)^2}-\frac{1}{(r+l)^2}\right]$
or $B =\frac{\mu_0 m}{4 \pi}\left[\frac{(r+l)^2-(r-l)^2}{(r-l)^2(r+l)^2}\right]$
$=\frac{\mu_0 m}{4 \pi}\left[\frac{2 \times 2 r l}{\{(r-l)(r+l)\}^2}\right]$
or $B =\frac{\mu_0}{4 \pi}\left[\frac{2 \times(m \times 2 l) \times r}{\left(r^2-l^2\right)^2}\right]$
But $m \times 2l$ = Magnetic dipole moment of the bar magnet = M
$\therefore \quad B =\frac{\mu_0}{4 \pi}\left[\frac{2 Mr }{\left(r^2-l^2\right)^2}\right]$ ...(3)
If distance r >> $l$ the magnet is called short bar magnet and in this case in equation (3) $l^2$ can be neglected in comparison to r². Hence magnetic field intensity for a short bar magnet is given by :
$B=\frac{\mu_0}{4 \pi}\left[\frac{2 Mr}{\left(r^2\right)^2}\right] \Rightarrow B =\frac{ \mu _0}{ 4 \pi}\left[\frac{ 2 M }{ r ^3}\right] \ldots$(4)
(b) Broad side on position : In this position the point of observation lies on the perpendicular bisector of the bar magnet. This position is called as broad side on position or transverse position.
Let in Fig. (b), P be a point in the broad side on position of the bar magnet NS of length $2l$ at a distance $r$ from its centre O and let $m$ be the pole strength of each of its magnetic poles. From Fig. 8.10 it is evident that the distance of the point P from each of the poles N and S is equal and is given as follows :
$NP = SP =\left(r^2+ l ^2\right)^{1 / 2}$
Image
The magnitude of magnetic field intensity $\overrightarrow{ B _1}$ at P due to north pole N is given by :
$B _1=\frac{\mu_0}{4 \pi}\left(\frac{m}{ NP ^2}\right)$
$\Rightarrow \quad B _1=\frac{\mu_0}{4 \pi}\left[\frac{m}{\left(\sqrt{\left.r^2+l^2\right)^2}\right.}\right]$
$\Rightarrow \quad B_1=\frac{\mu_0}{4 \pi}\left[\frac{ m }{\left( r ^2+ l ^2\right)}\right]$... (1)
The direction of $\overrightarrow{ B _1}$ at P is away from the north pole N.
The magnitude of magnetic field intensity $\overrightarrow{ B _2}$ at P due to south pole S is given by :
$B _2=\frac{\mu_0}{4 \pi}\left(\frac{m}{ SP ^2}\right)$
$\Rightarrow \quad B _1=\frac{\mu_0}{4 \pi}\left[\frac{m}{\left(\sqrt{\left(r^2+l^2\right)}\right)^2}\right]$
$\Rightarrow \quad B _2=\frac{\mu_0}{4 \pi}\left[\frac{m}{\left(r^2+l^2\right)}\right]$ ...(2)
The direction of vec $\overrightarrow{ B _2}$ at P is towards the south pole S. From equations (1) and (2) it is obvious that :
$B _1= B _2=\frac{\mu_0}{4 \pi}\left[\frac{m}{\left(r^2+l^2\right)}\right]$
In order to find out the resultant magnetic field at P due to the bar magnet; we have to resolve $\overrightarrow{ B _1}$ and $\overrightarrow{ B _2}$ into two rectangular components: one along equitorial line and the other along axial line.
If $\angle PSO =\theta$, then $\angle SPX ^{\prime}=\theta$ and $\angle ONP =\angle X ^{\prime} PA ^{\prime}$ $=\theta$
The components of vec $\overrightarrow{ B _1}$ and $\overrightarrow{ B _2}$ along equatorial line are $B _1 \sin \theta$ and $B _2 \sin \theta$ respectively. As $B _1= B _2$, hence these components will be equal in magnitude, but are opposite in direction. Therefore these will cancel each other. The components of $\overrightarrow{ B _1}$ and $\overrightarrow{ B _2}$ along axial line will be $B _1 \cos \theta$ and $B _2 \cos \theta$ respectively and these will be added up; because these are in the same direction.
Therefore the resultant magnetic field at P due to bar magnet is given by :
$B = B _1 \cos \theta+ B _2 \cos \theta$
[Along $PX ^{\prime}$ ]
$\Rightarrow \quad B =\frac{\mu_0}{4 \pi}\left[\left(\frac{m}{\left(r^2+l^2\right)}\right) \cos \theta\right]$ $+\frac{\mu_0}{4 \pi}\left[\left(\frac{m}{r^2+l^2}\right) \cos \theta\right]$
or $B =\frac{\mu_0}{4 \pi}\left(\frac{2 m}{r^2+l^2}\right) \cos \theta$
But from right angled triangle $\Delta SOP$, we have
$\cos \theta=\frac{ SO }{ SP }=\frac{l}{\left(r^2+l^2\right)^{1 / 2}}$
$\therefore \quad B =\frac{\mu_0}{4 \pi}\left[\frac{2 m}{\left(r^2+l^2\right)}\right] \times \frac{l}{\left(r^2+l^2\right)^{1 / 2}}$
$\Rightarrow \quad B =\frac{\mu_0}{4 \pi}\left[\frac{2 m l}{\left(r^2+l^2\right)^{3 / 2}}\right]$
$\Rightarrow \quad B =\frac{\mu_0}{4 \pi}\left[\frac{m \times 2 l}{\left(r^2+l^2\right)^{3 / 2}}\right]$
But $m \times 2 l=$ Magnetic dipole moment of the bar magnet = M.
$\therefore \quad B =\frac{\mu_0}{4 \pi}\left[\frac{ M }{\left(r^2+I^2\right)^{3 / 2}}\right]$ ...(3)
Direction of $\overrightarrow{ B }$ is along $PX ^{\prime}$ i.e. from north pole towards south pole parallel to the magnetic axis. Hence its direction is opposite to that of the magnetic dipole moment of the magnet which is directed from south pole towards north pole.
Hence equation in its vector form can be written as follows :
$\vec{B}=-\frac{\mu_0}{4 \pi}\left[\frac{\vec{M}}{\left(r^2+l^2\right)^{3 / 2}}\right]$ ... (4)
For a short bar magnet i.e. a bar magnet of very small length we have $l \ll r$, hence in equations (3) and (4) $l^2$ can be neglected in comparison to $r^2$. Therefore these equations can be written in the following form respectively.
$B=\frac{\mu_0}{4 \pi}\left(\frac{M}{r^3}\right)$ ... (5)
$\overrightarrow{B}=-\frac{\mu_0}{4 \pi}\left(\frac{\overrightarrow{M}}{r^3}\right)$ ... (6)
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Question 35 Marks
Define magnetic permeability and magnetic susceptibility. Prove that $\mu_{r} = (1+\chi)$.
Answer
Magnetic Permeability : It is the ability of a material to permit the passage of magnetic field lines through it. Hence it is the degree to which the magnetic field can penetrate the material.
For example, when a piece of iron is placed in a magnetising field, lines of force become much more concentrated in iron (where they become lines of induction) than in air. Thus iron is said to have greater permeability than air. The permeability of a medium may be defined as ratio of the lines of induction per unit normal area to the number of lines of force of the magnetising field per unit normal area. The former measures the magnetic induction B in the medium and the latter measures the strength of the magnetising field H. Hence permeability µ is defined as the ratio of the magnetising induction (B) in the specimen to the magnetising field (H). Thus $\mu=\frac{B}{H}$.
The ratio of magnetic permeability of the material $\mu$ and magnetic permeability of free space $\left(\mu_0\right)$ is called relative permeability of the substance denoted by $\mu_r$ i.e. $\mu_r=\mu / \mu_0$.
where $\mu_0 \times 4 \pi \times 10^{-7} Wb / Am$
Magnetic Susceptibility : For a large number of magnetic materials the intensity of magnetisation I is proportional to magnetising field strength, H. i.e.
$I \propto H$ or $I=\chi . H$
Where $\chi$ is a proportionality constant and is known as magnetic susceptibility of magnetic material.
Now $I =\chi H \Rightarrow \chi= I / H$
Hence the magnetisation per unit field strength is called the magnetic susceptibility. Since I and H have the same dimensions the quantity $\chi$ is dimensionless i.e. it has no unit.
The magnetic susceptibility of a specimen of a magnetic material measures the ease with which the specimen can be magnetised. Since I is equal to the magnetic moment per unit volume, hence the susceptibility as defined above is also called volume susceptibility of the material.
Derivation of $\mu _r=(1+\chi)$ : When a magnetic material is placed in a magnetising field of magnetic intensity $\overrightarrow{ H }$, the material gets magnetised. In this state the total magnetic induction $\overrightarrow{ B }$ is the sum of the magnetic induction $\overrightarrow{ B }_0$ in vaccum produced by the magnetic intensity $\overrightarrow{ H }$ and magnetic induction $\overrightarrow{ B }_m$ due to magnetisation of the material i.e. intensity of magnetisation.
Hence $\overrightarrow{ B }=\overrightarrow{ B }_0+\overrightarrow{ B }_m$
but $\overrightarrow{ B }=\mu \overrightarrow{ H }, B _0=\mu_0 \overrightarrow{ H }$ and $\overrightarrow{ B }_m=\mu_0 \overrightarrow{ I }$
$\mu H =\mu_0( H + I )$
$\Rightarrow \quad \mu H =\mu_0 H \left(1+\frac{ I }{ H }\right)$
$\Rightarrow \quad \frac{\mu}{\mu_0}=\left(1+\frac{ I }{ H }\right)$. But $\mu / \mu_0=\mu_r$ and $I / H =\chi$
$\therefore \quad \mu_r=(1+\chi)$
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Question 45 Marks
Define diamagnetic, paramagnetic and ferromagnetic substance with examples. Explaining the meaning of domain explain ferromagnetism.
Answer
Diamagnetic Substances : Diamagnetic sub-stances are those which acquire feeble magnetisation in a direction opposite to that of magnetising field i.e. for these materials $\overrightarrow{ I }$ is always opposite to $\overrightarrow{ H }$. Therefore when these substances are brought close to a pole of a powerful magnet they are somewhat repelled away from the material. Due to this fact the susceptibility $(\chi)$ of these substances is negative.
Examples : Bismuth, antimony, gold, water, alcohol, quartz, copper, lead, zinc, tin, silicon, nitrogen, sodium-chloride etc.
Paramagnetic Substances : Paramagnetic sub-stances are those which, when placed in a magnetic field, acquire a feeble magnetisation in the same sense as the applied field. i.e. for these substances $\overrightarrow{ I }$ is always in the direction of $\overrightarrow{ H }$. Therefore, when these substances are brought close to any pole of a powerful magnet, they are attracted towards the magnet. Due to this fact the susceptibility of these substances is positive.
Examples : Platinum, aluminium chromium, manganese copper sulphate, solution of salts of iron and nickel, oxygen, crow glass, radium, copper chloride etc.
Ferromagnetic Substances : Those substances which when placed in a magnetic field are strongly magnetised in the direction of magnetising field are called ferromagnetic substances. For such substances $\overrightarrow{ I }$ is always in the direction of $\overrightarrow{ H}$ and $\overrightarrow{ I }$ is much greater than $\overrightarrow{H }$. Therefore when these substances are brought close to any pole of a magnet, they are strongly attracted by the magnet. Due to this fact the susceptibility \[(\chi\] = I/H) of ferromagnetic substances is positive and is very much greater than 1.
Examples : Iron, Cobalt, Nickel, Magnetite (Fe3O4) alloys like alnico etc.
Explanation of Ferromagnetism : Like param-agnetic substance the atoms of ferromagnetic material also passess net magnetic moment and behave as a tiny magnetic dipole. The difference in paramagnetism and ferromagnetism is only that of intensify of magnetisation due to large magnetic moment in ferromagnetism substances. The atomic dipoles of ferromagnetic substance tend to align parallel to each other even in a weak magnetising field. The atomic dipoles are not independent of each other, in ferromagnetic substances but due to quantum mechanical mutual interaction the atoms in these substances form large number of small regions of size 10-4 cm to 1 cm containing 1017 to 1021 atomic dipoles. These regions are called domains. The atomic dipoles of one domain are aligned in the same direction but in different direction from the neighboring domain [Fig. (a)]. In this way each domain in the absence of any external magnetising field has net magnetic moment in a particular direction, but due to random orientation of different domains the resultant magnetic moment in the substance is zero. It is due to this fact that every piece of iron is not a magnet.
Image
In Fig. (b) the probable directions of magnetic moment of four domains of a ferromagnetic substance are shown. When the ferromagnetic substance is placed in an external magnetising field the ferromagnetism can be increased in the following two ways :
Image
(i) By the displacement of the boundaries of the domains : When the external magnetising field is weak then the ferromagnetic substance is magnetised by this mechanism. In this mechanism the boundaries of the domains containing atomic magnetic dipoles oriented in the direction of external field increase in size, whereas those oriented in opposite direction to the field are reduced in size [Fig. (c)]. Here the ferromagnetic substance is magnetised strongly in the direction of external magn-etising field.
(ii) By the rotation of the domains : When the external magnetising field is strong then the ferromagnetic substance is magnetised by this mechanism. In this mechanism under the effect of strong magnetising field all the domains rotate until their magnetic momenta are aligned in the direction of external magnetic field [Fig. (d)]. Here the ferromagnetic substance is magnetised strongly in the direction of external magnetising field.
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Question 55 Marks
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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5 Marks Questions - Physics STD 12 Science Questions - Vidyadip