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Physics 3 Marks Question

Magnetism And Matter · Uttarakhand Board (UBSE) STD 12 Science (Uttarakhand - English Medium). 40 questions.

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Question 13 Marks
A sample of paramagnetic salt contains $2.0 \times 10^{24}$ atomic dipoles each of dipole moment $1.5 \times 10^{-23} \mathrm{~J} \mathrm{~T}^{-1}$. The sample is placed under a homogeneous magnetic field of 0.64 T , and cooled to a temperature of 4.2 K . The degree of magnetic saturation achieved is equal to $15 \%$. What is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K ? (Assume Curie's law)
Answer
Number of atomic dipoles, $\mathrm{n}=2.0 \times 10^{24}$
Dipole moment of each atomic dipole, $\mathrm{M}=1.5 \times 10^{-23} \mathrm{~J} \mathrm{~T}^{-1}$
When the magnetic field, $\mathrm{B}_1=0.64 \mathrm{~T}$
The sample is cooled to a temperature, $T_1=4.2^{\circ} \mathrm{K}$
Total dipole moment of the atomic dipole, $\mathrm{M}_{\text {tot }}=\mathrm{n} \times \mathrm{M}$
$ =2 \times 10^{24} \times 1.5 \times 10^{-23} $
$=30 \mathrm{JT}^{-1}$
Magnetic saturation is achieved at 15%.
Hence, effective dipole moment, $\text{M}_1=\frac{15}{100}\times30=4.5\text{ J T}^{-1}$
When the magnetic field, $B_2= 0.98\ T$
Temperature, $T_2= 2.8°\ K$
Its total dipole moment $= M_2$
According to Curie’s law, we have the ratio of two magnetic dipoles as:
$\frac{\text{M}_2}{\text{M}_1}=\frac{\text{B}_2}{\text{B}_1}\times\frac{\text{T}_1}{\text{T}_2}$
$\therefore\ \text{M}_2=\frac{\text{B}_2\text{T}_1\text{M}_1}{\text{B}_1\text{T}_2}$
$=\frac{0.98\times4.2\times4.5}{2.8\times0.64}=10.336\text{ J T}^{-1}$
Therefore, $10.336\text{ J T}^{-1}$ is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K.
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Question 23 Marks
A short bar magnet of magnetic moment $5.25 \times 10^{–2}J\ T^{–1}$ is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45° with earth’s field on (a) its normal bisector and (b) its axis. Magnitude of the earth’s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.
Answer
Magnetic moment of the bar magnet, $M = 5.25 \times 10^{-2} J\ T^{-1}$
Magnitude of earth's magnetic field at a place, $H = 0.42 G = 0.42 \times 10^{-4}\ T$
(a) The magnetic field at a distance R from the centre of the rnagnet on the normal bisector· is given by the relation:
$\text{B}=\frac{\mu_0\text{M}}{4\pi\text{R}^2}$
Where,
$\mu_0$ = Permeability of free space $=4\pi\times10^{-7}\ \text{Tm A}^{-1}$
When the resultant field is inclined at 45° with earth's field, B = H
$\therefore\ \frac{\mu_0\text{M}}{4\pi\text{R}^3}=\text{H}=0.42\times10^{-4}$
$\text{R}^3=\frac{\mu_0\text{M}}{0.42\times10^{-4}\times4\pi}$
$=\frac{4\pi\times10^{-7}\times5.25\times10^{-2}}{4\pi\times0.42\times10^{-4}}=12.5\times10^{-5}$
$\therefore$ R = 0.05 m = 5 cm
(b) The magnetic field at a distanced R' from the centre of the magnet on its axis is given as:
$\text{B}'=\frac{\mu_02\text{M}}{4\pi\text{R}^3}$
The resultant field is inclined at 45° with earth's field.
$\therefore\ \text{B}'=\text{H}$
$\frac{\mu_02\text{M}}{4\pi(\text{R}')^3}=\text{H}$
$\text{(R}')^3=\frac{\mu_02\text{M}}{4\pi\times\text{H}}$
$=\frac{4\pi\times10^{-7}\times2\times5.25\times10^{-2}}{4\pi\times0.42\times10^{-4}}$
$\therefore\ \text{R}'=0.063\text{ m}=6.3\text{ cm}$
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Question 33 Marks
A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth’s magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null–point (i.e., 14 cm) from the centre of the magnet? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth’s magnetic field.)
Answer
Earth's magnetic field at the given place, H = 0.36 G
The magnetic field at a distanced, on the axis of the magnet is given as:
$\text{B}_1=\frac{\mu_0}{4\pi}\frac{2\text{M}}{\text{d}^3}=\text{H}\ \dots(\text{i})$
Where,
$\mu_0$ = Permeability of free space
M = Magnetic moment
The Magnetic field at the same distanced, on the equatorial line of the maqnet is given as:
$\text{B}_=\frac{\mu_0\text{M}}{4\pi\text{d}^3}=\frac{\text{H}}{2}\ \ [\text{Using equation (i)}]$
Total magnetic field, $B = B_1 + B_2$
$=\text{H}+\frac{\text{H}}{2}$
= 0.36 + 0.18 = 054 G
Hence, the magnetic field is 0.54 G in the direction of earth's rnaqnetlc field.
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Question 43 Marks
A bar magnet of magnetic moment $1.5\ J\ T^{–1}$ lies aligned with the direction of a uniform magnetic field of 0.22 T.
  1. What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction?
  2. What is the torque on the magnet in cases (i) and (ii)?
Answer
Given,
Magnetic moment of bar magnet, $M = 1.5\ JT^{-1}$
Uniform magnetic field, B = 0.22 T
  1. Work done on magnet
  1. Amount of work required to turn the magnet normal to the field direction is,
$\text{W}_{1}=-\text{MB}\big[\cos90^\circ-\cos0^\circ\big]$

= + MB

= MB = 1.5 × 0.22

= 0.33 N-m
  1. Amount of work required to turn the magnet in a direction opposite to the field direction is,
$\text{W}_{2}=-\text{MB}\big[\cos180^\circ-\cos0^\circ\big]$

= 2 MB

$=-\text{MB}\big[\cos\theta_{2}-\cos\theta_{1}\big]$

= 2 MB = 2 × 0.33

= 0.66 J
  1. Torgue acting on the magnet.
  1. $\tau=\text{MB}\sin 90^\circ$
= MB

= 0.33J.

It works in the direction that tends to align the magnetic moment vector along B.
  1. Torque acting on the magnet so as to align it's magnetic moment in a direction opposite to the field direction.
$\tau=\text{MB }\sin\theta$

Here, $\theta=180^\circ$

$\tau=1.5\times0.22\times\sin180^\circ$

= 1.5 × 0.22 × 0

= 0.
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Question 53 Marks
A circular coil of 16 turns and radius 10 cm carrying a current of 0.75 A rests with its plane normal to an external field of magnitude $5.0 \times 10^{–2} T$. The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of $2.0\ s^{–1}$. What is the moment of inertia of the coil about its axis of rotation?
Answer
Given,
Number of turns in the coil, N = 16
Radius of the coil, r = 10 cm = 0.1 m
Current passing through the coil, I = 0.75 A
External magnetic field, $B = 5. 0 x 10^{-2}T$
Frequency of oscillation, $\nu=2.0\text{ s}^{-1}$
Magnetic moment is given by, M= NIA
Therefore,
$\text{M}=\text{NI}\pi\text{r}^2$
$=16\times0.75\times\frac{22}{7}\big(0.1\big)^2$
$= 0377\ JT^{-1}$
Now, using formula $\nu=\frac{1}{2\pi}\sqrt{\frac{\text{MB}}{\text{I}}}$
$\therefore\ \text{V}^2=\frac{\text{MB}}{4\pi^2\text{I}}$
$\text{I}=\frac{\text{MB}}{4\pi^2\nu^2}$
Putting values, we get
$\text{I}=\frac{0.377\times5.0\times10^{-2}}{4\times\Big(\frac{22}{7}\Big)^2\times2^2}$
$= 1.2 \times 10^{-4}kg\ m^2$
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Question 63 Marks
A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 45° with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east.
  1. Determine the horizontal component of the earth’s magnetic field at the location.
  2. The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of 90° in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.
Answer
Number of turns in the circular coil, N = 30
Radius of the circular coil, r = 12 cm = 0.12 m
Current in the coil, I = 0.35 A
Angle of dip, $\delta=45^\circ$
  1. The magnetic field due to current I, at a distance r, is given as:
$\text{B}=\frac{\mu_02\pi\text{NI}}{4\pi\text{r}}$

Where,

$\mu_0$ = Permeability of free space = $4\pi\times10^{-7}\text{ TM A}^{-1}$

$\therefore\ \text{B}=\frac{4\pi\times10^{-7}\times2\pi\times30\times0.35}{4\pi\times0.12}$

$=5.49\times10^{-5}\text{ T}$

The compass needle points from west to East. Hence, the horizontal component of earth's rnagnetic field is given as:

$\text{B}_\text{H}=\text{B}\sin\delta$

$=5.49\times10^{-5}\sin45^\circ$ $=3.88\times10^{-5}\text{ T}=0.388\text{ G}$
  1. When the current in the coil is reversed and the coil is rotated about its vertical axis by an angle of 90°, the needle will reverse its original direction. In this case, the needle will point from East to West.
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Question 73 Marks
A short bar magnet has a magnetic moment of $0.48\ J\ T^{–1}$. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on.(a) the axis, (b) the equatorial lines (normal bisector) of the magnet.
Answer
Magnetic moment of the bar magnet, $M = 0.48\ J\ T^{-1}$
(a) Distance, d = 10 cm = 0.1 m
The rnagnetic field at distanced, from the centre of the magnet on the axis is given by the relation:
$\text{B}=\frac{\mu_0}{4\pi}\frac{2\text{M}}{\text{d}^3}$
Where,
$\mu_0=$ Permeability of free space $=4\pi\times10^{-7}\text{ TM A}^{-1}$
$\therefore\ \text{B}=\frac{4\pi\times10^{-7}\times2\times0.48}{4\pi\times(0.1)^3}$
$=0.96\times10^{-1}\text{ T}=0.96\text{ G}$
The rnagnetic field is along the s - N direction.
(b) The magnetic field at a distance of 10 cm (i.e., d = 0.1 m) on the equatorial line of the magnet is given as:
$\text{B}=\frac{\mu_0\times\text{M}}{4\pi\times\text{d}^3}$
$=\frac{4\pi\times10^{-7}\times0.48}{4\pi(0.1)^3}$
= 0.48 G
The magnetic field is along the N - s direction.
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Question 83 Marks
A closely wound solenoid of 2000 turns and area of cross-section $1.6 \times 10^{–4} m2,$ carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane.
  1. What is the magnetic moment associated with the solenoid?
  2. What is the force and torque on the solenoid if a uniform horizontal magnetic field of $7.5 \times 10^{–2}\ T$ is set up at an angle of 30º with the axis of the solenoid?
Answer
  1. Given,
Number of turns of coil. N = 2000

Area of cross-section of solenoid, $A = 1.6 x 10^{-4} m^2$

Current passing through the coil. I = 4 A

Magnetic moment. M = NIA

$= 2000 \times 4 \times 1.6 \times 10^{-4}$

$= 1.28 Am^2$​​​​​​​

The direction of $\overrightarrow{\text{M}}$ is along the axis of the solenoid in the direction related to the sense of current via the right handed screw rule.
  1. Uniform magnetic field applies, $B = 7.5 \times 10^{-2}\ T$
Angle between the axis of the solenoid and magnetic field, $\theta=30^\circ$

Since the magnetic field is uniform on the solenoid, force acting on the solenoid is 0.

Torque is given by,

$\tau=\text{MB }\sin\theta$

$=1.28\times7.5\times10^{-2}\sin 30^\circ$

$=1.28\times7.5\times10^{-2}\times\frac{1}{2}\text{ J}$

$=0.048\text{ J}.$

The direction of the torque is such that the solenoid tends to align the axis of the solenoid (magnetic moment vector) along the direction of magnetic field $\overrightarrow{\text{B}}.$
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Question 93 Marks
If the bar magnet in exercise 5.13 is turned around by 180°, where will the new null points be located?
Answer
The rnaqnettc field on the axis of the rnagnet at a distance $d_1 = 14\ cm,$ can be written as:
$\text{B}_1=\frac{\mu_02\text{M}}{4\pi(\text{d}_1)^3}=\text{H}\ \dots(1)$
Where,
M = Magnetic moment
$\mu_0$ = Permeability of free space
H = Horizontal component of the magnetic field at $d_1$
If the bar magnet is turned through 180°, then the neutral point will lie on the equatorial line.
Hence, the magnetic field at a distance dz, on the equatorial line of the magnet can be Written as:
$\text{B}_2=\frac{\mu_0\text{M}}{4\pi(\text{d}_2)^3}=\text{H}\ \dots(2)$
Equating equations (1) and (2), we get:
$\frac{2}{(\text{d}_1)^3}=\frac{1}{(\text{d}_2)^3}$
$\Big(\frac{\text{d}_2}{\text{d}_1}\Big)^3=\frac{1}{2}$
$\therefore\ \text{d}_2=\text{d}_1\times\Big(\frac{1}{2}\Big)^{\frac{1}{3}}$
= 14 × 0794 = 11.1 cm
The new null points will be located 11.1 cm on the normal bisector.
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Question 103 Marks
Deduce the expression for the torque $\overrightarrow{\tau}$ acting on a planar loop of area $\overrightarrow{\text{A}}$ and carrying current I placed in a uniform magnetic field $\overrightarrow{\text{B}}.$ If the loop is free to rotate, what would be its orientation in stable equilibrium?
Answer


Force on each perpendicular arm:
$?_1 = ?_2 = ? ? ?$
Moment of couple $= ? ? ?. ? \sin ?$
$? = ? ?? ? ????$
$? = I ?? ????  \overrightarrow{\tau} =\overrightarrow{\text{IA}}\times\overrightarrow{\text{B}}$
When the plane of the loop is perpendicular to the magnetic field, the loop will be in stable equilibrium.$(\overrightarrow{\text{A}}||\overrightarrow{\text{B}}) , \Rightarrow\theta = 0^{\circ}$
$\overrightarrow{\text{M}} = $ Equivalent magnetic moment of the planer loop $ = \overrightarrow{\text{IA}}$
$\therefore \text{Torque} =\overrightarrow{\text{M}}\times\overrightarrow{\text{B}} = \overrightarrow{\text{IA}}\times\overrightarrow{\text{B}}$
$\text{|??????|} = \text{IAB}\sin\theta$.
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Question 113 Marks
State the principle of working of a galvanometer.
A galvanometer of resistance G is converted into a voltmeter to measure upto V volts by connecting a resistance R1 in series with the coil. If a resistance R2 is connected in series with it, then it can measure upto V/2 volts. Find the resistance, in terms of R1 and R2, required to be connected to convert it into a voltmeter that can read upto 2 V. Also find the resistance G of the galvanometer in terms of R1 and R2.
Answer
Working principle: A current carrying coil experiences a torque when placed in a magnetic field which tends to rotate the coil and produces an angular deflection.
$V = I (G + R_1)$
$\frac{\text{V}}{2}= \text{I}(\text{G} +\text{R}_{2})$
$= > 2= \frac{\text{G} +\text{R}_{1}}{\text{G} +\text{R}_{2}}$
$ = > \text{G} = \text{R}_{1} - 2\text{R}_{2}$
Let $R_3$ be the resistance required for conversion into voltmeter of range 2V.
$\therefore 2V = I_g (G + R_3)$ also, $V = I_g (G + R_1)$
$\therefore2 =\frac{\text{G} + \text{R}_{3}}{\text{G} + \text{R}_{1}}$
$\therefore R_3 = G + 2R_1 = R_1– 2R_2 + 2R_1= 3R_1– 2R_2.$
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Question 123 Marks
Define the term magnetic inclination. Deduce the relation connecting the horizontal component and inclination with the help of a diagram.
Answer
Magnetic Inclination: It is the angle made by resultant magnetic field of earth with the horizontal. It is also called angle of dip. Relation: suppose $\overrightarrow{\text{Be}}$ is earth’s net magnetic field, $\theta$ is angle of dip. Resolving $\overrightarrow{\text{Be}}$ along horizontal and vertical directions; the horizontal component is H and vertical component is V. From fig. $\therefore\text{ H=Be}\cos\theta$ This is the required relation.
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Question 133 Marks
Consider a non-conducting ring of radius r and mass m that has a total charge q distributed uniformly on it. The ring is rotated about its axis with an angular speed $\omega.$
  1. Find the equivalent electric current in the ring.
  2. Find the magnetic moment $\mu$ of the ring.
  3. Show that $\mu=\frac{\text{q}}{2\text{m}}\text{l}$ where l is the angular momentum of the ring about its axis of rotation.
Answer
  1. $\text{i}=\frac{\text{q}}{\text{t}}=\frac{\text{q}}{\text{t}}=\frac{\text{q}}{\Big(\frac{2\pi}{\omega}\Big)}=\frac{\text{q}\omega}{2\pi}$
  2. $\mu=\text{n}\text{ ia}=\text{i A}[\because\text{n}=1]$
$=\frac{\text{q}\omega\pi\text{r}^2}{2\pi}$
  1. $\mu=\frac{\text{q}\omega\text{r}^2}{2}$
$\text{L}=\text{I}\omega=\text{mr}^2\omega,\ \frac{\mu}{\text{L}}$
$=\frac{\text{q}\omega\text{r}^2}{2\text{mr}^2\omega}=\frac{\text{q}}{2\text{m}}$
$\Rightarrow\mu=\Big(\frac{\text{q}}{2\text{m}}\Big)\text{L}$
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Question 143 Marks
Define magnetic susceptibility of a material. Name two elements, one having positive susceptibility and the other having negative susceptibility. What does negative susceptibility signify
Answer
Magnetic susceptibility of a material is defined as the ratio of the intensity of magnetisation (i) induced in the material to the magnetisation force (H) applied on it. Magnetic susceptibility is represented as: Substances such as copper, lead, etc. have negative susceptibility. These substances are called diamagnetic substances. Substances such as aluminium, calcium, etc. have positive susceptibility. These substances are called paramagnetic substances. Substances with negative susceptibility signify that they are repelled by magnets.
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Question 153 Marks
The horizontal component of earth’s magnetic field at a given place is $0.4 \times 10^{-4} Wb/m^2$ and angle of dip is 30°. Calculate the value of (i) Vertical component (ii) Total intensity of earth’s magnetic field.
Answer
  1. $=0.4\times10^-4\text{Wb/m}^2,\theta=30^\circ$
$\tan\theta=\frac{\text{V}}{\text{H}}$

$\Rightarrow\text{vertical component}\text{V}=\text{H}\tan\theta$

$=0.4\times10^{-4}\times\tan^{30^\circ}$

$=\frac{0.4\times10^{-4}}{\sqrt{3}}=0.23\times10^{-4}\text{Wb/m} ^2$
  1. Total intensity of earth's magnetic field
$\text{Be}=\sqrt{\text{H}^2+\text{V}^2}$

$=\sqrt{(0.4\times1^{-4})^2+\big(\frac{04.\times10^{-4}}{\sqrt{3}}\big)^2}$

$=.46\times1^{-4}\text{Wb/m}^2$
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Question 163 Marks
What are permanent magnets? What is an efficient way of preparing a permanent magnet? Write two characteristic properties of materials which are required to select them for making permanent magnets.
Answer
Permanent Magnets: The magnets prepared from ferromagnetic materials which retain their magnetic properties for a long time are called permanent magnets. An efficient way to make a permanent magnet is to place a ferromagnetic rod in a solenoid and pass a current. The magnetic field of the solenoid magnetises the rod.
The materials used for permanent magnet must have the following characteristic properties:
  1. High retentivity so that the magnet may cause strong magnetic field.
  2. High coercivity so that the magnetisation is not wiped out by strong external fields, mechanical ill-treatment and temperature changes. The loss due to hysteresis is immaterial because the magnet in this case is never put to cyclic changes.
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Question 173 Marks
Can we have magnetic hysteresis in paramagnetic or diamagnetic substances?
Answer
No, magnetic hysteresis is the lagging of intensity of magnetization (I) behind magnetising force (H). When diamagnetic and paramagnetic materials are placed in a magnetic field they get weekly magnetised. Also, they lose their magnetization as the magnetic field is removed (low retentively). Therefore, they do not form magnetic hysteresis curve.
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Question 183 Marks
A tightly-wound, long solenoid carries a current of 2.00A. An electron is found to execute a uniform circular motion inside the solenoid with a frequency of $1.00 \times 10^8$ rev/s. Find the number of turns per metre in the solenoid.
Answer
$\text{i} = 2\text{a}, \text{f} = 10^8\text{rev/sec,}$
$\text{n}= ?,$
$\text{m}_\text{e} = 9.1 × 10^{-31}\text{kg},$
$\text{q}_\text{e} = 1.6\times10^{-19}\text{c},$
$\text{B}=\mu_0\text{ni}\Rightarrow\text{n}=\frac{\text{B}}{\mu_0\text{i}}$
$\text{f}=\frac{\text{qB}}{2\pi\text{m}_\text{e}}\Rightarrow\text{B}=\frac{\text{f}2\pi\text{m}_\text{e}}{\text{q}_\text{e}}\Rightarrow\text{n}=\frac{\text{B}}{\mu_0\text{i}}=\frac{\text{f}2\pi\text{m}_\text{e}}{\text{q}_\text{e}\mu_0\text{i}}$
$=\frac{10^8\times9.1\times10^{31}}{1.6\times10^{-19}\times2\times10^{-7}\times2\text{A}}2=1421\ \text{turns/m}$
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Question 193 Marks
Three identical bar magnets are rivetted together at centre in the same plane as shown in Fig. This system is placed at rest in a slowly varying magnetic field. It is found that the system of magnets does not show any motion. The north-south poles of one magnet is shown in the Fig. Determine the poles of the remaining two.
Answer
When the net force on the system is zero and net torque on the system is also zero then the system is in stable equilibrium. This is possible only when the poles of the remaining two magnets are as shown in the figure given below.
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Question 203 Marks
Answer the following questions:
If χ-stands for the magnetic susceptibility of a given material, identity the class of materials for which $(\text{i})-1 \geq\text{X} < 0 (\text{ii}) 0< \text{X} < \varepsilon 0 (\varepsilon$ is a small positive number).
  1. Write the range of relative magnetic permeability of these materials.
  2. Draw the pattern of the magnetic field lines when these materials are placed on an strong magnetic field.
Answer
  1. Material is diamagnetic (ii) Material is paramagnetic. $\mu\text{r} = 1+ \text{X}$
  1. Range of relative magnetic permeability for diamagnetic is $0 \leq \mu < 1.$
  2. Range of relative magnetic permeability for paramagnetic is $1< \mu\text{r} <1+ \varepsilon$
  1.  


​​​​​​​
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Question 213 Marks
What should be the orientation of a magnetic dipole in a uniform magnetic field so that its potential energy is maximum?
Answer
Potential energy of a magnetic dipole in a uniform magnetic field $\text{U}= -\text{ MB}\cos \text{q};$ clearly the potential energy is maximum when $\cos \theta = - 1 \text{ or }\theta=\pi$ That is, potential energy is maximum when magnetic dipole with its magnetic moment $\vec{\text{M}}$ is oriented opposite to the direction of magnetic field (or angle between $\vec{\text{M}} \text{ and }\vec{\text{B}}\text{ is }180^\circ).$
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Question 223 Marks
How does the (i) pole strength and (ii) magnetic moment of each part of a bar magnet change if it is cut into two equal pieces transverse to length?
Answer
When a bar magnet of magnetic moment $(\vec{\text{M}}=\text{m}2\vec{\text{t}})$ is cut into two equal pieces transverse to its length.
  1. The pole strength remains unchanged (since pole strength depends on number of atoms in cross-sectional area).
  2. The magnetic moment is reduced to half (since M $\propto$ length and here length is halved).
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Question 233 Marks
What is the basic difference between the atom and molecule of a diamagnetic and a paramagnetic material? Why are elements with even atomic number more likely to be diamagnetic?
Answer
Atoms/ molecules of a diamagnetic substance contain even number of electrons and these electrons from the pairs of opposite spin; while the atoms/ molecules of a paramagnetic substance have excess of electrons spinning in the same direction. The elements with even atomic number Z has even number of electrons in its atoms/ molecules, so they are more likely to form electrons pairs of opposite spin and hence more likely to be diamagnetic.
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Question 243 Marks
A ball of superconducting material is dipped in liquid nitrogen and placed near a bar magnet.
  1. In which direction will it move?
  2. What will be the direction of it’s magnetic moment?
Answer
A superconducting material and nitrogen both are diamagnetic in nature.
When a diamagnetic material is dipped in liquid nitrogen, it again behaves as a diamagnetic material. Thus, superconducting material will again behave as a diamagnetic material. When this diamagnetic material is placed near a bar magnet, it will be feebly magnetised opposite to the direction of magnetising field.
  1. So it will move away from the magnet.
  2. Magnetic moment is from left to right and it is opposite to the direction of magnetic field.
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Question 253 Marks
From molecular view point, discuss the temperature dependence of susceptibility for diamagnetism, paramagnetism and ferromagnetism.
Answer
Diamagnetism is due to orbital motion of electrons developing magnetic moments opposite to applied field and hence is not much affected by temperature.
Paramagnetism & ferromagnetism is due to alignments of atomic magnetic moments in the direction of the applied field. As temperature increases, this aligment is disturbed and hence susceptibilities of both decrease as temperature increases.
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Question 263 Marks
A capacitor of capacitance $100\mu\text{F}$ is connected to a battery of 20 volts for a long time and then disconnected from it. It is now connected across a long solenoid having 4000 turns per metre. It is found that the potential difference across the capacitor drops to 90% of its maximum value in 2.0 seconds. Estimate the average magnetic field produced at the centre of the solenofd during this period.
Answer
$\text{C}=100\mu\text{F},$
$\text{Q}=\text{CV}=2\times10^{-3}\text{C},$
$\text{t}=2\ \text{sec},$
$\text{V}=20\text{V},$
$ \text{V}' = 18\text{V},$
$\text{Q}'=\text{CV}=1.8\times10^{-3}\text{C},$
$\therefore\text{i}=\frac{\text{Q}-\text{Q}'}{\text{t}}=\frac{2\times10^{-4}}{2}=10^{-4}\text{A}$ $\text{n} = 4000\ \text{turns/m}$
$\therefore\text{B}=\mu_0\text{ni}=4\pi\times10^{-7}\times4000\times10^{-4}$
$=16\pi\times10^{-7}\text{T}$
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Question 273 Marks
A proton has spin and magnetic moment just like an electron. Why then its effect is neglected in magnetism of materials?
Answer
We know that,
$\mu_\text{p}\approx\frac{\text{eh}}{2\text{m}_\text{p}}$ and $\mu_\text{e}\approx\frac{\text{eh}}{2\text{m}_\text{e}}\Big[\text{Here, h}=\frac{\text{h}}{2\pi}\Big]$
$\Rightarrow\ \mu\propto\frac{1}{\text{m}}$
As, $\text{m}_\text{p}> >\text{m}_\text{e}$ therefore, $\mu_\text{e}>>\mu_\text{p}.$
So, effect of proton is neglected.
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Question 283 Marks
Distinguish between diamagnetic and ferromagnetic materials in respect of their (i) intensity of magnetisation (ii) behaviour in non-uniform magnetic field and (iii) susceptibility.
Answer
S. No. S. No. S. No. Ferromagnetic
(i) Intensity of magnetisation Negative and very small Positive and very large.
(ii) Behaviour in nonuniform magnetic field Attracted towards a region of weaker magnetic field Attracted towards a region of stronger magnetic field.
(iii) Susceptibility Negative and small $0<\text{X}<\in\in$ small quantity Positive and large χ of the order of hundreds & thousands.
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Question 293 Marks
The magnetic intensity Hat the centre of a long solenoid carrying a current of 2.0A, is found to be 1500A/m. Find the number of turns per centimetre of the solenoid.
Answer
$\text{B}=\mu_0\text{ni},$
$\text{H}=\frac{\text{B}}{{\mu_0}}$
$\Rightarrow\text{H}=\text{ni}$
$\Rightarrow1500\text{A/m}=\text{n}\times2$
$​​\Rightarrow\text{n}=750\text{turnes/meter}$
$​​\Rightarrow\text{n}=7.5\text{turnes/cm}$
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Question 303 Marks
The magnetic field B inside a long solenoid, carrying a current of 5.00A, is $3.14 \times 10^{-2}T.$ Find the number of turns per unit length of the solenoid.
Answer
$\text{i} = 25\text{A},$
$\text{B} = 3.14\times10^{-2}\text{T},$
$\text{n} =?$
$\text{B}=\mu_0\text{ni}$
$\Rightarrow3.14\times10^{-2}=4\times\pi\times10^{-7}\text{n}\times5$
$\Rightarrow\text{n}=\frac{10^{-2}}{20\times10^{-7}}=\frac{1}{2}\times10^4$
$=0.5\times10^4=5000\ \text{turns/m}$
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Question 313 Marks
A rod is inserted as the core in the current-carrying solenoid of the previous problem.
  1. What is the magnetic intensity H at the centre?
  2. If the magnetization I of the core is found to be 0.12Nm, find the susceptibility of the material of the rod.
  3. Is the material paramagnetic, diamagnetic or ferromagnetic?
Answer
  1. H = 1500A/m
As the solenoid and the rod are long and we are interested in the magnetic intensity at the centre, the end effects may be neglected. There is no effect of the rod on the magnetic intensity at the centre.
  1. I = 0.12A/m
We know $\overrightarrow{\text{I}}=\text{X}\overrightarrow{\text{H}}$

X = Susceptibility

$\Rightarrow\text{X}=\frac{\text{I}}{\text{H}}=\frac{0.12}{1500}=0.00008=8\times10^{-5}$
  1. The material is paramagnetic.
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Question 323 Marks
Draw diagrams to depict the behaviour of magnetic field lines near a ‘bar’ of: (i) Copper (ii) Aluminium (iii) Mercury, cooled to a very low temperature (4.2K)
Answer
  1. Copper is diamagnetic.
  2. Aluminium is paramagnetic.

  1. Mercury cooled to low temperature (4.2K) is a super conductor. It behaves as a perfect diamagnetic, so no lines of force passes through it.
​​​​​​​
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Question 333 Marks
Consider the plane S formed by the dipole axis and the axis of earth. Let P be point on the magnetic equator and in S. Let Q be the point of intersection of the geographical and magnetic equators. Obtain the declination and dip angles at P and Q.
Answer
Let point P is in the plane, S needle is in north, so the declination is zero.

From figure,
For point P: Since point P lies in plane S formed by the dipole axis and the axis of the Earth, declination is zero.
For point Q: Since point Q lies on the magnetic equator, angle of dip is zero. Thus the angle of declination is 11.3º.
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Question 343 Marks
The variation of intensity of magnetisation I and the applied magnetic field intensity H for three magnetic materials X, Y and Z are as shown in the given graphs.

  1. Identify the materials X, Y and Z.
  2. Show graphically the variation of susceptibility with temperature for X.
  3. Put of Y and Z, which of the material will you prefer for making transformer cores and why?
Answer
  1. X - diamagnetic, Z - paramagnetic, Y - ferromagnetic.
  2. For X.
  1. Y will be preferred to be used in the transformer core because it has high permeability and low hysteresis loss.
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Question 353 Marks
Why is it not possible to make permanent magnets from paramagnetic materials?
Answer
Permanent magnets are made from the material that are easily magnetized and retain the magnetization even reverse magnetizing field is applied (high coercivity). Paramagnetic materials get small magnetization, if they are placed in magnetic field, they lose their magnetization easily if the reverse field is applied. Hence, they are not used to make permanent magnet.
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Question 363 Marks
When a compass needle be placed at magnetic north pole, how would it behave? If a dip needle be placed at the place, how would it behave?
Answer
Compass needle stays in horizontal north-south direction. At poles horizontal component H = 0; therefore there will be no effect of earth’s field on magnetic north pole and it can stay in any direction; on the other hand a dip needle points along the resultant magnetic field and at poles the resultant field is vertical; hence the needle becomes vertical.
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Question 373 Marks
A solenoid has a core of a material with relative permeability 400 . The windings of the solenoid are insulated from the core and carry a current of 2A. If the number of turns is 1000 per metre, calculate (a) $H$, (b) $M$, (c) $B$ and (d) the magnetising current $I_m$.
Answer
(a) The field $H$ is dependent of the material of the core, and is
$
H=n I=1000 \times 2.0=2 \times 10^3 A / m \text {. }
$

(b) The magnetic field $B$ is given by
$
\begin{aligned}
B & =\mu_r \mu_0 H \\
& =400 \times 4 \pi \times 10^{-7}\left( N / A ^2\right) \times 2 \times 10^3( A / m ) \\
& =1.0 T
\end{aligned}
$

(c) Magnetisation is given by
$
\begin{aligned}
M & =\left(B-\mu_0 H\right) / \mu_0 \\
& =\left(\mu_r \mu_0 H-\mu_0 H\right) / \mu_0=\left(\mu_r-1\right) H=399 \times H \\
& \cong 8 \times 10^5 A / m
\end{aligned}
$

(d) The magnetising current $I_M$ is the additional current that needs to be passed through the windings of the solenoid in the absence of the core which would give a $B$ value as in the presence of the core. Thus $B=\mu_r n\left(I+I_M\right)$. Using $I=2 A , B=1 T$, we get $I_M=794 A$.
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Question 383 Marks
(a) Magnetic field lines show the direction (at every point) along which a small magnetised needle aligns (at the point). Do the magnetic field lines also represent the lines of force on a moving charged particle at every point?
(b) If magnetic monopoles existed, how would the Gauss's law of magnetism be modified?
(c) Does a bar magnet exert a torque on itself due to its own field? Does one element of a current-carrying wire exert a force on another element of the same wire?
(d) Magnetic field arises due to charges in motion. Can a system have magnetic moments even though its net charge is zero?
Answer
(a) No. The magnetic force is always normal to $B$ (remember magnetic force $=q v \times B$ ). It is misleading to call magnetic field lines as lines offorce.
(b) Gauss's law of magnetism states that the flux of B through any closed surface is always zero $\int_S B \cdot \Delta s =0$.
If monopoles existed, the right hand side would be equal to the monopole (magnetic charge) $q_m$ enclosed by $S$. [Analogous to Gauss's law of electrostatics, $\int_S B \cdot \Delta s =\mu_0 q_m$ where $q_m$ is the (monopole) magnetic charge enclosed by $S$.
(c) No. There is no force or torque on an element due to the field produced by that element itself. But there is a force (or torque) on an element of the same wire. (For the special case of a straight wire, this force is zero.)
(d) Yes. The average of the charge in the system may be zero. Yet, the mean of the magnetic moments due to various current loops may not be zero. We will come across such examples in connection with paramagnetic material where atoms have net dipole moment through their net charge is zero.
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Question 393 Marks
Figure 5.4 shows a small magnetised needle $P$ placed at a point $O$. The arrow shows the direction of its magnetic moment. The other arrows show different positions (and orientations of the magnetic moment) of another identical magnetised needle $Q$.
(a) In which configuration the system is not in equilibrium?
(b) In which configuration is the system in (i) stable, and (ii) unstable equilibrium?
(c) Which configuration corresponds to the lowest potential energy among all the configurations shown?
Image
Answer
Potential energy of the configuration arises due to the potential energy of one dipole (say, $Q$ ) in the magnetic field due to other $( P )$. Use the result that the field due to $P$ is given by the expression [Eqs. (5.7) and (5.8)]:
$
\begin{array}{ll}
B _{ P }=-\frac{\mu_0}{4 \pi} \frac{ m _{ P }}{r^3} & \text { (on the normal bisector) } \\
B _{ P }=\frac{\mu_0 2}{4 \pi} \frac{ m _{ P }}{r^3} \quad \text { (on the axis) }
\end{array}
$
where $m _P$ is the magnetic moment of the dipole $P$.
Equilibrium is stable when $m _{ Q }$ is parallel to $B _{ P }$, and unstable when it is anti-parallel to $B _{ P }$.
For instance for the configuration $\Theta_3$ for which 3 is along the perpendicular bisector of the dipole $P$, the magnetic moment of $Q$ is parallel to the magnetic field at the position 3 . Hence $Q_3$ is stable. Thus,
(a) $PQ _1$ and $PQ _2$
(b) (i) $PQ _3, PQ _6$ (stable); (ii) $PQ _5, PQ _4$ (unstable)
(c) $PQ _6$
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Question 403 Marks
(a) What happens if a bar magnet is cut into two pieces: (i) transverse to its length, (ii) along its length?
(b) A magnetised needle in a uniform magnetic field experiences a torque but no net force. An iron nail near a bar magnet, however, experiences a force of attraction in addition to a torque. Why?
(c) Must every magnetic configuration have a north pole and a south pole? What about the field due to a toroid?
(d) Two identical looking iron bars A and B are given, one of which is definitely known to be magnetised. (We do not know which one.) How would one ascertain whether or not both are magnetised? If only one is magnetised, how does one ascertain which one? [Use nothing else but the bars A and B.]
Answer
(a) In either case, one gets two magnets, each with a north and south pole.
(b) No force if the field is uniform. The iron nail experiences a nonuniform field due to the bar magnet. There is induced magnetic moment in the nail, therefore, it experiences both force and torque. The net force is attractive because the induced south pole (say) in the nail is closer to the north pole of magnet than induced north pole.
(c) Not necessarily. True only if the source of the field has a net non-zero magnetic moment. This is not so for a toroid or even for a straight infinite conductor.
(d) Try to bring different ends of the bars closer. A repulsive force in some situation establishes that both are magnetised. If it is always attractive, then one of them is not magnetised. In a bar magnet the intensity of the magnetic field is the strongest at the two ends (poles) and weakest at the central region. This fact may be used to determine whether A or B is the magnet. In this case, to see which one of the two bars is a magnet, pick up one, (say, A) and lower one of its ends; first on one of the ends of the other (say, B), and then on the middle of B. If you notice that in the middle of $B, A$ experiences no force, then $B$ is magnetised. If you do not notice any change from the end to the middle of $B$, then A is magnetised.
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