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 Bihar Board Class 12th Physics Solutions Chapter 5 Magnetism and Matter

Bihar Board Class 12 Physics Magnetism and Matter Textbook Questions and Answers

Question 1.
Answer the following questions regarding earth’s magnetism:
(a) A vector needs three quantities for its specification. Name the three independent quantities conventionally used to specify the earth’s magnetic field.
(b) The angle of dip at a location in southern India is about 18°. Would you expect a greater or smaller dip angle in Britain?
(c) If you made a map of magnetic field lines at Melbourne in Australia, would the lines seen to go into the ground or come out of the ground?
(d) In which direction would a compass free to move in the vertical plane point to, if located right on the geomagnetic north or south pole?
(e) The earth’s field, it is claimed, roughly, approximates the field due to a dipole of magnetic moment 8 x 1022 JT-11 located at its centre. Check the order of magnitude of this number in some way.
(f) Geologists claim that besides the main magnetic N – S poles, there are several local poles on the earth’s surface oriented in different directions. How is such a thing possible at all?
Answer:
(a) The three independent quantities conventionally used to specify the earth’s magnetic field are magnetic declinatiion (θ), horizontal component of earth’s magnetic field (BH) and magnetic dip (δ).
(b) Yes, Great Britain will have a greater dip angle because it is located closer to the magnetic North Pole. The value of magnetic dip there is about 70°.
(c) The magnetic lines of force will seem to come out of ground at
Melbourne because it is situated in Southern hemisphere where north pole of earth’s magnetic field lies.
(d) At the magnetic poles earth’s magnetic field is exactly vertical and as the compass needle is free to rotate in horizontal plane, so the compass may point in any direction there.
(e) The magnetic field at a point on equatorial line of magnetic dipole (assuming earth has magnetic dipole) is given by
might be the ‘battery’ (i.e., the source of energy) to sustain these currents?
(d) The earth may have even reversed the directin of its field several times during its history of 4 to 5 billion years. How can geologists know about the earth’s field in such distant past?
(e) The earth’s field departs from its dipole shape substantially at large distances (greater than about 30,000 km). What agencies may be responsible for this distortion?
B = μ04π⋅⋅Mr3
Taking M =8 x 1022 JT-1 and r = 6.4 x 106 m = radius of earth.
∴ B = 4π×10−74π×8×1022(6.4×106)3
= 0.3 x 10-4T = 0.3G.
The above value is almost same as the earth’s magnetic field. This checks the magnitude of the dipole moment.
(f) Yes, this is possible. The earth’s magnetic field is only approximately a diple field. Thus, local N – S poles may exist oriented in different directions. This is possible due to deposits of magnetised materials.

Question 2.
Answer the following questions:
(a) The earth’s magnetic field varies from point to point in space. Does it also change with time? If so, on what time scale does it change
appreciably?
(b) The earth’s core is known to contain iron. Yet geologists do not regard this as a source of’the earth’s magnetism. Why?
(c) The charged currents in the outer conducting regions of the earth’s core are thought to be responsible for earth’s magnetism. What might be the ‘battery’ (i.e., the source of energy) to sustain these currents?
(d) The earth may have even reversed the directin of its field several times during its history of 4to 5 billion years. How can geologists know about the earth’s field in such distant past?
(e) The earth’s field departs from its dipole shape substantially at large distances (greater than about 30,000 km). What agencies may be responsible for this distortion?
(f) Interstellar space has an extremely weak magnetic field of the order of 10-12 T. Can such a weak field be of any significant consequence? Explain.
[Note : Exercise 5.2 is meant mainly to arouse your curiosity. Answers to some questions above are’ tentative or unknown. Brief answers wherever possible are given at the end. For details, you should consult a good text on geomagnetism.]
Answer:
(a) Yes, earth’s magnetic field changes with time for say, annual changes, daily changes, secular changes with period of about 960 years. Magnetic storms are considered as irregular changes. The time for an appreciable change is roughly few hundred years.
(b) Molten iron cannot retain magnetism because it is above its curie temperature.
(c) Radioactivity in the interior of earth but it is not certain.
(d) Earth’s magnetic field gets weakly recorded in certain rocks during solidification. Analysis of this rock magnetism gives us idea about geomagnetic history.
(e) If the distance is large, the magnetic field gets modified by the field produced by motion of ions in earth’s ionosphere.
(f) Tire deflection of a charged particle moving in a magnetic field is given by
Be v = mv2r or r = mvBe
If B is low, r is high, i.e. radius of curvature of path is very large. Thus, if distance is very large for eg. is space, the charged particle is hardly noticeable.

Question 3.
A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experience a torque of magnitude equal to 4.5 x 10 J. What is the magnitude of magnetic moment of the magnet ?
Answer:
Here, θ = 30°
B = uniform external magnetic field = 0.25 T
τ = torque = 4.5 x 10-2 ]
M = magnitude of magnetic moment of magnet = ? We know that

Question 4.
A short bar magnet of magnetic moment m = 0.32 JT-1 is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (i) stable and (ii) unstable equilibrium? What is the potential energy of the magifc in each case?
Answer:
Here m = magnetic moment of bar magnet = 0.32 J T-1
B = uniform magnetic field = 0.15 T
(i) When m→ is parallel to B→ , the magnet is in stable equilibrium,
∴ θ = 0°
∴ P.E. in this case is given by
U = m→ . B→ = – m B cos θ
= -0.32 x 0.15 x 1 =-0.048 J.

(ii)When m is anti  parallel to B , the magnet will be in unstable equilibrium,
.’. θ = 180°
Thus Potential energy in this case is given by
U = – m→ . B→ = – mB cos 180°
= – 0.32 x 0.15 x (-1) = + 0.048 J.

Question 5.
A closely wound solenoid of 800 turns and area of cross section 2.5 x 10-4 m2 carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?
Answer:
Here, N = number of turns = 800.
A = area of cross-section of solenoid = 2.5 x 10-4 m2.
I = current in the solenoid = 3.0 A.
Let M = associated magnetic moment = ?
We know that M for a current carrying solenoid is given by:
M = NIA
= 800 x 3.0 x 2.5 x 10-4
= 60 x 10-2 Am2
= 0.60 Am2 = 0.60 JT-1
Which acts along the axis of the solenoid in the direction related to the sense of flow of current according to right handed screw rule.

Question 6.
If the solenoid in exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?
Answer:
Here B = uniform horizontal magnetic field = 0.25 T
θ = angle made by m→ (i.e., the axis of solenoid with B→ = 30°
m→ = 0.60 Am2
τ = magnitude of Torque on the solenoid = ?

We know that x is given by
x = MB sin θ
= 0.60 x 0.25 x sin 30°
= 0.6 x 0.25 x 12
= 0.075 Nm.

Question 7.
A bar magnet of magnetic moment 1.5 JT-1 lies aligned with the direction of a uniform magnetic field of 0.22 T.
(a) 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?
(b) What is the torque on the magnet in cases (i) and (ii)?
Answer:
Here, M = magnetic moment of bar magnet =1.5 JT-1
B = uniform magnetic field = 0.22 T.
θ 1= 0 as the dipole is aligned with “g .
(a) (i) When the bar magnet is displaced to perpendicular position, then θ 2 = 90°
If W = work done so as to align m→ normal to B→ then using the relation,
W = MB (cos θ 1 – cos θ 2), we get
W = 1.5 x 0.22 (cos 0 – cos 90°)
= 1.5 x 0.22(1 – 0)
= 0.33 J.

(ii) When the magnet is displaced so as to align M→ opposite to B→ ,then
θ 2= 180°
τ2 = MB sin θ 2
= 1.5 x 0.22 x sin 180°
= 0.33 x (0) = 0

If W’ be the work done in this case, then
W’ = MB (cos θ 1 – cos θ 2)
= 1.5 x 0.22 (cos.θ – cos 180°)
= 0.33 [1 – (-1)]
= 0.33 x 2 =0.66 J.

(b) (i) Let τ1 be the torque on the magnet when it is aligned ⊥ ar to B→ , then using the relation,
τ = MB sin θ, We get
τ1 = MB sin θ1
τ1 = 1.5 x 0.22 x sin 90° (θ1 = 90°)
= 0.33 x 1 = 0.33 Nm-2.

(ii) Let τ2 be the torque in case (ii)
Here, θ2 = 180°
∴ τ2 = MB sin θ2
= 1.5 x 0.22 x sin 180°
= 0.33 x (0) = 0.

Question 8.
A closely wound solenoid of 2000 turns and area of crosssection 1.6 x 10-4 m2, carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane.
(a) What is the magnetic moment associated with the solenoid?
(b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 x 10-2 T is set up at an angle of 30° with the axis of the solenoid?
Answer:
Here, N = Number of turns in the solenoid = 2000
A = area of cross-section of solenoid = 1.6 x 10-4 m2
I = current in the solenoid = 4.0 A
(a) Let M = magnetic moment of the solenoid.
∴ Using the relation M = NLA, we get
M = 2000 x 4.0 x 1.6 x 10-4
= 1.28 JT-1.
The direction of M→ is along the axis of the solenoid in the direction related to the sense of current according to right-handed screw rule.

(b) Here θ = 30°
B→ = 7.5 x 10-2 T
Let F = force on the solenoid = ?
τ = torque on the solenoid = ?
Tire solenoid behaves as a barmagnet placed in a uniform magnetic field, so the force is
∴ F = m B→ + (- m B→ ) = 0.
where m = pole strength of the magnet.

Using the relation, t = MB sin θ, we get
τ = 1.28 x 7.5 x 10-2 x sin 30°
= 1.28 x 7.5 x 10-2 x 12 = 0.048 J.
The direction of the torque is such that it tends to align the axis of the solenoid (i.e,, magnetic moment vector M→ ) along B→.

Question 9.
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 x 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:
Here, N = 16, r = radius of coil = 10 cm = 0.10 m
I = current in the coil = 0.75 A.
BH = external.magnetic field = 5.0 x 10-2 T
v = frequency of oscillation of the coil = 2.0 s-1.
I = Moment of Inertia of the coil = ?
Let A = area of cross-section of the coil = πr2 = n (0.01)2 = 104 πm2.
If M be the magnetic moment of the coil, then
M = NIA = 16 x 0.75 x 10-4 it Am2.
Also let T be the time period of oscillation of the coil.
∴ T = 1υ = 12 = 0.5 s
Also we know that .

Question 10.
A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22° with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.
Answer:
Here
BH = horizontal component of earth’s magnetic field = 0.35 G.
ρ = angle of dip = 22°
Let B be the magnitude of earth’s magnetic field at the place = ?
We know that BH = B cos δ
B = BHcosδ=0.35cos22∘=0.350.9272
= 0.38 G.

Question 11.
At a certain location in Africa, a compass points 12° west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points 60° above the horizontal. The horizontal component of the earth’s field is measured to be 0.16 G. Specify the direction and magnitude of the earth’s field at the location.
Answer:
Here, 0 = angle of declination = 12°
δ = angle of dip = 60°
BH = horizontal component of earth’s magnetic field = 0.16 G.

Let B→ = earth’s magnetic field at that location = ?
Using the relation, BH = B cos δ, we get the magnitude of B→ given by,
B = BHcosδ=0.16cos60∘=0.16(12)
= 0.6 x 2 = 0.32 G
or B = 0.32 x 10-4 (∴ 1 G = 10-4 T)

Direction of B→ The earth’s field lies in a vertical plane 12c west of geographic meridian at an angle of 60c above the horizontal line.

Question 12.
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 (i) the axis (ii) the equatorial lines (normal bisector) of the magnet.
Answer:
Here, Mp magnetic moment of bar magnet = 0.48 J T-1.
r = distance of’the point from the centre of the magnet = 10 cm = 0.1 m.

(i) When the point lies on the axis :
Let B1 be the magnetic field at P.
∴ B1 = μ04π⋅2Mr3 = 10-7 x 2×0.48(0.1)3
= 0.96 x 10-4 T along S.N or NP
= 0.96 G along S-N direction.

Question 13.
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:
Here, B = Earths total magnetic field = 0.36 G. = 0.36 x 10-4 T.
δ = angle of dip = 0
r = distance of the null points on the axis of magnet = 0.14 m.
Let B’ be the fieLd due to the magnet at a point on the axial line.
∴ B’ = μ04π⋅2Mr3 ……..(i)
Also we know that at null points
B’ = BH ………(ii)
and BH = B cos δ = B cos 0 = B = 0.36 x 10-4T ………(iii)
∴ From (1), (ii) and (iii), we get

Let Beqt be the magnetic field due to the magnet at a point on the equitorial line at a distance 0.14 m from its centre.

Let Bt be the total magnetic field at point on the normal bisector at 0.14 m from the centre of magnet. Then, Bt = Bmagnet + Bearth
or Bt = Beqt + BH
= 0.18+ 0.36 = 0.54 G
in the direction of earth’s field.

Question 14.
If the bar magnet in Exercise 5.13 is turned around by 180°, where will the new null-points be located?
Answer:
Where the magnet is turned through 180° i.e. reversed i.e., north pole is pointing north, the neutral points will lie on the equitorial line of the magnet. If r be the distance of the neutral points on the equitorial line, then the magnetic field Beqt is given by

Question 15.
A short bar magnet of magnetic moment 5.25 x 10-2 J I-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:
Here, M = 5.25 x 102 JT-1
BH = 0.42 G = 0.42 x 10-4 T.

(a) On its normal bisectr :
Let r be the distance of point Pon the equitorial line at which the resultant magnetic field B→ is inclined at 45°with the earth’s field BH. For it to happen, Beqt must be equal to BH.
i.e Beqt = BH

(b) On its axis :
Let r be the distance of the point Pon the axis of the magnet where the resultant field is inclined at 45 with the earth’s magnetic field.

Question 16.
Answer the following questions :
(a) Why does a paramagnetic sample display greater magnetisation (for the same magnetising field) when cooled?
(b) Why is diamagnetism, in contrast, almost independent of temperature?
(c) If a toroid uses bismuth for its core, will the field in the core be (slightly) greater or (slightly) less than when the core is empty?
(d) Is the permeability of a ferromagnetic material independent of the magnetic field? If not, is it more for lower or higher fields?
(e) Magnetic field lines are always nearly normal to the surface of a ferromagnet at every point. (This fact is analogous to the static electric field lines being normal to the surface of a conductor at every point.) Why?
(f) Would the maximum possible magnetisation of a paramagnetic sample be of the same order of magnitude as the magnetisation of a ferromagnet?
Answer:
(a) On cooling, the tendency of thermal agitations to disrupt the alignment of magnetic dipoles decreases in case of paramagnetic materials. Thus, they display greater amount of magnetism.
(b) On placing a sample of diamagnetic material in magnetic field., the magnetisation (induced dipolement) is opposite to direction of magnetising field. Thus, it is not affected by temperature.
(c) The field in the core will be slightly less than when the core is emptly as Bismuth is diamagnetic.
(d) No; because permeability of ferromagnetic material is dependent on applied magnetic field. Graph for B and H is shown. B is large for smaller value of H, thus permeability μ (= B/H) is greater for lower fields.

(e) The field lines meet the material normally as μr »1.
(f) Yes. A paramagnetic substance sample with satured magnetism will have same order of magnetisation of a ferromagnet. However, a saturated magnetisation will require very high field. There will be a minor difference in the strength of the atomic dipoles of paramagnetic and ferromagnetic materials.

Question 17.
Answer the following questions :
(a) Explain qualitatively on the basis of domain picture the irreversibility in the magnetisation curve of a ferromagnet.
(b) The hysteresis loop of a soft iron piece has a much smaller area than that of a Carbon Steel piece. If the material is to go through repeated cycles of magnetisation, which piece will dissipate greater heat energy?
(c) A system displaying a hysteresis loop such as a ferromagnet, is a device for storing memory? Explain the meaning of this statement.
(d) What kind of ferromagnetic material is used for coating magnetic tapes in a cassette player, or for building ‘memory stores’ in a modem computer?
(e) A certain region of space is to be shielded from magnetic fields. Suggest a method.
Answer:
(a) In a ferromagnetic material, atoms form a very large number of small effective regions called domains. Each domain has a linear dimension ≈ 1000 Å. Within each domain, a special interaction called exchange coupling renders dipole moments of all atoms in a particular direction. Thus each domain is a strong magnet without any external magnetic field but the ferromagnetic substance does not behave as a magnet in the absence of external magnetic field as the magnetic moments of different domains are randomly oriented.
(b) The carbon steel piece will dissipate greater heat energy as heat lost per cycle is proportional to the area of the hysteresis loop.
(c) The ferromagnetic substance shows that it remains magnetised even after the removal of external magnetic field. It means that magnetism is stored as a memory in the ferromagnet. Thus, a system displaying hysteresis is a device for storing memory.
(d) Ceramics are used for coating magnetic tapes in cassette players or for building. Memory stores in modem computers. Ceramics are double oxides of barium and iron. Ceramics are also called ferrites.
(e) It is done by surrounding the region with soft iron rings. Magnetic field lines will be drawn into rings and enclosed space will be free of magnetic fields. But this shielding is only approximate, unlike the perfect electric shielding of a cavity in a conductor placed in an external electric field.

Question 18.
A long straight horizontal cable carries a current of 2.5 A in the direction 10° south of west to 10° north of east. The magnetic meridian of the place happens to be 10° west of the geographic meridian. The earth’s magnetic field at the location is 0.33 G, and the angle of dip is zero. Locate the line of neutral points (ignore the thickness of the cable)? (At neutral points, magnetic field due to a current carrying cable is equal and opposite to the horizontal component of earth’s magnetic field.)
Answer:
The magnetic n.s line is 10° west of geographical N.S. line straight horizontal cable XY is 10°N of east.

Here, B = earth’s total magnetic field = 0.33 G = 0.33 x 10-4 T
I = 2.5 A = Current in the cable,
angle of dip, δ = 0
If BH be the horizontal component of earth’s field, then
BH = B cos δ = B cos0° = B = 0.33 G = 0.33 x 10-4T
At points below the cable, the direction of the magnetic field produced is same as that of BH, while at a point above the cable, the direction of the field produced is opposite to that of BH. So the neutral point lies above the cable.
Let r be the distance of the neutral points from the cable.
If Bmrg be the magnetic field due to the straight current carrying cable, then
Bmrg = μ04π⋅2Ir
At netural points

i.e., the neutral point lies on a line parallel to the cable and above it at 1.52 cm.

Question 19.
A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in the same direction east to west. The earth’s magnetic field at the place is 0.39 G, and the angle of dip is 35°. The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable?
Answer:
Here, earth’s magnetic field, B = 0.39 G = 0.39 x 10-4 T
δ = angle of dip = 35°
θ = angle of magnetic declination = 0°.
n = number of wires = 4
1 = current = 1.0 A.
r = distance = 4 cm each = 0.04 m.
If Bv and Bu be the vertical and horizontal components of earth’s total field, then
BH = B cos δ = 0.39 cos 35° = 0.39 x 0.8192 = 0.3195 G
and Bv = B sin δ = 0.39 sin 35° = 0.39 x 0.5736 = 0.2237 G.
Let B1 be the field due to a single wire, then
B1 = μ04π⋅2Ir

∴ If B’ be the total magnetic field produced by the four wires of the cable, then

At a point below cable :
At a point say Q, 4. cm below the wire, the horizontal component of earth’s magnetic field BH and the field due to current are in opposite direction, so the net horizontal field RH is given ‘by
RH = BH – B’ = 0.3195 – 0.2
= 0.1195 G = 0.12 G
Net vertical component, Rv = Bv = 0.22 37 G = 0.22 G.

Above the Cable :
Say at point. P, 4 cm above the wire, the horizontal component of earth’s field and the field due to the current are in the same direction.
∴ Resultant horizontal field RH is given by
RH = BH + B’ = 0.3195 + 0.2 = 0.5195 = 0.52 G
and resultant vertical field is
Rv = Bv = 0.22 G.

If R’ be their resultant field above the cable, then

Direction of R→ :
If θ1 be the angle made by R′→ with RH, ther

Question 20.
A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12cm. 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.
(a) Determine the horizontal component of the earth’s magnetic field at the location.
(b) 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:
Here, n = no. of turns in the coil = 30
r = radius of circular coil = 12 m = 12 x 10-2 m.
I = current in the coil = 0.35 A.
(a) The horizontal component of the earth’s magnetic field (BH) acts along the magnetic N-S line. As the coil is placed in a vertical plane making angle 45° with the magnetic meridian, the magnetic field B→ produced by the coil on passing current through it, will be along normal to the plane of the coil i.e., in a direction making angle of 45° with the WE direction. The compass needle placed at the centre of the coil points along W to E.

From the figure, it follows that
BH = B sin 45° = B = 12√ ……….(i)
The magnetic field produced at the centre of the coil is given by
B = μ04π⋅2πIIr ……..(ii)
From (i) and (ii),

(b) The needle will reverse its original direction i.e., it will point east to west to the plane of coil makes an angle of 45° with the magnetic meridian on the other side, so the needle rotates.

Question 21.
A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60°, and one of the fields has a magnitude of 1.2 x 10-2 T. If the dipole comes to stable equilibrium at an angle of 15° with this field, what is the magnitude of the other field ?
Answer:
The two fields B1−→ and B2−→ are shown in the figure here in which a magnet is placed s.t.

Let θ1 and θ2 be the inclination of the dipole with B1−→ and B2−→ respectively.
θ1 = 15°, θ2 = 45°.
If τ1 and τ2 be the torques on the dipole due to B1−→ and B2−→ respectively, then
Using the relation, τ = MB sin θ, we get
τ1 = MB1 sin θ1 and
τ2= MB2 sin θ2
As the dipole is in equilibrium, the torques on the dipole due to B1−→ and B2−→ are equal and opposite . i.e.,
τ1 = τ2
or MB1 sin θ1, = MB2 sin θ2
or B2 = B1sinθ1

Question 22.
A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of 0.40 G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm.
(me = 9.11 x 1031 kg, e = 1.60 x 1019 C).
Note : [Data in this exercise are so chosen that the answer will give you an idea of the effect of earth’s magnetic field on the motion of electron beam from the electron gun to the screen in a TV set.]
Answer:
Here, Energy = E = 18 KeV = 18 x 1.6 x 10-16 J.
(∴ 1 KeV = 103ev = 103x 1.6x 1019J)
B = horizontal magnetic field = 0.40 G = 0.40 x 10-4 J.
m = 9.11 x 10-31 kg, e = 1.6 x 10-19C.
x = 30 cm = 0.30 m.
As the magnetic field is normal to the velocity, the charged particle
follows circular path in magnetic field. The centrepetal force mv2r required for this purpose is provided by force on electron due to magnetic field, i.e., BeV
Here mv2r = Be V


Let y be the deflection at the end of the path, then

sin θ = 0.3011.3=3113
= 0.02654
θ = sin (0.0264) 15
cos θ = 0.9997.
Up and down deflection of the beam = r (1 – cos θ)
= 11.3(1 – 0.9997)
= 11.3 x 0.0003
= 3.89 = 4 mm.

Question 23.
A sample of paramagnetic salt contains atomic dipoles each of dipole moment 1.5 x 10-23 J 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:
Here,no. of dipoles, n= 2 x 1024
Dipole moment of each dipole, M’ = 1.5 x 1023JT-1.
B1 = homogeneous magnetic field = 0.64 T.
T1 = initial temp, of sample = 4.2 K.
T2 = Final temp, of sample = 2.8 K.
M2 = total dipole moment of the sample at T2 = ?
B2 = Final magnetic field at T2 = 0.98 T.
Let M1 be the total initial dipole moment of the sample at magnetic.
Degree of saturation, D = 15%

According to Curie’s law,
Xm = CT = IH
I = C x HT
But I ∝ magnetic moment, M and H ∝ magnetic induction, B
∴ From (i), Magnetic moment = CxBT

Question 24.
A Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core of relative permeability 800. What is the magnetic field B in the core for a magnetizing current of 1.2 A?
Answer:
Here, a = mean radius of Rowland,
I = magnetising current = 1.2 A, ring = 15 cm = 0.15 m.
N = Total number of turns = 3500.
µr = relative permeability = 800
l = length over which the wire is wound.
= 2πa = 2π x 0.15 = 0.30 π.
If n be the number of turns per unit length, then
n = Nl=35000.30π=35×10430π
∴ The magnetic field B→ in the core is given by,

Question 25.
The magnetic moment vectors ps and pr associated with the intrinsic spin angular momentum S and orbital angular momentum 1, respectively, of an electron are predicted by quantum theory (and verified experimentally to a high accuracy) to be given by:
µs→ = (e/m) S→ , µl→ = (e/2m) l→
Which of these relations is in accordance with the result expected classically? Outline the derivation of the classical result.
Answer:
Out of these two relations, µs→ = −e2 m l→ is in accordance
with classical Physics and can be derived as follows :
We know that electrons revolving around the nucleus of an atom in circular orbits behave as tiny current loops having angular momentum
l→ given in magnitude as :
l→ = mvr …….(i)
where m = mass of an electron
v = its orbital velocity
r = radius of the circular orbit.
or v r = lm ……(ii)
l→ acts along the normal to the plane of the orbit in upward direction.
The orbital motion of electron is taken as equivalent to the flow of conventional current I given by
I = eT=e(2πrv)=ev2πr


where – ve sign shows that the electron is negatively charged. The eqn. (iii) shows that pe and l→ are opposite to each other i.e., antiparallel and both being normal to the plane of the orbit as shown in the figure :
∴ µl→ = −e2 m⋅l⃗ 

μsS is contrast to μll⃗  is em i.e., twice the classically expected value. This latter result is an outstanding consequence of modern quantum theory and cannot be obtained classically.