Magnetism And Matter Class 12th Physics Part I CBSE Solution

Class 12^th Physics Part I CBSE Solution

Exercises

1. A vector needs three quantities for its specification. Name the three independent…
2. The angle of dip at a location in southern India is about 18°. Would you expect a greater…
3. If you made a map of magnetic field lines at Melbourne in Australia, would the lines seem…
4. In which direction would a compass free to move in the vertical plane point to, if located…
5. The earth’s field, it is claimed, roughly approximates the field due to a dipole of…
6. Geologists claim that besides the main magnetic N-S poles, there are several local poles…
7. Answer the following question: The earth’s magnetic field varies from point to point in…
8. The earth’s core is known to contain iron. Yet geologists do not regard this as a source…
9. The charged currents in the outer conducting regions of the earth’s core are thought to be…
10. The earth may have even reversed the direction of its field several times during its…
11. The earth’s field departs from its dipole shape substantially at large distances (greater…
12. Interstellar space has an extremely weak magnetic field of the order of 10-12 T. Can such…
13. A short bar magnet placed with its axis at 30° with a uniform external magnetic field of…
14. A short bar magnet of magnetic moment M = 0.32 JT-1 is placed in a uniform magnetic field…
15. A closely wound solenoid of 800 turns and area of cross section 2.5 × 10-4 m^2 carries a…
16. If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform…
17. A bar magnet of magnetic moment 1.5 J T-1 lies aligned with the direction of a uniform…
18. A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10-4 m^2 , carrying…
19. A circular coil of 16 turns and radius 10 cm carrying a current of 0.75 A rests with its…
20. A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has…
21. At a certain location in Africa, a compass points 12° west of the geographic north. The…
22. A short bar magnet has a magnetic moment of 0.48 J T-1. Give the direction and magnitude…
23. A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic…
24. If the bar magnet in exercise 5.13 is turned around by 180°, where will the new null…
25. A short bar magnet of magnetic moment 5.25 × 10-2 J T-1 is placed with its axis…

Additional Exercises

1. Why does a paramagnetic sample display greater magnetization (for the same magnetizing…
2. Why is diamagnetism, in contrast, almost independent of temperature? Answer the following…
3. If a toroid uses bismuth for its core, will the field in the core be (slightly) greater or…
4. Is the permeability of a ferromagnetic material independent of the magnetic field? If not,…
5. Magnetic field lines are always nearly normal to the surface of a ferromagnet at every…
6. Would the maximum possible magnetization of a paramagnetic sample be of the same order of…
7. Explain qualitatively on the basis of domain picture the irreversibility in the…
8. The hysteresis loop of a soft iron piece has a much smaller area than that of a carbon…
9. ‘A system displaying a hysteresis loop such as a ferromagnet, is a device for storing…
10. What kind of ferromagnetic material is used for coating magnetic tapes in a cassette…
11. A certain region of space is to be shielded from magnetic fields. Suggest a method. Answer…
12. A long straight horizontal cable carries a current of 2.5 A in the direction 10° south of…
13. A telephone cable at a place has four long straight horizontal wires carrying a current of…
14. A compass needle free to turn in a horizontal plane is placed at the centre of circular…
15. A magnetic dipole is under the influence of two magnetic fields. The angle between the…
16. A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected…
17. A sample of paramagnetic salt contains 2.0 × 10^24 atomic dipoles each of dipole moment…
18. A Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core…
19. The magnetic moment vectors μs and μl associated with the intrinsic spin angular momentum…

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Exercises

Question 1.

Answer the following question regarding earth’s magnetism:

A vector needs three quantities for its specification. Name the three independent quantities conventionally used to specify the earth’s magnetic field.

Answer:

The three independent quantities used to specify the earth’s magnetic field are the following.

1) Magnetic declination: It is the angle between the geographic north and the magnetic north at a place.

Note: If the compass deflects right of geographic north, then declination is positive and if deflection is towards the left of geographic north then it is a negative declination

2) Angle of Dip: It is the angle between the horizontal plane and the magnetic axis, as observed in the compass.

3) The horizontal component of the earth’s magnetic field

Question 2.

Answer the following question regarding earth’s magnetism:

The angle of dip at a location in southern India is about 18°. Would you expect a greater or smaller dip angle in Britain?

Answer:

The angle of dip will be greater than 18° in Britain. The angle of dip increases from equator towards pole. Britain is farther from equator as compared to India.

Note: The angle of dip increases from equator towards pole. At equator angle of dip is zero while at poles it is 90 degrees. The magnetic field lines are perpendicular at poles.

Question 3.

Answer the following question regarding earth’s magnetism:

If you made a map of magnetic field lines at Melbourne in

Australia, would the lines seem to go into the ground or come out of the ground?

Answer:

The magnetic field lines would come out of the ground at Melbourne in Australia. The magnetic field lines emanate from magnetic north pole (near Geographic south) and merge at magnetic South Pole (near geographic north). Australia is near the magnetic north pole so, magnetic field lines will seem to be emerging.

Question 4.

Answer the following question regarding earth’s magnetism:

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?

Answer:

The compass will get aligned in vertical direction (up-down) if is held vertically at the North Pole. This is due to the fact that magnetic field is perpendicular at the Poles and the magnetic needle of the compass aligns to it.

Note: There will be no predictable motion in horizontal plane due to very weak horizontal component of magnetic field.

Question 5.

Answer the following question regarding earth’s magnetism:

The earth’s field, it is claimed, roughly approximates the field due to a dipole of magnetic moment 8 × 10²² J T^–1 located at its centre. Check the order of magnitude of this number in some way.

Answer:

Given:

Magnetic moment, m = 8 × 10²² JT^-1

Radius of earth, r = 6.4 × 10⁶m

Our task is to check the order of m, we shall attempt to do so by finding the magnetic field of earth using the available information. If we are close to the standard measured value then the order of m is correct.

We know that magnetic field strength,

 …(1)

Where, �₀ = permeability of free space

�₀ = 4π × 10^-7 TmA^-1

m = magnetic moment

r = radius

Putting values in equation (1) we get,

⇒ B = 0.3 × 10^-4T

⇒ B = 0.3 G

Which is a close value to the strength of earth’s magnetic field. Hence the order of magnetic moment is correct.

Question 6.

Answer the following question regarding earth’s magnetism:

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:

There may be several local north and south poles, such a thing is possible due to deposits of minerals which are magnetic in nature.

Question 7.

Answer the following question:

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?

Answer:

The earth’s magnetic field changes with time due to many reasons, changes is the speed of convection current in core is one of them. These changes are very slow and will take hundreds of years to show substantial effects.

Question 8.

Answer the following questions:

The earth’s core is known to contain iron. Yet geologists do not regard this as a source of the earth’s magnetism. Why?

Answer:

The earth’s core contains iron in molten state which is not ferromagnetic in nature hence geologists do not regard this as a source of earth’s magnetism.

Fact: After Curie temperature (T_c) materials lose permanent magnetism. This is due to inability is magnetic domains to align properly. After this temperature magnetic materials attain paramagnetic nature.

Question 9.

Answer the following questions:

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?

Answer:

The charged currents here refer to the convection currents in the core of earth, which is responsible for earth’s magnetic nature. This current is sustained partly by the Coriolis force due to rotation of earth, which rotates the molten core.

Question 10.

Answer the following questions:

The earth may have even reversed the direction 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?

Answer:

Earth changed its magnetic orientation several times in history. The magnetic orientation of the earth were weekly recorded in the rocks during solidification.

Question 11.

Answer the following questions:

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?

Answer:

Earth’s atmosphere has a layer called Ionosphere, which mainly constitutes of electrons and other charges particles. A magnetic field is associated with these charged particles when they are in motion. So we can say that Ionosphere is one of the agencies which can distort the shape of earth’s dipole at greater altitudes.

Question 12.

Answer the following questions:

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.

Answer:

The magnetic field has a tendency to affect the path of charged particles in circular motion. Since Interstellar space is Vast hence a minor change in path due to magnetic field can become substantial deviation from the destination.

Question 13.

A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10^–2 J. What is the magnitude of magnetic moment of the magnet?

Answer:

Given:

Angle between axis of bar magnet and external magnetic field, θ = 30°

Strength of external magnetic field, B = 0.25 T

Torque experienced by bar magnet, Τ = 4.5 × 10^-2J

We know that, when a Bar magnet is placed in a uniform magnetic field it experiences a Net torque given by,

T = M × B …(1)

T = M × B × Sinθ

Where,

M = magnetic moment

T = Torque on bar magnet

θ = angle between magnetic field and magnetic moment

From equation (1) we can write,

M = 0.36 JT^-1

Hence, the magnitude of moment of the Bar magnet is 0.36JT^-1.

Question 14.

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 (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?

Answer:

Given:

Magnetic moment of magnet, M = 0.32JT^-1

Strength of external magnetic field, B = 0.15T

(a) When the magnetic moment is aligned (0°) with the magnetic field then we consider this as stable equilibrium, if the magnet is rotated, then it will have the tendency to come back in this position.

θ = 0°

We know that,

U = - M.B

U = -M × B × Cos θ …(1)

Where, U = potential energy

M = magnetic moment

B = magnetic field

θ = Angle between filed and magnetic moment

By putting the given values in the equation (1), we have,

U = -0.32 × 0.15 × Cos 0°

U = -0.048 J

Potential energy of the system in stable equilibrium is -0.048 J.

(b) Whereas, in case of unstable equilibrium the moment is at 180° with the magnetic field. If the magnet is rotated then it will never come back in the initial position.

θ = 180°

Now, by putting values in equation (1), we get,

U = -0.32 × 0.15 × Cos 180°

U = 0.048 J

Potential energy of the system in unstable equilibrium is 0.048J.

Note: The direction of the magnetic moment is from south pole to north pole inside the magnet.

Question 15.

A closely wound solenoid of 800 turns and area of cross section 2.5 × 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:

Given:

Number of turns, N = 800

Area of cross-section, A = 2.5 × 10^-4 m²

Current, I = 3A

The solenoid acts like a bar magnet because the magnetic axis of the solenoid is along its length, which is analogous to the magnetic axis of Bar magnet.

We know that,

Magnetic moment, M = N × I × A …(1)

Now putting the values in equation 1, we get

M = 800 × 3A × 2.5 × 10^-4m²

M = 0.6 JT^-1

Question 16.

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:

Magnetic field strength, B = 0.25 T

Magnetic moment, M = 0.6 JT⁻¹

The angle θ, between the axis of the solenoid and the direction of the applied field is 30°.

Therefore, the torque acting on the solenoid is given as:

τ = MBsinθ 
= 0.6 JT^-1 × 0.25 T × sin 30^o

= 0.6 JT^-1 × 0.25 T × (1/2)

(because sin 30 = 1/2)

= 0.075 J
= 7.5 × 10^-2 J

The magnitude of torque is 7.5 × 10^-2 J.

Question 17.

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.

(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:

(a) Magnetic moment, M = 1.5 J T⁻¹

Magnetic field strength, B = 0.22 T

(i) Initial angle between the axis and the magnetic field, θ₁ = 0°

Final angle between the axis and the magnetic field, θ₂ = 90°

The work required to make the magnetic moment normal to the direction of magnetic field is given as:

W = -MB(cosθ₂ - cosθ₁)

= -1.5 J T⁻¹ x0.22 T(cos 90^o-cos 0^o)

= -0.33 Jx(0-1)

= 0.33 J

(ii) Initial angle between the axis and the magnetic field, θ₁ = 0°

Final angle between the axis and the magnetic field, θ ₂ = 180°

The work required to make the magnetic moment opposite to the direction of magnetic field is given as:

W = -MB(cos θ₂ - cos θ₁)

= -1.5 J T⁻¹ x.22 T(cos 180^o-cos 0^o)

= -0.33 Jx(-1-1)

= 0.66 J

(b) For case i) θ = θ₂ = 90^o

τ = MBsinθ = 1.5 J T⁻¹ x 0.22 T x 1 = .33 J

(Remember the torque and the work applied by the torque are not the same thing)

For case ii) θ = θ₂ = 180^o

τ = MBsinθ = 1.5 J T⁻¹ x.22 Tx0 = 0 J

Question 18.

A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 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 × 10^–2 T is set up at an angle of 30° with the axis of the solenoid?

Answer:

Number of turns on the solenoid, n = 2000

Area of cross-section of the solenoid, A = 1.6 × 10^-4 m²

Current in the solenoid, I = 4 A

(a) The magnetic moment along the axis of the solenoid is calculated as:

M = n.A.I = 2000 × 1.6 × 10^-4 m² x 4A

= 1.28 Am²

(b) Magnetic field, B = 7.5 × 10^-2 T

Angle between the magnetic field and the axis of the solenoid, θ = 30°

τ = MBsinθ = 1.28 JT^-1 × 7.5x10^-2 T × sin30^o

= 4.8 × 10^-2 N-m

= 4.8x10^-2 J

But the Force is zero as the Magnetic field is uniform.

Question 19.

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 × 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:

Number of turns in the circular coil, N = 16

Radius of the coil, r = 10 cm = 0.1 m

Cross-section of the coil, A = πr² = π × (0.1)² m²

Current in the coil, I = 0.75 A

Magnetic field strength, B = 5.0 × 10⁻² T

Frequency of oscillations of the coil, f = 2.0 s⁻¹

Magnetic moment, M = NIA = NI × πr²

= 16 × 0.75 A × π × (0.1)² m²

= 0.377 J T⁻¹

Now the frequency is given by,

I = Moment of Inertia of the coil (Remember this equation)

So, 

Or, 

⇒ I = 1.19 x 10^-4 kgm²

The moment of inertia of the coil about its axis of rotation is 1.19 × 10^-4 kgm²

Question 20.

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:

Horizontal component of earth’s magnetic field, B_H = 0.35 G

Angle made by the needle with the horizontal plane = Angle of dip = δ 22°

Earth’s magnetic field strength = B

We know, B_H = B cosδ

B = B_H/cos δ = 0.35/(cos 22^o) = 0.377 G

The strength of earth’s magnetic field at the given location is 0.377 G.

Question 21.

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:

Angle of declination, θ = 12°

Angle of dip, δ = 60°

Horizontal component of earth’s magnetic field, B_H = 0.16 G

Earth’s magnetic field at the given location = B

We know that,

B_H = B cosδ

B = B_H/cos δ = 0.16 G/(cos 60^o) = 0.32 G

Earth’s magnetic field lies in the vertical plane, 12° West of the geographic meridian, making an angle of 60° (upward) with the horizontal direction. Its magnitude is 0.32 G.

Question 22.

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 magnetic field at distance d, from the centre of the magnet on the axis is given by the relation:

μ₀ = permeability of the free space = 4 x 10^-7

⇒ B₁ = 0.96 x 10^-4 T

∴ B₁ = 0.96 G

The magnetic 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:

 = 0.48 G

The magnetic field is along the N − S direction.

Question 23.

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, B = 0.36 G

The magnetic field at a distance d, on the axis of the magnet is given as:

The magnetic field at the same distance d, on the equatorial line of the magnet is given

as:

(NOTE: Refer the pictures of the previous question)

= B₁/2

Hence the total magnetic field is B₁ + B_{2 =} B₁ + B₁/2 = (0.36 + 0.18) G = 0.54 G

The magnetic field is 0.54 G in the direction of earth’s magnetic field.

Question 24.

If the bar magnet in exercise 5.13 is turned around by 180°, where will the new null points be located?

Answer:

The magnetic field on the axis of the magnet at a distance d₁ = 14 cm, can be written as:

 …………(1)

Where,

M = Magnetic moment

 = Permeability of free space

If the bar magnet is turned through 180°, then the neutral point will lie on the equatorial line.

 …………(2)

Equating (1) and (2) we get,

Or, 

Or, 

Or, d₂ = 14x 0.794 = 11.1cm

The new null points will be located 11.1 cm on the normal bisector.

Question 25.

A short bar magnet of magnetic moment 5.25 × 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 × 10⁻² J T⁻¹

Magnitude of earth’s magnetic field at a place, H = 0.42 G = 0.42 × 10⁻⁴

(a) The magnetic field at a distance R from the centre of the magnet on the normal bisector is given by the relation:

When the resultant field is inclined at 45° with earth’s field, B = H (Note: This is the key point of the question, try to sort out this kind of data from the question when you are reading it, so you need not read it again.)

 = H = 0.42 × 10⁻⁴ T

R³ = 

⇒ 

⇒ R³ = 12.5 x 10^-5 m³

⇒ R = 0.05 m = 5 cm

(b) The magnetic field at a distance R₁ from the centre of the magnet on its axis is given as:

The resultant field is inclined at 45° with earth’s field.

So, B₁ = H

R₁³ = 

⇒ R₁³ = 25 x10^-5 m³

⇒ R_{1 =} 0.063m = 6.3cm

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Additional Exercises

Question 1.

Answer the following question:

Why does a paramagnetic sample display greater magnetization (for the same magnetizing field) when cooled?

Answer:

At high temperatures, due to the random thermal motion of molecules in a paramagnetic sample, the alignments of dipoles get disrupted. On reducing the temperature random thermal motion reduces, thus disruption is reduced. Hence, a paramagnetic sample displays greater magnetization when cooled.

Question 2.

Answer the following question:

Why is diamagnetism, in contrast, almost independent of temperature?

Answer:

The magnetism in a diamagnetic substance is due to induced dipole moment (opposite to the magnetizing field). Hence, the internal motion of the atoms, which depends on the temperature, does not affect the diamagnetism of a material.

Question 3.

Answer the following question:

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?

Answer:

A toroid using bismuth core has a magnetic field slightly greater than a toroid whose core is empty because bismuth is a diamagnetic substance.

Question 4.

Answer the following question:

Is the permeability of a ferromagnetic material independent of the magnetic field? If not, is it more for lower or higher fields?

Answer:

The permeability of ferromagnetic materials is inversely dependent on the applied magnetic field, i.e. it is lower for a higher field and vice versa.

Question 5.

Answer the following question:

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?

Answer:

The magnetic field lines are always nearly normal to the surface of ferromagnetic materials at every point because permeability of ferromagnetic material is always greater than one.

Question 6.

Answer the following question:

Would the maximum possible magnetization of a paramagnetic sample be of the same order of magnitude as the magnetization of a ferromagnet?

Answer:

Yes, the maximum possible magnetization of a paramagnetic sample can be of the same order of magnitude as the magnetization of a ferromagnet but this requires very strong magnetic fields.

Question 7.

Answer the following question:

Explain qualitatively on the basis of domain picture the irreversibility in the magnetization curve of a ferromagnet.

Answer:

The relation between B (i.e. the external magnetic field) and H (i.e. magnetic intensity) in ferromagnetic materials is shown in the following curve.

From the above curve it is clear that magnetization persists even when the external field is removed, this depicts the irreversibility of magnetization in a ferromagnet.

Question 8.

Answer the following question:

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 magnetization, which piece will dissipate greater heat energy?

Answer:

After going through repeated cycles of magnetization, a carbon steel piece dissipates greater heat energy than a soft iron piece, because the dissipated heat energy is directly proportional to the area of a hysteresis loop and the carbon steel piece has a greater hysteresis curve area than a soft iron piece.

Question 9.

Answer the following question:

‘A system displaying a hysteresis loop such as a ferromagnet, is a device for storing memory?’ Explain the meaning of this statement.

Answer:

The record of hysteresis loop cycles during magnetization (i.e. value of magnetization) corresponds to bits of information. Therefore, the material displaying hysteresis loops like a ferromagnet can be used for storing information.

Question 10.

Answer the following question:

What kind of ferromagnetic material is used for coating magnetic tapes in a cassette player, or for building ‘memory stores’ in a modern computer?

Answer:

Ferromagnetic material like Ceramic is used for coating magnetic tapes in cassette players and for building memory stores in modern computers.

Question 11.

Answer the following question:

A certain region of space is to be shielded from magnetic fields. Suggest a method.

Answer:

The certain region of space can be surrounded by soft iron coil to shield it from magnetic field.

Question 12.

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:

Current in the cable (I) = 2.5 A

Earth’s magnetic field at the location (H) = 0.33 G = 0.33 × 10^-4T

Angle of dip (δ) = 0

Let the line of neutral points be at a distance “r” meters from the horizontal cable.

The magnetic field at the neutral point due to current carrying cable,

 ,

where μ₀ = Permeability of free space = 4π × 10^-7 TmA¹

Horizontal component of earth’s magnetic field (H’) = H × cosδ

At neutral points, magnetic field due to a current-carrying cable is equal and opposite to the horizontal component of earth’s magnetic field.

∴ H’ = H_N

⇒ H × cosδ = 

⇒ 0.33 × 10^-4 T × cos0^o = 

⇒ r = 

⇒ r = 15.15 × 10^-3m

⇒ r = 1.515cm

Question 13.

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:

Current in the wires (I) = 1.0 A (east to west)

The earth’s magnetic field at the place (H) = 0.39G = 0.39 × 10^-4T

The angle of dip (δ) = 35°

Magnetic declination (θ) ∼ 0^o

Number of wires in the cable (n) = 4

The horizontal component of earth’s magnetic field (H’) = H × cosδ

Magnetic fields due to current carrying cables (B) = 

where μ₀ = Permeability of free space = 4π × 10^-7TmA¹

The resultant magnetic fields at point 4.0 cm below the cables

R = 4cm = 0.04m

Resultant horizontal magnetic fields (H_h) = H’- B

⇒ H_h = H’- B

⇒ H_h = 

⇒ H � �_h = (0.39 × 10^-4T × cos35^o)- 

⇒ H_h = 0.39 × 10^-4T × 0.819 - 4 × 2 × 25 × 10^-7T

⇒ H_h = 0.319 × 10^-4T-0.2 × 10^-4T

⇒ H_h = 0.119 × 10^-4

Or H_h ∼ 0.12 × 10^-4 T or 0.12 Gauss

Resultant vertical magnetic field (H � �_{v �}) = vertical component of earth’s magnetic field = H × sinδ

= 0.39 × 10^-4T × sin35^o

= 0.39 × 10^-4T × 0.573

= 0.22 × 10^-4T

∴ Resultant magnetic field = 

= 0.25 × 10^-4T

Question 14.

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.

(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:

(a) Number of turns in the coil (n) = 30

Radius of coil (r) = 12cm = 0.12m

Current in the coil (I) = 0.35A

Angle of dip (δ) = 45^o

Magnetic fields due to current carrying coils (B) = 

where, μ₀ = Permeability of free space = 4π × 10^-7TmA¹

B = 

= 5.49 × 10^-5 T

∴ Horizontal component of the earth’s magnetic field = B sinδ

= 5.49 × 10^-5 T × sin45^o

= 5.49 × 10^-5 T × 0.707

= 3.88 × 10^-5 T

(b) When 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, then the needle will reverse its original direction (i.e. the needle will point from East to West).

Question 15.

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 × 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:

Magnitude of one of the magnetic fields (B₁) = 1.2 × 10^–2 T

The angle between the magnetic field directions (θ) = 60°

The angle between the dipole and the magnetic field B₁ is θ₁ = 15°

Let magnitude of second of the magnetic fields = B₂

∴ The angle between the dipole and the magnetic field B₂ is θ₂ = θ-θ₁ = 45°

Let magnetic moment of the dipole be M

Thus, at rotational equilibrium,

Torque due to field B₁ = Torque due to field B₂

⇒ M × B₁sinθ₁ = M × B₂ sinθ₂

⇒ 

⇒ B₂ = 9 × 10^-3 T

Thus, the magnitude of the other magnetic field = 4.39 × 10^-3T

Question 16.

A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of 0.04 G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm (m_e = 9.11 × 10^–31 kg). [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 the electron beam from the electron gun to the screen in a TV set.]

Answer:

Energy of the electron beam(E) = 18keV = 18000eV = 18000 × 1.6 × 10^-19 = 2.88 × 10^-15J

Mass of electron (m_e) = 9.11 × 10^–31 kg

Horizontal magnetic field (B) = 0.04 G = 0.04 × 10^-4T

Distance of deflection from the beam (r) = 30 cm = 0.3m

Let velocity of electrons be v.

∴ Energy of the beam = Kinetic energy of electrons

⇒ E = 1/2 m_ev²

⇒ 

⇒ 

∴ v = 7.95 × 10⁷m/s

∵ The electron beam deflects along a circular path in a magnetic field.

Let the radius of the circular path be r m.

The force due to magnetic field on the electron = qvBsinθ = e × v × B × sin 90°

Here, q = charge on the electron,

θ = angle between velocity and magnetic field.

∵ The force due to magnetic field on electron = centripetal force on electron

⇒ 

⇒ 

⇒ 

⇒ r = 11.3 m

Let the deflection of electron beam in upward and downward directions be x = r(1-cosΦ),

where Φ = angle of deflection.

∵ 

⇒  = 0.026

⇒ Φ = sin^-1(0.026) = 1.521°

Thus, Deflection(x) = r(1- cosΦ) = 11.3 × (1- cos(1.521)°)

= 11.3 × (1- 0.90)

= 11.3 × 0.01

= 1.1310^-3m

Question 17.

A sample of paramagnetic salt contains 2.0 × 10²⁴ atomic dipoles each of dipole moment 1.5 × 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:

Number of atomic dipoles (n) = 2 × 10²⁴

Initially, Dipole moment of each atomic dipole (M) = 1.5 × 10^-23JT^-1

Magnetic field (B₁) = 0.64 T and Temperature (θ₁) = 4.2K

∵ Degree of magnetic saturation = 15%

∴ Effective Dipole moment (M₁) = 15% of Total dipole moment

⇒ M₁ =  × (no. of atomic dipole × dipole moment of each)

⇒ M₁ =  × 2 × 10²⁴ × 1.5 × 10^-23JT^-1 = 4.5 JT^-1

When, Magnetic field (B₂) = 0.98 T and Temperature (θ₂) = 2.8 K

Let the dipole moment be M_2.

According to the Curie’s Law, magnetization of a paramagnetic material is directly proportional to applied magnetic field and inversely proportional to temperature.

∴ Ratio of magnetic dipole moments (i.e. M₁:M₂) = (B₁θ₂/B₂θ₁)

⇒  ⇒ M₂ = M₁ ×  ⇒ M₂ = 4.5 JT^-1 ×  = 10.336 JT^-1

Thus, total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K = 10.336 JT^-1

Question 18.

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 magnetising current of 1.2 A?

Answer:

Radius of ring (r) = 15cm = 0.15m

Number of turns in the ring (n) = 3500

Relative permeability of the ferromagnetic core (μ_r) = 800

Current in the Rowland ring (I) = 1.2A

Magnetic Field due to circular coil (B) = 

Where, μ₀ = Permeability of free space = 4π × 10^-7TmA^-1

∴ B =  = 4.48T

Question 19.

The magnetic moment vectors μ_s and μ_l associated with the intrinsic spin angular momentum S and orbital angular momentum l, respectively, of an electron are predicted by quantum theory (and verified experimentally to a high accuracy) to be given by:

μ_s = – (e/m) S,

μ_l = – (e/2m)l

Which of these relations is in accordance with the result expected classically? Outline the derivation of the classical result.

Answer:

We know,

It follows from the definition of μ₁ and l.

μ_{1 =} i × A = 

l = mvr = m

Where, r is the radius of the circular path traced by the electron around the nucleus, with the mass of electron m and charge (-e) and time period T.

Divide the above equations-

⇒ 

⇒ 

Clearly μ₁ and l will be antiparallel.

On the other hand,  is obtained on the basis of quantum mechanics.