NCERT · Class 12 · Physics

Chapter 5: Magnetism and Matter

Practice each question, then open the answer or worked solution.

Questions

25 questions

Q5.1Written answer

Answer the following:

(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 in Melbourne, Australia, would the lines seem 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, is 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 the magnetic moment 8 × 10²² JT^-1 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?

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Worked solution

(a) The three independent conventional quantities used for determining the earth’s magnetic field are as follows:

(i) Magnetic declination

(ii) Angle of dip

(iii) The horizontal component of the earth’s magnetic field

(b) The angle of dip at a point depends on how far the point is located with respect to the north pole or the south pole. Hence, as the location of Britain on the globe is closer to the magnetic north pole, the angle of dip would be greater in Britain (About

\(\begin{array}{l}70°\end{array} \)

(c) It is assumed that a huge bar magnet is submerged inside the earth with its north pole near the geographic south pole and its south pole near the geographic north pole.

Magnetic field lines originate from the magnetic north pole and terminate at the magnetic south pole. Hence, in a map depicting the earth’s magnetic field lines, the field lines at Melbourne, Australia, would seem to move away from the ground.

(d) If a compass is placed in the geomagnetic north pole or the south pole, then the compass will be free to move in the horizontal plane while the earth’s field is exactly vertical to the magnetic poles. In such a case, the compass can point in any direction.

(e) Magnetic moment, M =

\(\begin{array}{l}8\times 10^{22}\;J\;T^{-1}\end{array} \)

Radius of earth, r =

\(\begin{array}{l}6.4\times 10^{6}\; m\end{array} \)

Magnetic field strength, B =

\(\begin{array}{l}\frac{\mu _{0}M}{4\pi r^{3}}\end{array} \)

Where,

\(\begin{array}{l}\mu_{0}\end{array} \)

\(\begin{array}{l}4\pi\times 10^{-7}\;TmA^{-1}\end{array} \)

Therefore, B =

\(\begin{array}{l}\frac{4\pi\times 10^{-7}\times 8\times 10^{22}}{4\pi \times (6.4\times 10^{6})^{3}} = 0.3\; G\end{array} \)

This quantity is of the order of magnitude of the observed field on earth.

(f) Yes, there are several local poles on the earth’s surface oriented in different directions. A magnetised mineral deposit is an example of a local N-S pole.

Q5.2Written answer

Answer the following:

(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 the 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 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 a 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.

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Worked solution

(a) The earth’s magnetic field varies with time, and it takes a couple of hundred years to change by an obvious sum. The variation in the earth’s magnetic field with respect to time can’t be ignored.

(b) The iron core at the earth’s centre cannot be considered as a source of the earth’s magnetism because it is in its molten form and is non-ferromagnetic.

(c) The radioactivity in the earth’s interior is the source of energy that sustains the currents in the outer conducting regions of the earth’s core. These charged currents are considered to be responsible for the earth’s magnetism.

(d) The earth’s magnetic field reversal has been recorded several times in the past, about 4 to 5 billion years ago. These changing magnetic fields were weakly recorded in rocks during their solidification. One can obtain clues about the geomagnetic history from the analysis of this rock magnetism.

(e) Due to the presence of the ionosphere, the earth’s field deviates from its dipole shape substantially at large distances. The earth’s field is slightly modified in this region because of the field of single ions. The magnetic field associated with them is produced while in motion.

(f) A remarkably weak magnetic field can deflect charged particles moving in a circle. This may not be detectable for a large radius path. With reference to the gigantic interstellar space, deflection can alter the passage of charged particles.

Q5.3Written answer

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

\(\begin{array}{l}4.5\times 10^{-2}\; J\end{array} \)

. What is the magnitude of the magnetic moment of the magnet?

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Worked solution

Magnetic field strength, B = 0.25 T

Torque on the bar magnet, T =

\(\begin{array}{l}4.5\times 10^{-2}\; J\end{array} \)

The angle between the bar magnet and the external magnetic field,

\(\begin{array}{l}\theta = 30°\end{array} \)

Torque is related to magnetic moment (M) as:

\(\begin{array}{l}T = MB sin\theta\end{array} \)

\(\begin{array}{l}∴ M = \frac{T}{B sin\theta }\end{array} \)

\(\begin{array}{l}\frac{4.5\times 10^{-2}}{0.25\times sin 30°} = 0.36\; J\; T^{-1}\end{array} \)

Hence, the magnetic moment of the magnet is

\(\begin{array}{l}0.36\; J\; T^{-1}\end{array} \)

Q5.4Written answer

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?

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Worked solution

Moment of the bar magnet, M =

\(\begin{array}{l}0.32\; J\; T^{-1}\end{array} \)

External magnetic field, B = 0.15 T

(a) The bar magnet is aligned along the magnetic field. This system is considered to be in stable equilibrium. Hence, the angle

\(\begin{array}{l}\theta\end{array} \)

\(\begin{array}{l}0°\end{array} \)

Potential energy of the system =

\(\begin{array}{l}-MB cos\theta\end{array} \)

\(\begin{array}{l}-0.32\times 0.15\; cos 0°\end{array} \)

\(\begin{array}{l}-4.8\times 10^{-2}\; J\end{array} \)

(b) The bar magnet is oriented

\(\begin{array}{l}180°\end{array} \)

\(\begin{array}{l}\theta = 180°\end{array} \)

Potential energy =

\(\begin{array}{l}-MB cos\theta\end{array} \)

\(\begin{array}{l}-0.32\times 0.15\; cos 180°\end{array} \)

\(\begin{array}{l}4.8\times 10^{-2}\; J\end{array} \)

Q5.5Written answer

A closely wound solenoid of 800 turns and an area of cross-section 2.5 × 10^–4 m²carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?

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Worked solution

Number of turns in the solenoid, n = 800

Area of cross-section, A =

\(\begin{array}{l}2.5\times 10^{-4}\;m^{2}\end{array} \)

Current in the solenoid, I = 3.0 A

A current-carrying solenoid behaves like a bar magnet because a magnetic field develops along its axis, i.e., along with its length.

The magnetic moment associated with the given current-carrying solenoid is calculated as:

M = n I A

\(\begin{array}{l}800\times 3\times 2.5 \times 10^{-4}\end{array} \)

\(\begin{array}{l}0.6\; J\; T^{-1}\end{array} \)

Q5.6Written answer

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 the torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?

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Worked solution

Magnetic field strength, B = 0.25 T

Magnetic moment, M = 0.6

\(\begin{array}{l}T^{-1}\end{array} \)

The angle

\(\begin{array}{l}\theta\end{array} \)

\(\begin{array}{l}30°\end{array} \)

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

\(\begin{array}{l}\tau = MB sin \theta\end{array} \)

\(\begin{array}{l}0.6\times 0.25\; sin 30°\end{array} \)

\(\begin{array}{l}7.5\times 10^{-2}\end{array} \)

Q5.7Written answer

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)?

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Worked solution

(a) Magnetic moment, M =

\(\begin{array}{l}1.5\; J\; T^{-1}\end{array} \)

Magnetic field strength, B = 0.22 T

(i) Initial angle between the axis and the magnetic field,

\(\begin{array}{l}\theta_{1} = 0°\end{array} \)

The final angle between the axis and the magnetic field,

\(\begin{array}{l}\theta_{2} = 90°\end{array} \)

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

W =

\(\begin{array}{l}- MB (cos\theta_{2} – cos\theta_{1})\end{array} \)

\(\begin{array}{l}- 1.5\times 0.22 (cos 90° – cos 0°)\end{array} \)

= – 0.33 (0 – 1)

= 0.33 J

(ii) Initial angle between the axis and the magnetic field,

\(\begin{array}{l}\theta_{1} = 0°\end{array} \)

The final angle between the axis and the magnetic field,

\(\begin{array}{l}\theta_{2} = 180°\end{array} \)

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

W =

\(\begin{array}{l}- MB (cos\theta_{2} – cos\theta_{1})\end{array} \)

\(\begin{array}{l}- 1.5\times 0.22 (cos 180° – cos 0°)\end{array} \)

= – 0.33 (– 1 – 1)

= 0.66 J

(b)
For case (i):

\(\begin{array}{l}\theta = \theta_{2} = 90°\end{array} \)

∴ Torque,

\(\begin{array}{l}\tau = MB sin\theta\end{array} \)

\(\begin{array}{l}MB sin 90°\end{array} \)

= 1.5 × 0.22 sin 90⁰

= 0.33 J

The torque tends to align the magnitude moment vector along B.

For case (ii):

\(\begin{array}{l}\theta = \theta_{2} = 180°\end{array} \)

\(\begin{array}{l}∴\end{array} \)

\(\begin{array}{l}\tau = MB sin\theta\end{array} \)

\(\begin{array}{l}MB sin 180°\end{array} \)

= 0 J

Q5.8Written answer

A closely wound solenoid of 2000 turns and area of cross-section

\(\begin{array}{l}1.6\times 10^{-4}\; m^{2}\end{array} \)

, carrying 4.0 A current, is suspended through its centre, thereby 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?

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Worked solution

Number of turns on the solenoid, n = 2000

Area of cross-section of the solenoid, A =

\(\begin{array}{l}1.6\times 10^{-4}\; m^{2}\end{array} \)

Current in the solenoid, I = 4.0 A

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

M = nAI

\(\begin{array}{l}2000\times 4\times1.6\times10^{-4}\end{array} \)

= 1.28 Am²

(b) Magnetic field, B =

\(\begin{array}{l}7.5\times 10^{-2}\; T\end{array} \)

The angle between the magnetic field and the axis of the solenoid,

\(\begin{array}{l}\theta = 30°\end{array} \)

Torque,

\(\begin{array}{l}\tau = MB sin\theta\end{array} \)

\(\begin{array}{l}1.28\times 7.5\times 10^{-2}\; sin30°\end{array} \)

\(\begin{array}{l}0.048 J\end{array} \)

Since the magnetic field is uniform, the force on the solenoid is zero. The torque on the solenoid is

\(\begin{array}{l}0.048 J\end{array} \)

Q5.9Written answer

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?

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Worked solution

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 =

\(\begin{array}{l}\pi r^{2} = \pi\times (0.1)^{2} \;m^{2}\end{array} \)

Current in the coil, I = 0.75 A

Magnetic field strength, B =

\(\begin{array}{l}5.0\times 10^{-2}\; T\end{array} \)

Frequency of oscillations of the coil, v =

\(\begin{array}{l}2.0\; s^{-1}\end{array} \)

∴ Magnetic moment, M = NIA =

\(\begin{array}{l}N I \pi r^{2}\end{array} \)

\(\begin{array}{l}16\times 0.75\times n\times (0.1)^{2}\end{array} \)

\(\begin{array}{l}0.377\; J \;T^{-1}\end{array} \)

Frequency is given by the relation:

\(\begin{array}{l}v = \frac{1}{2\pi }\sqrt{\frac{MB}{I}}\end{array} \)

Where,

I = Moment of inertia of the coil

Rearranging the above formula, we get:

\(\begin{array}{l}∴ I = \frac{MB}{4 \pi^{2} v^{2}}\end{array} \)

\(\begin{array}{l}\frac{0.377\times 5\times 10^{-2}}{4\pi^{2}\times (2)^{2}}\end{array} \)

\(\begin{array}{l}1.2\times 10^{-4}\; kg\; m^{2}\end{array} \)

Hence, the moment of inertia of the coil about its axis of rotation is

\(\begin{array}{l}1.19\times 10^{-4}\; kg\; m^{2}\end{array} \)

Q5.10Written answer

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.

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Worked solution

The horizontal component of earth’s magnetic field,

\(\begin{array}{l}B_{H} = 0.35\; G\end{array} \)

Angle made by the needle with the horizontal plane = Angle of dip =

\(\begin{array}{l}\delta = 22°\end{array} \)

Earth’s magnetic field strength = B

We can relate B and

\(\begin{array}{l}B_{H}\end{array} \)

\(\begin{array}{l}B_{H} = Bcos\delta\end{array} \)

\(\begin{array}{l}∴ B = \frac{B_{H}}{cos\delta}\end{array} \)

\(\begin{array}{l}\frac{0.35}{cos\; 22°}\end{array} \)

Hence, the strength of the earth’s magnetic field at the given location is 0.38 G.

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Chapter 5: Magnetism and Matter

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