NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter: Topics Covered, Question with Solutions PDF

Class 12 Physics Chapter 4 - Moving Charges and Magnetism NCERT Solutions

Q.5.1 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 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, 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 × 1022 J T–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?

Ans.5.1 Earth’s magnetic field can be specified by three following independent quantities

Magnetic declination

Angle of dip

Horizontal component of earth’s magnetic field.

The angle of dip at a point depends on how far the point is located with respect to North pole or South pole. The angle of dip will be more in Britain than Southern India as Britain is closer to Magnetic North pole than South India to the Magnetic South pole.

It is a hypothesis that a huge bar magnet is embedded deep in Earth’s ground with its north pole near magnetic south pole of earth and south pole is near magnetic north pole of earth. Magnetic field lines emanate from a magnetic north pole and terminate at a magnetic south pole. Hence, in a map depicting earth’s magnetic field lines, the field lines at Melbourne, Australia would seem to come out of the ground.

If a compass is located on the geomagnetic North pole or South pole, then the compass will be free to move in the horizontal plane while earth’s field is exactly vertical to the magnetic poles. In such case, the compass can point in any direction.

Magnetic moment, M = 8 $\times {10}^{22}$ J $T^{-1}$

Radius of the Earth, r = 6.4 $\times {10}^6$ m

Magnetic field strength is given by the expression:

B = $\frac{{\mu }_{0}M}{4\pi r^3}$ . where ${\mu }_0$ = Permeability of free space = 4 $\pi \times {10}^{-7}$ T m $A^{-1}$

Hence, B = $\frac{4\pi \times {10}^{-7}\times 8\mathrm{}\times {10}^{22}}{4\pi \times {(6.4\mathrm{}\times {10}^6)}^3}$ = 0.305 G

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

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

Q.5.2 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 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?

(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.]

Ans.5.2 Earth’s magnetic field changes with time. It takes a few hundred years to change by an appreciable amount. The variation in earth’s magnetic field with the time cannot be neglected.

Earth’s core contains molten iron. This form of iron is not ferromagnetic. Hence this is not considered as a source of earth’s magnetism.

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

The change of earth’s magnetic field got weakly recorded in rocks during their solidification. One can get a clue after analysing this rock magnetism.

It departs because of the presence of ionosphere. In this region, earth’s field gets modified because of the field of single ions. While in motion, these ions produce the magnetic field associated with them.

An extremely weak magnetic field can bend charged particles moving in a circle. This may not be noticeable for a large radius path. With reference to the gigantic interstellar space, the deflection can affect the passage of charged particles.

Ans.5.3 Magnetic field strength, B = 0.25 T

Torque on the bar magnet, $\tau$ = 4.5 $\times {10}^{-2}$ J

Angle between the bar magnet and the external magnetic field, $\theta$ = 30 $°$

From the relation T = MB $\sin \theta$ , where M = Magnetic moment, we get

M = $\frac{\tau }{B\sin \theta }$ = $\frac{4.5\mathrm{}\times {10}^{-2}}{0.25sin30°}$ = 0.36 J/T

Hence the magnetic moment is 0.36 J/T

Ans.5.4 Moment of the bar magnet, M = 0.32 J/T

Magnetic field, B = 0.15 T

The bar magnet is aligned along the magnetic field. This system is considered as being in stable equilibrium. Hence, the angle $\theta$ , between the bar magnet and the magnetic field is 0 $°$ .

Potential energy of the system = -MBcos $\theta$ = - 0.32 $\times 0.15\times \cos 0°$ = -4.8 $\times {10}^{-2}$ J

When the bar magnet is oriented 180 $°$ to the magnetic field, it becomes unstable equilibrium.

Potential energy = - MBcos $\theta$ = - 0.32 $\times 0.15\times \cos 180°$ = 4.8 $\times {10}^{-2}$ J