Magnetism and Matter

Physics

NCERT

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 $18^o$ . 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 $8 * 1022JT^{-1}$ 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?

Solution :

(a) The three independent quantities conventionally use for specifying earth’s magnetic field are:$\\$ (i) Magnetic declination,$\\$ (ii) Magnetic inclination or angle of dip, and$\\$ (ii) Horizontal component of 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. The angle of dip would be greater in Britain (it is about $70^o$ ) than in southern India because the location of Britain on the globe is closer to the magnetic North Pole.$\\$ (c) It is hypothetically considered that a huge bar magnet is dipped inside earth with its north pole near the geographic South Pole and its south pole near the geographic North Pole. 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.$\\$ (d) 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 a case, the compass can point in any direction. $\\$ (e) Magnetic moment,$M=8*10^{22}J T^{-1}$$\\$ Radius of earth, $r = 6.4 * 10^6 m$$\\$ Magnetics field strength,$B=\dfrac{\mu_0M}{4 \pi r^3}$$\\$ Where,$\\$ $\mu_0 =$Permeability of free space$=4\pi*10^{-7}TmA^{-1}$$\\$ $\therefore B=\dfrac{4 \pi * 10^{-7}*8*10^{22}}{4 \pi*(6.4*10^6)^3}\\ =0.3G$$\\$ This quantity is of the order of magnitude of the observed field on earth.$\\$ (f) Yes, there are several local poles on earth’s surface oriented in different directions. $A$ magnetized mineral deposit is an example of a local $N-S$ pole.

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 current?$\\$ (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. Answer 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.]$\\$

Solution :

(a) 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.$\\$ (b) 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.$\\$ (c) 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. BUT THIS IS NOT CERTAIN.$\\$ (d) Earth reversed the direction of its field several times during its history of $4$ to $5$ billion years. These magnetic fields got weakly recorded in rocks during their solidification. One can get clues about the geomagnetic history from the analysis of this rock magnetism.$\\$ (e) Earth’s field departs from its dipole shape substantially at large distances (greater than about $30,000 km$) because of the presence of the 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.$\\$ (f) 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.

3   A short bar magnet placed with its axis at $30^o$ 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?

Solution :

Magnetic field strength, $B =0.25 T$$\\$ Torque on the bar magnet, $T = 4.5 * 10^{- 2 }J$$\\$ Angle between the bar magnet and the external magnetic field, $\theta=30^o$$\\$ Torque is related to magnetic moment $(M)$ as:$\\$ $T=MB \sin \theta \\ \therefore M=\dfrac{T}{B\sin \theta}\\ =\dfrac{4.5 * 10^{-2}}{0.25* \sin 30^o}\\ =0.36JT^{-1}$$\\$ Hence, the magnetic moment of the magnet is $0.36J T^{- 1}$.

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

Solution :

Moment of the bar magnet, $M = 0.32 J T^{- 1}$$\\$ External magnetic field, $B = 0.15 T$$\\$ (a) When the bar magnet is aligned along the magnetic field, it is considered as being in stable equilibrium. Hence, the angle $\theta$ , between the bar magnet and the magnetic field is $0^o$ .$\\$ Potential energy of the system $=- MB \cos \theta$$\\$ $=-0.32*0.15 \cos 0^o\\ =-4.8 * 10^{-2} J$$\\$ (b) When the bar magnet is aligned opposite to the magnetic field, it is considered as being in unstable equilibrium, $\theta =180 ^o$$\\$ Potential energy $=- MB \cos \theta $$\\$ $=- 0.32 * 0.15 \cos 180^o\\ = 4.8 * 10^{- 2} J$

5   A closely wound solenoid of $800$ turns and area of cross section $2.5*10^{-4} m^2$ 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?

Solution :

Number of turns in the solenoid, $n = 800$$\\$ Area of cross-section, $A = 2.5 * 10^{-4} m ^2$$\\$ Current in the solenoid, $I=3.0 A$$\\$ A current-carrying solenoid behaves as a bar magnet because a magnetic field develops along its axis, i.e., along its length.$\\$ The magnetic moment associated with the given current-carrying solenoid is calculated as:$\\$ $M = nI A\\ =800*3*2.5*10^{-4}\\ 0.6 JT^{-1}$

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 $30^o$ with the direction of applied field?

Solution :

Magnetic field strength,$B = 0.25T$$\\$ Magnetic moment, $M = 0.6T^{- 1}$$\\$ The angle, $\theta $ between the axis of the solenoid and the direction of the applied field is $30^o$ .$\\$ Therefore, the torque acting on the solenoid is given as:$\\$ $\tau =MB \sin \theta \\ =0.6 * 0.25 \sin30^o\\ =7.5 * 10^{- 2} J$

7   A bar magnet of magnetic moment $1.5J T^{- 1}$ 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 filed direction?$\\$ (b) What is the torque on the magnet in cases (i) and (ii)?

Solution :

(a) Magnetic moment, $M = 1.5JT^{-1}$$\\$ Magnetic field strength, $B = 0.22T$$\\$ (i) Initial angle between the axis and the magnetic field, $\theta _1 = 0^o$$\\$ Final angle between the axis and the magnetic field, $\theta_ 2 = 90 ^o$$\\$ The work required to make the magnetic moment normal to the direction of magnetic field is given as:$\\$ $W =- MB ( \cos \theta _2 - \cos \theta_ 1 )\\ =- 1.5* 0.22 (\cos90^o - \cos0^o)\\ =- 0.33 ( 0 -1 )\\ =0.33J $$\\$ (ii) Initial angle between the axis and the magnetic field, $\theta_ 1 = 0 ^o$$\\$ Final angle between the axis and the magnetic field, $\theta_ 2 = 180 ^o$$\\$ The work required to make the magnetic moment opposite to the direction of magnetic field is given as:$\\$ $W=-MB (\cos \theta _2 - \cos \theta_ 1 )\\ =- 1.5*0.22 ( \cos180 - \cos0^o)\\ =-0.33(-1-1)\\ =0.66J$$\\$ (b)For case (i)$\\$ $\theta =\theta_2=90^o\\ \therefore \text{Torque} , \tau =MB \sin \theta \\ =1.5* 0.22 \sin 90^o\\ =0.33J$$\\$ For case(ii):$\\$ $ \theta=\theta_2=180^o\\ \therefore \text{Torque,} \tau = MB \sin\theta\\ =MB\sin180^o=0J$

8   A closely wound solenoid of $2000$ turns and area of cross-section $1.6 * 10 ^{-4} m^ 2$ , 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^o$ with the axis of the solenoid?

Solution :

Number of turns on the solenoid, $n = 2000$$\\$ Area of cross-section of the solenoid, $A =1.6 * 10 ^{- 4} m^ 2$$\\$ Current in the solenoid,$ I = 4A$$\\$ (a) The magnetic moment along the axis of the solenoid is calculated as:$\\$ $M=nAI\\ =2000*1.6*10^{-4}*4\\ =1.28 Am^2$$\\$ (b) Magnetic field, $B = 7.5 * 10^{- 2} T$$\\$ Angle between the magnetic field and the axis of the solenoid, $\theta 30 ^o$$\\$ Torque, $\tau= MB \sin \theta \\ = 1.28 * 7.5 * 10 ^{- 2} \sin30^o\\ =4.8 * 10^{-2} Nm$$\\$ Since the magnetic field is uniform, the force on the solenoid is zero. The torque on the solenoid is $4.8 * 10^{- 2} Nm .$$\\$

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*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 $2.0s^{- 1}$ . What is the moment of inertia of the coil about its axis of rotation?

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= \pi r^ 2 =\pi *( 0.1 ) m^ 2$$\\$ Current in the coil, $I = 0.75A $$\\$ Magnetic field strength, $B = 5.0 * 10^{- 2} T$$\\$ Frequency of oscillations of the coil, $v = 2.0s^{- 1}$$\\$ $\therefore $ Magnetic moment, $M = NIA = NI \pi r ^2$$\\$ $= 16 * 0.75 * \pi * (0.1)^ 2\\ =0.377 J T^{-1}$$\\$ Frequency is given by the relation:$\\$ $v=\dfrac{1}{2 \pi}\sqrt{\dfrac{MB}{I}}$$\\$ Where,$\\$ I=Moment of inertia of the coil$\\$ $\therefore I=\dfrac{MB}{4\pi^2v^2}\\ =\dfrac{0.377*5*10^{-2}}{4 \pi^2*(2)^2}\\ =1.19*10^{-4}kgm^2$$\\$ Hence, the moment of inertia of the coil about its axis of rotation is $1.19*10^{-4}kgm^2$

10   A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at $22^o$ 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.

Solution :

Horizontal component of earth’s magnetic field, $B _H = 0.35G$$\\$ Angle made by the needle with the horizontal plane = Angle of dip$=\delta22 ^o$$\\$ Earth’s magnetic field strength = B$\\$ We can relate B and B H as: $B _H = B \cos \delta $$\\$ $\therefore B=\dfrac{B_H}{\cos \delta}\\ =\dfrac{0.35}{\cos 22^o}=0.377 G$$\\$ Hence, the strength of earth’s magnetic field at the given location is $0.377 G$