NCERT Solutions Class 12 Physics Chapter 5

Q.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 × 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.

(a) The three independent quantities used for specifying earth’s magnetic field are: – Magnetic declination (θ), Angle of dip (δ), and Horizontal component of earth’s magnetic field (H).

(b)The angle of dip at a point depends on the location of the point w.r.t. the North Pole or the South Pole. We expect greater angle of dip in Britain (it is about 70°) than in southern India because the location of Britain is closer to the magnetic North Pole.

(c)As Melbourne is situated in southern hemisphere, which behaves as the North pole of earth’s magnet, therefore, field lines due to the earth’s magnetism would seem to come out of the ground at Melbourne.

(d)At the poles, the angle of dip is 90^o; therefore, earth’s magnetic field at the poles is exactly in vertical direction. If a compass is located on the geomagnetic North Pole or South Pole, then the compass will be free to rotate in the horizontal plane while earth’s field is exactly vertical to the magnetic poles. In this situation, the compass can point in any direction.

$\begin{array}{l}\left(\text{e}\right)\text{Here,\hspace{0.17em}}\mathrm{m}\text{agnetic moment},\text{M}=\text{8}\times \text{1}{0}^{\text{22}}{\text{\hspace{0.17em}JT}}^{-\text{1}}\\ \text{Radius of the\hspace{0.17em}}\mathrm{E}\text{arth},\text{r}=\text{6}.\text{4}\times \text{1}{0}^{\text{6}}\text{m}\\ \text{Magnetic field strength\hspace{0.17em}of\hspace{0.17em}Earth},\mathrm{B}=\frac{{\mathrm{\mu }}_{\mathrm{o}}\mathrm{M}}{4{\mathrm{\pi r}}^3}\\ \text{Here},\\ {\mathrm{\mu }}_{\mathrm{o}}=\text{Permeability of free space}\\ =4\mathrm{\pi }\times \text{1}{0}^{\text{-7}}{\text{\hspace{0.17em}TmA}}^{\text{-1}}\\ \therefore \mathrm{B}=\frac{4\mathrm{\pi }\times \text{1}{0}^{\text{-7}}\times \text{8}\times \text{1}{0}^{\text{22}}}{4\mathrm{\pi }\times {(\text{6}.\text{4}\times \text{1}{0}^{\text{6}})}^3}\\ =0.3\mathrm{G}\\ \text{This value is of the order of magnitude of the}\\ \text{observed values\hspace{0.17em}of\hspace{0.17em}field on Earth}.\end{array}$

(f) There are several local poles on earth’s surface oriented in different directions. This is because of the presence of deposits of magnetized materials. The magnetised mineral deposit is an example of a local N-S pole.

Q.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.

(a) The earth’s magnetic field changes with time. Earth takes a few hundred years to change by an appreciable amount. The change in earth’s magnetic field with the time cannot be neglected.
(b) Earth’s core contains iron in the molten form which is is not ferromagnetic. Therefore, it cannot be considered as a source of earth’s magnetism.
(c) The radioactivity in the interior of Earth is the source of energy that sustains the currents in the outer conducting regions of the core of Earth. These charged currents are considered to be responsible for earth’s magnetism, but it is not certain.
(d) Earth reversed the direction of its magnetic field several times during its history of 4 to 5 billion years. These magnetic fields got weakly recorded in certain rocks during their solidification. The analysis of this rock magnetism can reveal the geomagnetic history of Earth.
(e) Earth’s magnetic field departs from its dipole shape substantially at large distances (greater than about 30,000 km) due to the presence of the ionosphere. In this region, earth’s field gets modified due to the magnetic field produced by the motion of single ions.
(f) An extremely weak magnetic field can deflect charged particles moving in a circular path. This may be less noticeable for a large radius path. Therefore, over the gigantic interstellar space, the deflection can affect the passage of charged particles.

Q.3 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.)

Ans.

\begin{array}{l}\text{The}\text{Earth}’\text{s magnetic field at the given place},\text{H}\\ =0.\text{36 G}\\ Asthenullpo\mathrm{int}slieontheaxisofthemagnet,\\ \therefore \text{Magnetic field at a distance d},\text{on the axis of the}\\ \text{magnet}\text{is}\text{given}\text{by,}\\ {\text{B}}_{\text{1}}=\frac{{\mu }_o}{4\pi }\frac{2M}{d^3}\\ =H\to (i)\\ \text{Here},=\text{Permeability of free space}\\ \text{M}=\text{Magnetic moment}\\ \text{Magnetic field at the same distance d},\text{on the}\\ \text{equatorial line of the magnet}\text{is}\text{given}\text{by,}\\ {\text{B}}_{\text{1}}=\frac{{\mu }_o}{4\pi }\frac{M}{d^3}\\ =\frac{H}{2}\to (ii)\\ \text{Total magnetic field},B=B_1+B_2\\ =H+\frac{H}{2}\\ =0.\text{36}+\frac{0.\text{36}}{2}\\ =0.54G\\ \therefore \text{The total}\text{magnetic field =}0.\text{54 G}\\ \text{It}\text{is}\text{along the direction of earth}’\text{s magnetic field}.\end{array}

Q.4 A short bar magnet of magnetic moment m = 0.32 J T⁻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?

Ans.

\begin{array}{l}\text{Here,}\text{magnetic}m\text{oment of bar magnet},\text{M}=0.{\text{32 JT}}^{-\text{1}}\\ \text{External magnetic field},\text{B}=0.\text{15 T}\\ \left(\text{a}\right)\text{In stable equilibrium,}t\text{he bar magnet is aligned}\\ \text{along the direction}\text{magnetic field}\\ \therefore \text{Angle between}\text{the bar magnet and the}\text{magnetic}\\ \text{field,}\theta \text{=}0°.\\ \text{Potential energy=-MBcos}\theta \\ =-0.32\times 0.\text{15}\times {\text{cos0}}^{\text{0}}\\ =-4.8\times {10}^{-2}J\\ \left(\text{b}\right)\text{In}\text{unstable}\text{equilibrium,}t\text{he bar magnet is oriented}\\ \text{at}\text{18}0°\text{to the magnetic field}.\\ \therefore \theta =\text{18}0°\\ \text{Potential energy}=-\text{MB cos}\theta \\ =-0.32\times 0.\text{15}\times {\text{cos180}}^{\text{0}}\\ =-0.32\times 0.\text{15}\times -1\\ =4.8\times {10}^{-2}J\end{array}

Q.5 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^o with the axis of the solenoid?

Ans.

Here, number of turns on the solenoid, n = 2000
Cross-sectional area of the solenoid, A = 1.6 × 10⁻⁴ m²
Current flowing in the solenoid, I = 4 A
(a)Magnetic moment associated with solenoid is given by,
M = nAI = 2000 × 1.6 × 10⁻⁴ × 4 = 1.28 Am²
(b)Here, magnetic field, B = 7.5 × 10⁻² T
As the magnetic field is uniform,
Force on the solenoid = 0

\begin{array}{l}\text{Angle between the magnetic field and the axis of the}\\ \text{solenoid},\theta =\text{3}0°\\ \therefore \text{Torque},\tau =MB\sin \theta \\ =\text{1}.\text{28}\times \text{7}.\text{5}\times \text{1}{0}^{-\text{2}}\times \sin \text{3}0°\\ =\text{1}.\text{28}\times \text{7}.\text{5}\times \text{1}{0}^{-\text{2}}\times \frac{1}{2}\\ =4.8\times \text{1}{0}^{-\text{2}}Nm\\ \therefore F\text{orce on the solenoid =0}\\ \text{Torque on the solenoid =}4.8\times \text{1}{0}^{-\text{2}}Nm\end{array}

Q.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?

Ans.

Here, applied magnetic field, B = 0.25 T
Magnetic moment, M = 0.6 T⁻¹
Angle between the axis of the solenoid and the direction of the applied field, θ=30°

$\begin{array}{l}\text{Torque acting on the solenoid\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}}\mathrm{as},\\ \mathrm{\tau }=\mathrm{MBsin\theta }\\ \therefore \mathrm{\tau }=0.6\times 0.25\sin {30}^{\mathrm{o}}\\ =0.075\mathrm{J}.\end{array}$

Q.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)?

Ans.

(a) Here, magnetic moment, M = 1.5 J T⁻¹
Magnetic field strength, B = 0.22 T

$\begin{array}{l}\left(\text{i}\right)\text{Initial angle between the axis\hspace{0.17em}of\hspace{0.17em}bar\hspace{0.17em}magnet and the}\\ \text{magnetic field},{\mathrm{\theta }}_{\text{1}}=0\mathrm{°}\\ \text{Final angle between the axis of\hspace{0.17em}bar\hspace{0.17em}magnet\hspace{0.17em}and the}\\ \text{magnetic field},{\mathrm{\theta }}_{\text{2}}=\text{9}0\mathrm{°}\\ \text{Work required to align the magnetic moment of\hspace{0.17em}the\hspace{0.17em}}\\ \mathrm{bar}\mathrm{magnet}\mathrm{normal}\mathrm{to}\mathrm{the}\mathrm{direction}\mathrm{of}\mathrm{magnetic}\mathrm{field},\\ \mathrm{W}=-\mathrm{MB}({\mathrm{cos\theta }}_2-{\mathrm{cos\theta }}_1)\\ =-1.5\times 0.22(\cos \text{9}0\mathrm{°}-\cos 0\mathrm{°})\\ =-0.33(0-1)=0.33\mathrm{J}\\ \left(\text{ii}\right)\text{Initial angle between the axis of\hspace{0.17em}bar\hspace{0.17em}magnet\hspace{0.17em}and}\\ \text{the magnetic field},{\mathrm{\theta }}_{\text{1}}=0\mathrm{°}\\ \text{Final angle between the axis\hspace{0.17em}of\hspace{0.17em}bar\hspace{0.17em}magnet and the}\\ \text{magnetic field},{\mathrm{\theta }}_{\text{2}}=\text{18}0\mathrm{°}\\ \text{Work required to align the magnetic moment of\hspace{0.17em}the\hspace{0.17em}}\\ \mathrm{bar}\mathrm{magnet}\mathrm{opposite}\mathrm{to}\mathrm{the}\mathrm{direction}\mathrm{of}\mathrm{magnetic}\mathrm{field},\\ \mathrm{W}=-\mathrm{MB}({\mathrm{cos\theta }}_2-{\mathrm{cos\theta }}_1)\\ =-1.5\times 0.22(\cos 180\mathrm{°}-\cos 0\mathrm{°})\\ =-0.33(-1-1)=0.66\mathrm{J}\\ \left(\text{b}\right)\mathrm{Torque}\mathrm{on}\mathrm{the}\mathrm{magnet}\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{the}\mathrm{relation},\\ \mathrm{\tau }=\mathrm{MBsin\theta }\\ \text{For case}\left(\text{i}\right):\\ \mathrm{\theta }={90}^{\mathrm{o}}\\ \therefore \text{Torque},\mathrm{\tau }=1.5\times 0.22\sin {90}^{\mathrm{o}}\\ =0.33\mathrm{J}\\ \text{For case}\left(\text{ii}\right):\\ \mathrm{\theta }={180}^{\mathrm{o}}\\ \therefore \text{Torque},\mathrm{\tau }=1.5\times 0.22\sin {180}^{\mathrm{o}}\\ =0\mathrm{J}\end{array}$

Q.8 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^o with the axis of the solenoid?

Ans.

Here, number of turns on the solenoid, n = 2000
Cross-sectional area of the solenoid, A = 1.6 × 10⁻⁴ m²
Current flowing in the solenoid, I = 4 A
(a)Magnetic moment associated with solenoid is given by,
M = nAI = 2000 × 1.6 × 10⁻⁴ × 4 = 1.28 Am²
(b)Here, magnetic field, B = 7.5 × 10⁻² T
As the magnetic field is uniform,
Force on the solenoid = 0

\begin{array}{l}\text{Angle between the magnetic field and the axis of the}\\ \text{solenoid},\theta =\text{3}0°\\ \therefore \text{Torque},\tau =MB\sin \theta \\ =\text{1}.\text{28}\times \text{7}.\text{5}\times \text{1}{0}^{-\text{2}}\times \sin \text{3}0°\\ =\text{1}.\text{28}\times \text{7}.\text{5}\times \text{1}{0}^{-\text{2}}\times \frac{1}{2}\\ =4.8\times \text{1}{0}^{-\text{2}}Nm\\ \therefore F\text{orce on the solenoid =0}\\ \text{Torque on the solenoid =}4.8\times \text{1}{0}^{-\text{2}}Nm\end{array}

Q.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 2.0 s⁻¹. What is the moment of inertia of the coil about its axis of rotation?

Ans.

Here, number of turns in the circular coil, N = 16
Radius of coil, r = 10 cm
= 0.1 m
Cross-sectional area of the coil, A = πr²
= π × (0.1)² m²
Current in the circular coil, I = 0.75 A
Magnetic field strength, B = 5.0 × 10⁻² T

$\begin{array}{l}\text{Frequency of oscillations of the coil},\text{v}=\text{2}.0{\text{s}}^{-\text{1}}\\ \therefore \text{Magnetic moment},\text{M}=\text{NIA}\\ \text{=NI}{\mathrm{\pi r}}^2\\ =\text{16}\times 0.\text{75}\times \mathrm{\pi }\times {(0.\text{1})}^{\text{2}}\\ =0.{\text{377 JT}}^{-\text{1}}\\ \mathrm{Frequency}\mathrm{is}\mathrm{given}\mathrm{by},\mathrm{\nu }=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{\mathrm{MB}}{\mathrm{I}}}\\ \mathrm{Here},\mathrm{I}=\mathrm{M}.\mathrm{I}.\mathrm{of}\mathrm{the}\mathrm{circular}\mathrm{coil}\\ \therefore \mathrm{I}=\frac{\mathrm{MB}}{4{\mathrm{\pi }}^2{\mathrm{\nu }}^2}\\ =\frac{0.\text{377}\times 5\times {10}^{-2}}{4{\mathrm{\pi }}^2{\left(2\right)}^2}\\ =1.19\times {10}^{-4}{\mathrm{kgm}}^2\\ \therefore \mathrm{M}\text{oment of inertia of the coil about its axis}\\ \text{of rotation is\hspace{0.17em}}1.19\times {10}^{-4}{\mathrm{kgm}}^2.\end{array}$

Q.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.

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}}\mathrm{h}\text{orizontal component of thearth}’\text{s magnetic field},\\ {\text{B}}_{\text{H}}=0.\text{35 G}\\ \text{Angle subtended by the needle with the horizontal}\\ \text{plane}=\text{Angle of dip}=\mathrm{\delta }\\ ={22}^{\mathrm{o}}\\ \text{Let\hspace{0.17em}strength\hspace{0.17em}}\mathrm{of}\mathrm{the}\text{\hspace{0.17em}Earth}’\text{s magnetic field}=\text{B}\\ {\text{The\hspace{0.17em}relation\hspace{0.17em}between\hspace{0.17em}B and B}}_{\text{H}}\mathrm{is}\mathrm{given}\text{\hspace{0.17em}as}:\\ {\text{B}}_{\text{H}}=\text{B}\mathrm{cos\theta }\\ \therefore \text{B}=\frac{{\text{B}}_{\text{H}}}{\mathrm{cos\delta }}\\ =\frac{0.\text{35}}{\cos {22}^{\mathrm{o}}}\\ =0.377\mathrm{G}\\ \therefore \text{The strength of the\hspace{0.17em}}\mathrm{E}\text{arth}’\text{s magnetic field at the}\\ \text{given location =\hspace{0.17em}}0.\text{377 G}.\end{array}$

Q.11 At a certain location in Africa, a compass points 12^o 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^o 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.

Ans.

\begin{array}{l}\text{Here,}a\text{ngle of declination},\theta =\text{12}°\\ \text{Angle of dip},\delta ={60}^o\\ \text{Horizontal component of the}E\text{arth}’\text{s magnetic field},{\text{B}}_{\text{H}}\\ =0.\text{16 G}\\ \text{Let}\text{Earth}’\text{s magnetic field at the given location}=\text{B}\\ \text{The}\text{relation}{\text{between B and B}}_{\text{H}}\text{is}\text{given}\text{as}:\\ {\text{B}}_{\text{H}}=B\cos \delta \\ \therefore B=\frac{{\text{B}}_{\text{H}}}{\cos \delta }\\ =\frac{0.\text{16}}{\cos {60}^o}\\ =0.32G\\ \\ \text{The}\text{Earth}’\text{s magnetic field lies in the vertical plane},\text{12}°\\ \text{west of geographic meridian}\text{at an angle of 6}0°\\ \left(\text{upward}\right)\text{with the horizontal direction}.\\ \text{The magnitude of}\text{Earth}’\text{s magnetic field}\text{is}0.\text{32 G}.\end{array}

Q.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 (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.

Ans.

Here, magnetic moment of bar magnet, M = 0.48 JT⁻¹

Distance, d = 10 cm

= 0.1 m

\begin{array}{l}\text{(a)}\text{Magnetic field on the axis of}\text{the}\text{magnet}\text{is given by}\\ \text{the relation}:\\ B=\frac{{\mu }_o}{4\pi }\frac{2M}{d^3}\\ \text{Here},{\mu }_o=\text{Permeability of free space}=4\pi \times {10}^{-7}Tm{A}^{-1}\\ \therefore B=\frac{4\pi \times {10}^{-7}\times 2\times 0.48}{4\pi \times {\left(0.1\right)}^3}\\ =0.96\times {10}^{-4}T\\ =0.96G\\ \text{It}\text{is along the S}-\text{N direction}.\\ \left(\text{b}\right)\text{The magnetic field on the equatorial line of}\text{the}\\ \text{magnet}\text{is given as}:\\ B=\frac{{\mu }_o}{4\pi }\frac{M}{d^3}={10}^{-7}\times \frac{0.48}{{\left(0.1\right)}^3}\\ =0.48\times {10}^{-4}T\\ \text{It}\text{is along the N}-\text{S direction}.\end{array}

Q.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.)

Ans.

\begin{array}{l}\text{The}\text{Earth}’\text{s magnetic field at the given place},\text{H}\\ =0.\text{36 G}\\ Asthenullpo\mathrm{int}slieontheaxisofthemagnet,\\ \therefore \text{Magnetic field at a distance d},\text{on the axis of the}\\ \text{magnet}\text{is}\text{given}\text{by,}\\ {\text{B}}_{\text{1}}=\frac{{\mu }_o}{4\pi }\frac{2M}{d^3}\\ =H\to (i)\\ \text{Here},=\text{Permeability of free space}\\ \text{M}=\text{Magnetic moment}\\ \text{Magnetic field at the same distance d},\text{on the}\\ \text{equatorial line of the magnet}\text{is}\text{given}\text{by,}\\ {\text{B}}_{\text{1}}=\frac{{\mu }_o}{4\pi }\frac{M}{d^3}\\ =\frac{H}{2}\to (ii)\\ \text{Total magnetic field},B=B_1+B_2\\ =H+\frac{H}{2}\\ =0.\text{36}+\frac{0.\text{36}}{2}\\ =0.54G\\ \therefore \text{The total}\text{magnetic field =}0.\text{54 G}\\ \text{It}\text{is}\text{along the direction of earth}’\text{s magnetic field}.\end{array}

Q.14 If the bar magnet in exercise 5.13 is turned around by 180^o, where will the new null points be located?

Ans.

\begin{array}{l}\text{Magnetic field on the axis of the magnet at a distance}\\ {\text{d}}_{\text{1}}=\text{14 cm},\text{can be written}\text{as}:\\ B_1=\frac{{\mu }_o}{4\pi }\frac{2M}{{\left(d_1\right)}^3}\\ =H\\ \text{Here},\text{M}=\text{Magnetic moment}\\ {\mu }_o=\text{Permeability of free space}\\ \text{H}={\text{Horizontal component of magnetic field at d}}_{\text{1}}\\ \text{If bar magnet is turned around}\text{by 18}0°,\text{then the}\\ \text{neutral point will lie on the equatorial}\text{line}.\\ \therefore \text{Magnetic field on the equatorial line}{\text{at a distance d}}_{\text{2}}\\ \text{of the magnet can be}\text{written as}:\\ B_2=\frac{{\mu }_o}{4\pi }\frac{M}{{\left(d_2\right)}^3}\\ =H\to (ii)\\ \text{Equating equation}\left(i\right)\text{and equation}\left(ii\right),\text{we obtain}\\ \frac{2}{{\left(d_1\right)}^3}=\frac{1}{{\left(d_2\right)}^3}\\ \therefore {\left(\frac{d_2}{d_1}\right)}^3=\frac{1}{2}\\ \therefore d_2=d_1\times {\left(\frac{1}{2}\right)}^{\frac{1}{3}}\\ =14\times 0.794\\ =11.1cm\\ \therefore \text{New null points will be located 11}.\text{1 cm on the normal}\\ \text{bisector}.\end{array}

Q.15 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^o 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.

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}}\mathrm{m}\text{agnetic moment of the bar magnet},\\ \text{M}=\text{5}.\text{25}\times \text{1}{0}^{-\text{2}}{\text{J T}}^{-\text{1}}\\ \text{Magnitude of earth}’\text{s magnetic field at the\hspace{0.17em}given place},\text{H}\\ =0.\text{42 G}=0.\text{42}\times \text{1}{0}^{-\text{4}}\text{T}\\ \left(\text{a}\right)\text{Magnetic field at a distance R from the centre}\\ \text{of the magnet on the normal bisector is given by,}\\ \text{B=}\frac{{\mathrm{\mu }}_{\mathrm{o}}}{4\mathrm{\pi }}\frac{\mathrm{M}}{{\mathrm{R}}^3}\\ \text{Here},\\ {\mathrm{\mu }}_{\mathrm{o}}=\text{Permeability of free space}=\text{4}\mathrm{\pi }\times \text{1}{0}^{-\text{7}}{\text{Tm A}}^{-\text{1}}\\ \text{When the resultant field is inclined at 45}\mathrm{°}\text{with earth}’\text{s}\\ \text{field},\mathrm{then}\text{B}=\text{H}\mathrm{}\\ \therefore \frac{{\mathrm{\mu }}_{\mathrm{o}}}{4\mathrm{\pi }}\frac{\mathrm{M}}{{\mathrm{R}}^3}=\mathrm{H}\\ =0.42\times {10}^{-4}\\ {\mathrm{R}}^3=\frac{{\mathrm{\mu }}_{\mathrm{o}}\mathrm{M}}{0.42\times {10}^{-4}\times 4\mathrm{\pi }}\\ =\frac{4\mathrm{\pi }\times \text{1}{0}^{-\text{7}}\times \text{5}.\text{25}\times \text{1}{0}^{-\text{2}}}{4\mathrm{\pi }\times 0.42\times {10}^{-4}}\\ =12.5\times \text{1}{0}^{-5}\\ \therefore \mathrm{R}=0.05\mathrm{m}\\ =5\mathrm{cm}\\ \left(\text{b}\right)\text{Magnetic field at a distance\hspace{0.17em}}\mathrm{R}‘\text{from the centre}\\ \text{of the magnet on its axis is\hspace{0.17em}given by}:\\ \text{B’}=\frac{{\mathrm{\mu }}_{\mathrm{o}}}{4\mathrm{\pi }}\frac{2\mathrm{M}}{{(\mathrm{R}‘)}^3}\\ \mathrm{Since}\mathrm{t}\text{he resultant field is inclined at 45}\mathrm{°}\text{with earth}’\text{s}\\ \text{field,}\\ \therefore \text{B’}=\mathrm{H}\\ \frac{{\mathrm{\mu }}_{\mathrm{o}}}{4\mathrm{\pi }}\frac{2\mathrm{M}}{{(\mathrm{R}‘)}^3}=\mathrm{H}\\ \therefore {(\mathrm{R}‘)}^3=\frac{{\mathrm{\mu }}_{\mathrm{o}}}{4\mathrm{\pi }}\frac{2\mathrm{M}}{\mathrm{H}}\\ =\frac{4\mathrm{\pi }\times \text{1}{0}^{-\text{7}}\times 2\times \text{5}.\text{25}\times \text{1}{0}^{-\text{2}}}{4\mathrm{\pi }\times 0.42\times {10}^{-4}}\\ =25\times {10}^{-5}\\ \therefore \mathrm{R}‘=0.063\mathrm{m}\\ =6.3\mathrm{cm}\end{array}$

Q.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 magnetization of a ferromagnet?

Ans.

(a)Due to the random thermal motion of molecules, the alignments of dipoles get disrupted at high temperatures. On cooling, this disruption is reduced due to the reduced random thermal motion. That is why, a paramagnetic sample displays greater magnetisation when cooled.
(b) In a diamagnetic substance, the induced dipole moment is always opposite to the magnetising field. Therefore, the random motion of the atoms (which is related to temperature) does not affect the diamagnetism of a material.
(c)As bismuth is a diamagnetic substance, therefore, a toroid with a bismuth core has a magnetic field slightly greater than a toroid whose core is empty.
(d)The permeability of a ferromagnetic material is not independent of the applied magnetic field. From the hysteresis curve it is clear that, μ is greater for a lower field.
(e)The permeability of a ferromagnetic material is always greater than one. That is why the magnetic field lines are always nearly normal to the surface of such materials at every point.
(f)Yes, maximum possible magnetisation of a paramagnetic sample can be of the same order of magnitude as the magnetisation of a ferromagnet. However, this requires very high magnetising fields for saturation, which are hard to achieve.

Q.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 modern computer?
(e) A certain region of space is to be shielded from magnetic fields. Suggest a method.

Ans.

The hysteresis curve (B-H curve) of a ferromagnetic material is shown in the given figure.
(a) From the given curve, it can be observed that magnetisation persists even when the external field is removed. This indicates the irreversibility in the magnetization curve of a ferromagnet.
(b) The energy dissipated per cycle is directly proportional to the area of the hysteresis loop. Since the carbon steel piece has a greater hysteresis curve area, therefore, it dissipates greater heat energy.
(c) The value of magnetisation of a ferromagnet depends not only on the magnetising field, but also on the history of magnetisation (i.e., how many cycles of magnetisation it has undergone). Hence, value of magnetisation of a specimen is the record of memory of the cycles of magnetization it has undergone. The system displaying such a hysteresis loop can be used for storing information.
(d) Ferrites are the ferromagnetic materials used for coating magnetic tapes in cassette players and for building memory stores in modern computers. Some common ferrites are FeFe₂O₄, CoFe₂O₄, and MnFe₂O₄ etc.
(e) A certain region of space can be shielded from magnetic fields by surrounding it by soft iron ring. In such arrangement, the magnetic field lines are drawn into the ring and the enclosed region will become free of magnetic field.

Q.18 A long straight horizontal cable carries a current of 2.5 A in the direction 10^o south of west to 10° north of east. The magnetic meridian of the place happens to be 10o 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.)

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}}\mathrm{c}\text{urrent in the wire},\text{I}=\text{2}.\text{5 A}\\ \text{Angle of dip at the given location},\mathrm{\delta }=0^{\mathrm{o}}\\ \text{Earth}’\text{s magnetic field\hspace{0.17em}at\hspace{0.17em}the\hspace{0.17em}given\hspace{0.17em}location},\text{B}\\ =0.\text{33 G}\\ =0.\text{33}\times \text{1}{0}^{-\text{4}}\text{T}\\ \text{Horizontal component of earth}’\text{s magnetic field is}\\ \text{given by,}\\ {\text{B}}_{\text{H}}=\text{B cos}\mathrm{\delta }\\ \text{=\hspace{0.17em}}0.\text{33}\times \text{1}{0}^{-\text{4}}\times \text{cos}{0}^{\mathrm{o}}\\ =0.\text{33}\times \text{1}{0}^{-\text{4}}\mathrm{T}\\ \text{Magnetic field at the neutral point at a distance R}\\ \text{from the cable is given by,}\\ {\text{B}}_{\text{H}}=\frac{{\mathrm{\mu }}_{\mathrm{o}}\mathrm{I}}{2\mathrm{\pi R}}\\ \text{Here},{\mathrm{\mu }}_{\mathrm{o}}=\text{Permeability of free space}\\ =4\mathrm{\pi }\times \text{1}{0}^{-7}{\mathrm{TmA}}^{-1}\\ \therefore \mathrm{R}=\frac{{\mathrm{\mu }}_{\mathrm{o}}\mathrm{I}}{2\mathrm{\pi }{\text{B}}_{\text{H}}}\\ =\frac{4\mathrm{\pi }\times \text{1}{0}^{-7}\times 2.5}{2\mathrm{\pi }\times 0.\text{33}\times \text{1}{0}^{-\text{4}}}\\ =15.15\times \text{1}{0}^{-3}\mathrm{m}\\ =1.51\mathrm{cm}\\ \therefore \mathrm{N}\text{eutral points parallel to and above the\hspace{0.17em}cable are}\\ \text{located at a normal\hspace{0.17em}distance of 1}.\text{51 cm}.\end{array}$

Q.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^o. The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable?

Ans.

\begin{array}{l}\text{Here,}\text{Number of horizontal wires},\text{n}=\text{4}\\ \text{Current flowing}\text{in each wire},\text{I}=\text{1}.0\text{A}\\ \text{Earth}’\text{s magnetic field at the}\text{given}\text{place},\text{R}=0.\text{39 G}\\ =0.\text{39}\times \text{1}{0}^{-\text{4}}\text{T}\\ \text{Angle of dip at the given}place,\delta =\text{35}°\\ \text{Angle of declination}\text{at}\text{the}\text{given}\text{place},\theta \sim 0°\\ \text{At}\text{a point 4 cm below the cable}:\\ \text{Distance},\text{r}=\text{4 cm}\\ =0.0\text{4 m}\\ \\ \text{The horizontal component of earth}’\text{s magnetic field,}\\ {\text{H}}_{\text{h}}=\text{Rcos}\delta -\text{B}\\ \text{Here},\\ \text{B}=\text{Magnetic field at 4 cm because}\text{of current I in the}\\ \text{four}\text{wires}=\text{4}\times \frac{{\mu }_{o}I}{2\pi r}\\ {\mu }_o=\text{Permeability of free space}=\text{4}\pi \times \text{1}{0}^{-\text{7}}{\text{Tm A}}^{-\text{1}}\\ \therefore B=\text{4}\times \frac{\text{4}\pi \times \text{1}{0}^{-\text{7}}\times 1}{2\pi \times 0.0\text{4}}\\ =0.\text{2}\times \text{1}{0}^{-\text{4}}\text{T}\\ =0.\text{2 G}\\ \therefore H_h=0.\text{39 cos 35}°-0.\text{2}\\ =0.\text{39}\times 0.\text{819}-0.\text{2}\approx 0.\text{12 G}\\ \text{Vertical component of earth}’\text{s magnetic field,}\\ {\text{H}}_{\text{v}}=\text{Rsin}\delta \\ =0.\text{39 sin 35}°\\ =0.\text{22 G}\\ \text{The angle made by the field with its horizontal}\\ \text{component is given by,}\theta \text{=}{\text{tan}}^{\text{-1}}\frac{{\text{H}}_{\text{v}}}{H_h}\\ ={\text{tan}}^{\text{-1}}\frac{0.\text{22}}{0.\text{12}}\\ ={61.39}^o\\ \\ \text{Resultant magnetic}\text{field at the point is given by,}\\ {\text{H}}_{\text{1}}=\sqrt{{\left({\text{H}}_{\text{v}}\right)}^2+{\left({\text{H}}_h\right)}^2}\\ =\sqrt{{\left(0.22\right)}^2+{\left(0.12\right)}^2}\\ =0.25G\\ \text{At a point 4 cm above the cable}:\\ \text{Horizontal component of earth}’\text{s magnetic fieldis}\text{given}\\ \text{by,}\\ {\text{H}}_{\text{h}}=\text{Rcos}\delta +\text{B}\\ =0.\text{39 cos 35}°+0.\text{2}\\ =0.\text{52 G}\\ \text{Vertical component of earth}’\text{s magnetic field}\text{is}\text{given}\text{by,}\\ {\text{H}}_{\text{v}}=\text{Rsin}\delta \\ =0.\text{39 sin 35}°\\ =0.\text{22 G}\\ \text{Angle},\theta ={\tan }^{-1}\frac{{\text{H}}_{\text{v}}}{{\text{H}}_{\text{h}}}\\ \text{=}{\tan }^{-1}\frac{0.22}{0.\text{52}}\\ \text{=22}.\text{9}°\\ \text{Resultant magnetic}\text{field}\text{at}\text{the}\text{point}\text{is}\text{given}\text{by,}\\ {\text{H}}_2=\sqrt{{\left({\text{H}}_{\text{v}}\right)}^2+{\left({\text{H}}_h\right)}^2}\\ =\sqrt{{\left(0.22\right)}^2+{\left(0.52\right)}^2}\\ =0.56G\end{array}

Q.20 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^o 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^o in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.

Ans.

Here, number of turns in the circular coil, N = 30
Radius of circular coil, r = 12 cm = 0.12 m
Current in the circular coil, I = 0.35 A
Angle of dip, δ = 45°

$\begin{array}{l}\left(\text{a}\right)\text{Magnetic field due to current I},\text{at a distance r},\\ \text{is given by\hspace{0.17em}the\hspace{0.17em}relation},\\ \text{B}=\frac{{\mathrm{\mu }}_{\mathrm{o}}2\mathrm{\pi NI}}{4\mathrm{\pi r}}\\ \text{Here},{\mathrm{\mu }}_{\mathrm{o}}=\text{Permeability of free space}\\ =\text{4}\mathrm{\pi }\times \text{1}{0}^{-\text{7}}{\text{Tm A}}^{-\text{1}}\\ \therefore \text{B}=\frac{\text{4}\mathrm{\pi }\times \text{1}{0}^{-\text{7}}\times 2\mathrm{\pi }\times 30\times 0.35}{\text{4}\mathrm{\pi }\times 0.12}\\ =\text{5}.\text{49}\times \text{1}{0}^{-\text{5}}\text{\hspace{0.17em}T}\\ \text{As\hspace{0.17em}}\mathrm{t}\text{he compass needle points from West to East}.\\ \therefore \text{The horizontal component of\hspace{0.17em}earth}’\text{s magnetic}\\ \text{field is given as}:\\ {\text{B}}_{\text{H}}=\text{Bsin}\mathrm{\delta }\\ =\text{5}.\text{49}\times \text{1}{0}^{-\text{5}}\text{sin 45}\mathrm{°}\\ =\text{3}.\text{88}\times \text{1}{0}^{-\text{5}}\text{T}\end{array}$

(b) When the current in the coil is reversed and the coil is rotated by an angle of 90^o anticlockwise, the needle will reverse its original direction (i.e., the needle will point from East to West).

Q.21 A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60^o, 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^o with this field, what is the magnitude of the other field?

Ans.

Magnitude of one of the magnetic fields, B₁=1.2 × 10⁻² T

Magnitude of the other magnetic field = B₂

Angle between the two fields, θ = 60°

In equilibrium, the angle between the dipole and field B₁, θ₁ = 15°

Angle between the dipole and field B₂, θ₂ = θ − θ₁

= 60° − 15°

= 45°

\begin{array}{l}\text{At equilibrium},\text{torques between the two fields must}\\ \text{balance each other}.\\ \therefore {\text{Torque due to field B}}_{\text{1}}={\text{Torque due to field B}}_{\text{2}}\\ {\text{MB}}_{\text{1}}\text{sin}{\theta }_{\text{1}}={\text{MB}}_{\text{2}}\text{sin}{\theta }_{\text{2}}\\ \text{Here},\text{M}=\text{Magnetic moment of the dipole}\\ \therefore {\text{B}}_{\text{2}}=\frac{{\text{B}}_{\text{1}}\text{sin}{\theta }_{\text{1}}}{\text{sin}{\theta }_{\text{2}}}\\ =\frac{\text{1}.\text{2}\times \text{1}{0}^{-\text{2}}\times \text{sin}{15}^o}{\text{sin}{45}^o}\\ =4.39\times \text{1}{0}^{-3}T\\ \therefore \text{The magnitude of the other magnetic field is}\\ \text{4}.\text{39}\times \text{1}{0}^{-\text{3}}\text{T}.\end{array}

Q.22 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 × 10⁻¹⁹ 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 the electron beam from the electron gun to the screen in a TV set.]

Ans.

Here, energy of the electron beam, E = 18 keV

= 18 ×10³ eV

Charge on electron, e = 1.6 × 10⁻¹⁹ C

E = 18 × 10³ × 1.6 × 10⁻¹⁹ J

Magnetic field, B = 0.04 G

Mass of electron, m_e = 9.11 × 10⁻¹⁹ kg

Distance up to which the electron beam travels, d = 30 cm = 0.3 m

\begin{array}{l}\text{Kinetic energy of the electron beam is}\text{given}\text{by,}\\ E=\frac{1}{2}m{v}^2\\ \therefore v=\sqrt{\frac{2E}{m}}\\ =\sqrt{\frac{2\times 18\times {10}^3\times 1.6\times {10}^{-19}\times {10}^{-15}}{9.11\times {10}^{-31}}}\\ =0.795\times {10}^{8}m{s}^{-1}\\ \text{The electron beam bends along a circular path of}\\ \text{radius},\text{r}.\\ \text{The force due to magnetic field balances the}\\ \text{centripetal force of the}\text{circular path}.\\ \text{Let the up and down deflection of the electron beam,}\text{x}\\ \text{=r}\left(1-\cos \theta \right)\\ \text{Here},\theta =\text{Angle of declination}\\ \text{sin}\theta \text{=}\frac{d}{r}\\ =\frac{0.3}{11.3}\\ \therefore \theta ={\text{sin}}^{-1}\left(\frac{0.3}{11.3}\right)={1.521}^o\\ \therefore x=11.3\left(1-\cos {1.521}^o\right)\\ =0.0039m\\ =3.9mm\\ \therefore \text{The up and down deflection of the beam is 3}.\text{9 mm}.\end{array}

Q.23 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)

Ans.

Here, number of atomic dipoles, n = 2.0 × 10²⁴

Dipole moment of each dipole, M = 1.5 × 10⁻²³ J T⁻¹

For the magnetic field, B₁ = 0.64 T

The sample is cooled to the temperature, T₁ = 4.2° K

Total dipole moment of the sample, M_tot = n × M

= 2 × 10²⁴ × 1.5 × 10⁻²³

= 30 JT⁻¹

As, magnetic saturation achieved = 15%.

\begin{array}{l}\therefore \text{Effective dipole moment},\\ M_1=\frac{15}{100}\times 30\\ =4.5J{T}^{-1}\\ \text{When the magnetic field},{\text{B}}_{\text{2}}=0.\text{98 T}\\ \text{Temperature},{\text{T}}_{\text{2}}=\text{2}.\text{8}°\text{K}\\ \text{Total dipole moment}\text{of}\text{the}\text{sample}={\text{M}}_{\text{2}}\\ \text{According to Curie}’\text{s law},\text{the ratio of two magnetic}\\ \text{dipoles is}\text{given}\text{as}:\\ \frac{{\text{M}}_{\text{2}}}{M_1}=\frac{B_2}{B_1}\times \frac{{\text{T}}_1}{{\text{T}}_{\text{2}}}\\ \therefore {\text{M}}_{\text{2}}=\frac{B_2{\text{T}}_1{M}_1}{B_1{\text{T}}_{\text{2}}}\\ =\frac{0.\text{98}\times 4.2\times 4.5}{\text{2}.\text{8}\times 0.64}\\ =10.336J{T}^{-1}\\ \therefore T\text{otal dipole moment of the sample for}\text{a magnetic}\\ \text{field of}0.\text{98 T and a temperature of 2}.\text{8 K=10}\text{.336}{\text{JT}}^{\text{-1}}\end{array}

Q.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 magnetising current of 1.2 A?

Ans.

\begin{array}{l}\text{Here},\text{mean radius of the Rowland ring},\text{r}=\text{15 cm}\\ =0.\text{15 m}\\ \text{Number of turns on the ferromagnetic core},\text{N}=\text{35}00\\ \text{Relative permeability of the core material},{\mu }_{\text{r}}=\text{8}00\\ \text{Magnetising current},\text{I}=\text{1}.\text{2 A}\\ \text{Magnetic field is given by the relation}:\\ \text{B=}\frac{{\mu }_{\text{r}}{\mu }_{o}IN}{2\pi r}\\ \text{Here},{\mu }_0=\text{Permeability of free space}\\ \therefore B=\frac{\text{8}00\times \text{4}\pi \times \text{1}{0}^{-\text{7}}\times \text{1}.\text{2}\times \text{35}00}{2\pi \times 0.15}\\ \text{=4}\text{.48 T}\\ \therefore \text{The magnetic field in the core is 4}.\text{48 T}.\end{array}

Q.25

$\begin{array}{l}{\text{The magnetic moment vectors μ}}_{\text{s}}{\text{and μ}}_{\text{l}}\\ \text{associated with the intrinsic spin angular}\\ \text{momentum}\overrightarrow{\text{S}}\text{and orbital angular momentum}\overrightarrow{\text{l}}\text{,}\\ \text{respectively, of an electron are predicted by}\\ \text{quantum theory (andverified experimentally}\\ \text{to a high accuracy) to be given by:}\\ {\text{μ}}_{\text{s}}\text{= –}\left(\text{e/m}\right)\overrightarrow{\text{S}}\text{,}\\ {\text{μ}}_{\text{l}}\text{= –}\left(\text{e/2m}\right)\overrightarrow{\text{l}}\\ \text{Which of these relations is in accordance with}\\ \text{the result expected classically?Outline the}\\ \text{derivation of the classical result.}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Out}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{two}\mathrm{relations},\mathrm{the}\mathrm{magnetic}\mathrm{moment}\\ \mathrm{associated}\mathrm{with}\mathrm{orbital}\mathrm{angular}\mathrm{momentum}\mathrm{is}\mathrm{valid}\mathrm{with}\\ \mathrm{classical}\mathrm{physics}.\\ \text{Magnetic moment associated with orbital angular}\\ \text{momentum is given by,\hspace{0.17em}}{\mathrm{\mu }}_{\text{l}}=-\left(\frac{\mathrm{e}}{2\mathrm{m}}\right)\mathrm{l}\\ \text{For current i and area of cross}-\text{section A},\mathrm{the}\mathrm{relation}\\ \mathrm{for}\mathrm{magnetic}\mathrm{moment}\mathrm{is}\mathrm{given}\mathrm{by},\\ {\mathrm{\mu }}_{\text{l}}=\mathrm{iA}\\ \mathrm{From}\mathrm{the}\mathrm{definitions}\mathrm{of}{\mathrm{\mu }}_{\text{l}}\mathrm{and}\mathrm{A},\mathrm{it}\mathrm{follows}\mathrm{that}\\ {\mathrm{\mu }}_{\text{l}}=\left(\frac{-\mathrm{e}}{\mathrm{T}}\right){\mathrm{\pi r}}^2\to \left(\mathrm{i}\right)\\ \text{Here},\text{e}=\text{Charge of the electron\hspace{0.17em}}\\ \text{r}=\text{Radius of the orbit}\\ \text{T}=\text{Time period\hspace{0.17em}of\hspace{0.17em}revolution\hspace{0.17em}of\hspace{0.17em}electron}\\ \text{Orbital\hspace{0.17em}}\mathrm{a}\text{ngular momentum},\text{l}=\text{mvr}\\ \mathrm{l}=\mathrm{m}\times \frac{2\mathrm{\pi r}}{\mathrm{T}}\times \mathrm{r}\to \left(\mathrm{ii}\right)\\ \\ \text{Here},\text{m}=\text{Mass of electron}\\ \text{v}=\text{Velocity of electron}\\ \text{r=\hspace{0.17em}Radius\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}orbit}\\ \text{Dividing equation}\left(\mathrm{i}\right)\text{by equation}\left(\mathrm{ii}\right),\text{we get}:\\ \frac{{\mathrm{\mu }}_{\text{l}}}{\mathrm{l}}=-\left(\frac{\mathrm{e}}{2\mathrm{m}}\right)\\ \therefore {\mathrm{\mu }}_{\text{l}}=-\left(\frac{\mathrm{e}}{2\mathrm{m}}\right)\mathrm{l}\\ \therefore \mathrm{The}\mathrm{magnetic}\mathrm{moment}\mathrm{associated}\mathrm{with}\mathrm{orbital}\mathrm{angular}\\ \mathrm{momentum}\mathrm{is}\mathrm{valid}\mathrm{with}\mathrm{classical}\mathrm{physics}.\end{array}$