NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter
Q. Consider the two idealised systems: (i) a parallel plate capacitor with large plates and small separation and (ii) a long solenoid of length L >> R, radius of cross-section. In (i) E is ideally treated as a constant between plates and zero outside. In (ii) magnetic field is constant inside the solenoid and zero outside. These idealised assumptions, however, contradict fundamental laws as below:
- (a) case (i) contradicts Gauss’ law for electrostatic fields.
- (b) case (ii) contradicts Gauss’ law for magnetic fields.
- (c) case (i) agrees with $$\oint\text{E.dl}=0$$
- (d) case (ii) contradicts$$\oint\text{H.dl}=\text{I}_{em}$$
- Ans. (b) case (ii) contradicts Gauss’ law for magnetic fields.
(ii) In which configuration is the system in (a) stable and (b) unstable equilibrium ?
(iii) Which configuration corresponds to the lowest potential energy among all the configuration shown ?
Ans. Potential energy of the configuration arises due to the potential energy of one dipole (say, Q) in the magnetic field due to (P). The field due to P is given by the expression
$$\text{B}_\text{P}=-\frac{\mu_0}{4\pi}\frac{m_p}{r^{3}}\\\text{(on the normal bisector)}\\\text{B}_\text{P} =\frac{\mu_02}{4\pi}\frac{m_p}{r^{3}}\space\text{(on the axis)}$$
where mP is the magnetic moment of the dipole
P.
Equilibrium is stable when mQ is parallel to BP, and unstable when it is anti-parallel to BP. For the configuration Q3 for which Q is along the perpendicular bisector of the dipole P, the magnetic moment of Q is parallel to the magnetic field at the position 3. Hence, Q3 is stable. Thus,
Q. Draw magnetic field lines when a (i) diamagnetic, (ii) paramagnetic substance is placed in an external magnetic field. Which magnetic property distinguishes this behaviour of the field lines due to the two substances ?
Ans. Figure 1 shows magnetic field lines when (i) diamagnetic materials are placed in external magnetic field.
Figure 2 shows magnetic field lines, when (ii) paramagnetic materials are placed in external magnetic field.
Magnetic dipole moment of materials are responsible for this diamagnetic material has zero magnetic moment.
Paramagnetic material has non-zero magnetic moment.
- Q. Answer the following questions :
- (i) Why does a paramagnetic sample display greater magnetisation (for the same magnetising field, when cooled ?
- (ii) Why is diamagnetism, in contrast almost independent of temperature ?
- (iii) 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?
- (iv) Is the permeability of a ferromagnetic material independent of the magnetic field? If not, is it more for lower or higher fields ?
- (v) 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 ?
- (vi) Would the maximum possible magnetisation of a paramagnetic sample be of the same order of magnitude as the magnetisation of a ferromagnet ?
- Ans. (i) Due to the random thermal motion of molecules, the alignments of dipoles get disrupted at high temperatures. On cooling, this disruption is reduced. Hence, a paramagnetic sample displays greater magnetisation when cooled.
- (ii) The induced dipole moment in a diamagnetic substance is opposite to the magnetising field always. Hence, the internal motion of the atoms does not affect the diamagnetism of a material.
- (iii) Bismuth is a diamagnetic substance. Hence, a toroid with a bismuth core has a magnetic field slightly lesser than a toroid whose core is empty.
- (iv) The permeability of ferromagnetic materials is dependent on the applied magnetic field. It is greater for a lower field w.r.t. higher magnetic field.
- (v) With the refraction of magnetic field lines as µ_{r} increases the normal component of magnetic field at surface of magnetised material increases.
- For ferromagnetic materials µ_{r} >> 1. Therefore, the tangential magnetisation near the surface is very small.
- Thus, magnetic lines on the surface are nearly normal.
- (vi) The maximum possible magnetisation of a paramagnetic sample can be of the same order of magnitude as the magnetisation of a ferromagnet. This requires high magnetising fields for saturation.
Ans. The time period of oscillation is
$$\text{T}=\frac{6.70}{10}=0.67s\\\text{Relation between moment of inertia and magnetic field}\\\text{B}=\frac{4\pi^2I}{mT^{2}}\\=\frac{4×(3.14)^2×7.5×10^{-6}}{6.7×10^{-2}×(0.67)^{2}}\\=0.01\text{T}$$
Q. 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 (i) the axis, (ii) the equatorial lines (normal bisector) of the magnet.
Ans. Given, Magnetic moment (M) = 0.48 J T^{– 1}
(i) Distance (d) = 10 cm = 0.1 m
The magnetic field at distance (d), from the centre of the magnet along the axis is given by
$$\text{B}=\frac{\mu_0}{4\pi}\frac{2M}{d ^{3}}\\\text{Where},\mu_0=\text{Permeability of free space}\\=4\pi×10^{-7}\text{TmA}^{-1}\\\therefore\space\text{B}=\frac{4\pi×10^{-7}×2×0.48}{4\pi×(0.1)^{3}}\\=0.96×10^{-4}T=0.96G\\\text{The magnetic field is along the S–N direction.}\\\text{(ii) The magnetic field at a distance of 10 cm along the equatorial line of the magnet is given as :}\\\text{B}=\frac{\mu_0×M}{4\pi×d^3}=\frac{4\pi×10^{-7}×0.48}{4\pi×(0.1)^{3}}\\=0.48G\\\text{The magnetic field is along the N–S direction.}$$
Q. 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.
- (i) Determine the horizontal component of the earth’s magnetic field at the location.
- (ii) 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.
- Ans. Given, Number of turns in coil (N) = 30
- Radius of the circular coil (r) = 12 cm = 0.12 m
- Current through coil (I) = 0.35 A
- Angle of dip (δ) = 45°
- (i) The magnetic field due to current I, at a distance r, is given by
$$\text{B}=\frac{\mu_02\pi\text{NI}}{4\pi r}\\\text{Where},\\\mu_0=\text{Permeability of free space =}4\pi×10^{-7}\\\text{B}=\frac{4\pi×10^{-7}×2\pi×30×0.35}{4\pi×0.12}\\\text{B}=5.49×10^{-5}$$
The compass needle points from west to East. Thus, the horizontal component of earth’s magnetic field is given by
B_{H} = B sin δ
= 5.49 × 10^{– 5} sin 45°
= 3.88 × 10^{– 5} T = 0.388 G
(ii) When the current through the coil is reversed and the coil gets rotated about its vertical axis by an angle of 90°, the needle will reverse its original direction. In this situation the needle will point from East to West.
Q. 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 currentcarrying cable is equal and opposite to the
horizontal component of earth’s magnetic field.)
Ans. Given, Current in the wire, I = 2.5 A
Angle of dip at the given location on earth,
δ = 10°
Earth’s magnetic field,
B_{E} = 0.33 G = 0.33 × 10^{– 4} T
The horizontal component of earth’s magnetic field is given as :
B_{H} = B_{E} cos δ
= 0.33 × 10^{– 4}× cos 10°
≅ 0.32 × 10– 4 T
The magnetic field at the neutral point at a distance R from the cable is given by the relation :
Neutral points = B_{net}= 0°
$$\text{B}_\text{H}=\frac{\mu_0I}{2\pi R}$$
Where,
$$\text{B}_\text{H}=\frac{\mu_0I}{2\pi R}$$
$$=\frac{4\pi×10^{-7}×2.5}{2\pi×0.33×10^{-4}}\\=15.15 × 10^{\normalsize – 3} m\\= 1.51\text{cm}\\\text{Hence, a set of neutral points parallel to and above the cable are located at a normal distance of 1.51 cm.}$$
$$=\frac{4\pi×10^{-7}×2.5}{2\pi×0.33×10^{-4}}\\=15.15 × 10^{\normalsize – 3} m\\= 1.51\text{cm}\\\text{Hence, a set of neutral points parallel to and above the cable are located at a normal distance of 1.51 cm.}$$
Q. A domain in ferromagnetic iron is in the form of a cube of side length 1 mm. Estimate the number of iron atoms in the domain and the maximum possible dipole moment and magnetisation of the domain. The molecular mass of iron is 55 g/mole and its density is 7.9 g/cm3. Assume that each iron atom has a dipole moment of 9.27 × 10– 24 A m2.
Ans. The volume of the cubic domain is
V = (10^{– 6} m)^{3} = 10^{– 18} m^{3}
= 10^{– 12} cm^{3}
Its mass is volume × density = 7.9 g cm^{– 3} × 10^{– 12} cm^{3} = 7.9 × 10^{– 12} g
It is given that Avogadro number (6.023 × 10^{23}) of iron atoms have a mass of 55 g. Hence, the number of atoms in the domain is
$$\text{N}=\frac{7.9×10^{\normalsize-12}×6.023×10^{23}}{55}\\\text{The maximum possible dipole moment m}_\text{max} \text{is achieved for the (unrealistic) case when all the atomic moments are perfectly aligned. Thus,}\\\text{m}_{max}=(8.65×10^{10})×(9.27×10^{-24})\\=8.0×10^{-13}\text{A m}^2\\\text{The consequent magnetisation is}\\\text{M}_{max}=\frac{m_{max}}{\text{Domain volume}}\\= 8.0 × 10^{\normalsize– 13} \text{Am}^2/10^{\normalsize– 18} m^{3}\\=8.0×10^{5}\text{Am}^{-1}$$
Ans. (i) The field H is dependent of the material of the core, and is
H = nI
= 1000 × 2.0 = 2 × 10^{3} A/m.
(ii) The magnetic field B is given by
B = µ_{r}µ_{0}H
= 400 × 4π × 10^{– 7}(N/A^{2})
× 2 × 10^{3} (A/m)
= 1.0 T
(iii) Magnetisation is given by
$$\text{M =}\frac{B-\mu_0H}{\mu_0}=\frac{\mu_r\mu_0H-\mu_0H}{\mu_0}\\=(\mu_r-1)H\\=399×H\\≅ 8×10^{5}A/m$$
$$\text{B =}\space\mu_r\mu_0(I+I_m).\\\text{using}\space\text{I}=2A,B=1T,\\\text{B}=\mu_r\mu_0nI\\=\mu_0n(I+I_m)\\\mu_rI=(I+I_m)\\\mu_rI=I+I_m\\I_m=(\mu_r-1)I\\=798A\\\text{Here}\space\mu_r\space\text{and}\space\chi\space \text{refer to the relative permeability and susceptibility.}$$
Q. Answer the following questions :
- (i) Explain qualitatively on the basis of domain picture the irreversibility in the magnetisation curve of a ferromagnet.
- (ii) 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 agnetisation, which piece will dissipate greater heat energy ?
- (iii) A system displaying a hysteresis loop such as a ferromagnet, is a device for storing memory ? Explain the meaning of this statement.
- (iv) What kind of ferromagnetic material is used for coating magnetic tapes in a cassette player or for building ‘memory stores’ in a modern computer ?
- (v) 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 figure.
- (i) It can be observed from the curve that magnetisation remains, even when the external field is removed. This reflects the irreversibility of a ferromagnet.
- (ii) The dissipated heat energy is directly proportional to the area of a hysteresis (B-H) loop. A carbon steel piece has a greater hysteresis curve area. Therefore, it dissipates greater heat energy.
- (iii) A ferromagnet has very high saturation, because of the high value of µ_{r}. Therefore, when it is magnetised it remains magnetised for long time even if external field is removed. Thus, it is said to have strong magnetic storage.
- (iv) Ferrites (MnFe_{2}O_{4,} FeFe_{2}O_{4}) is used for coating magnetic tapes in cassette players and for building memory stored in modern computers.
- (v) A certain region of space can be shielded from magnetic fields if it is surrounded by strong diamagnetic endosures.