# NCERT Solutions for Class 12 Physics Chapter 5 - Magnetism and Matter

**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 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?**

**Sol.** (a) There are three independent quantities required to specify the earth’s magnetic field which are:

Magnetic declination, angle of dip and horizontal component of earth’s magnetic field, they are called the magnetic elements of the earth.

(b) We can expect a greater value of angle of dip in Britain because Britain is located close to north pole in Britain the value of angle of dip is about 70°.

(c) Melbourne is situated in southern hemisphere where the north pole of earth’s magnetic field lies. So, the magnetic field lines seem to come out of the ground as magnetic lines of force emerges from North pole and enter in south pole.

(d) At the poles, the earth’s magnetic field is exactly vertical. The compass needle is always free to rotate in horizontal plane only so it may point out in any direction.

(e) Dipole of magnetic moment M = 8 × 10^{22} J/T.

We consider that at a point on equatorial line of short magnetic dipole for which distance d = R (radius of earth).

Radius of earth R = 6400 km = 6.4 × 10^{6} m

$$\text{Magnetic field, B =}\frac{\mu_{0}}{4\pi}.\frac{\text{M}}{d^{3}}\\=10^{\normalsize-7×}\frac{8×10^{22}}{(6.4×10^{6})^{3}}$$

= 0.31 × 10^{–4}T = 0.31G

This value is same as that of earth’s magnetic field.

(f) The earth’s magnetic field is only due to the dipole field. As there are several local N-S poles may exist oriented in different directions, so they may nullify the effect of each other. These local N-S poles may occurs due to the deposition of magnetized minerals.

**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 year. How can geologists known about the earth’s field in such distance past?**

**(e) ****The ****earth’s field departs from its dipole shape substantially at large distances (greater than about 30000 km). What agencies may be responsible for this distortion?**

**(f ****) Interstellar space has on extremely weak magnetic field of the order of 10–12T. Can such a weak field be of any significant consequence? Explain.**

**[Note ****Q. 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.]**

**Sol.** (a) Yes, the earth’s magnetic field varies from point to point in space with time. It may change daily, annually or secularly with period of the order of about 1000 year. It may change irregularly during magnetic storms etc. The time scale for appreciable change it about few hundred years.

(b) The earth’s core contains iron in the molten state. The molten iron is not ferromagnetic material in nature thus, it cannot be treated as a source of earth’s magnetism.

(c) The source of energy to sustain these currents may be the radioactive material in the interior of the earth.

(d) On solidification of certain rocks, it is recorded that the field was too weak. The analysis of these rocks may give the history of direction of field.

(e) The reason for this distortion may be the motion of ions in the earth’s ionosphere. The earth’s magnetic field may get changed by the field due to the motion of ions in earth’s atmosphere.

(f) When a charged particle moves in a magnetic field, it moves along a circular path.

The necessary centripetal force is provided by the magnetic force.

$$\text{i.e,}\space\text{Bev}=\frac{mv^{2}}{r}\\\text{or}\space r=\frac{\text{mv}}{\text{Be}}$$

As B is less, r is more. So, in the interstellar space, they move in a circular path of a large radius. Thus, the deflection in their paths becomes negligible.

**3. 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 4.5 × 10 ^{–2}J. What is the magnitude of magnetic moment of the magnet?**

**Sol.** Given, magnetic field, B = 0.25 T

The torque t = 4.5 × 10^{–2}J

Angle between magnetic moment M and magnetic field B, θ = 30°.

Torque experienced on a magnet placed in external magnetic field

$$τ=\vec{\text{M}}×\vec{\text{N}}\\τ=\text{MB sin θ}\space\bigg(\because\vec{\text{A}}×\vec{\text{B}}=\text{AB}\space\text{sin}\space\theta\bigg)\\4.5 × 10^{\normalsize–2} = \text{M} × 0.25 × \text{sin} 30°\\\text{M}=\frac{4.5×10^{-2}}{0.25×\text{sin}\space30\degree}\\=\frac{4.5×10^{-2}×2}{0.25×1}\space\bigg(\because\text{sin}30\degree=\frac{1}{2}\bigg)$$

= 0. 36 J/T

Thus, the magnitude of magnetic moment is 0.36 J/A.

**4. A short bar magnet of magnetic moment ****m = 0.32 J/T 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 magnetic in each case?**

**Sol.** Here, magnetic moment, m = 0.32 J/T

The magnitude of magnetic field B = 0.15T

(a) In stable equilibrium, the angle between magnetic moment (M) and magnetic field (B) is θ = 0°

[∵ In this position, it will be in a direction parallel to magnetic field thus no torque will act on it.]

∵ The potential energy of the magnet

$$\text{U}=-\vec{m}.\vec{B}\\=-\text{= – mB cos θ}\\(\because \vec{\text{A}}×\vec{\text{B}}=\text{AB cos θ})\space\\=-0.32×0.15\space\text{cos}\space0\degree\\=-4.8\space×10^{\normalsize-2}\text{J}$$

(b) In unstable equilibrium, the angle between the magnetic moment and magnetic field is 180°. Hence, maximum torque will act on it.

θ = 180°

Potential energy of the magnet

U = – mB cos 180°

= – 0.32 × 0.15 (–1) = 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 the associated magnetic moment?**

**Sol.** Given, total number of turns n = 800

Area of solenoid A = 2.5 × 10^{–4} m^{2}

Current through solenoid I = 3A

When a current passes through a solenoid, then a magnetic field is produced. Using Maxwell’s right hand grip rule, the direction of magnetic field is along the axis of the solenoid. Using the formula of magnetic moment.

M = nIA

M = nIA

= 800 × 3 × 2.5 × 10^{–4}

= 0.6T

**6. If the solenoid in Q. 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?**

**Sol.**

From the Q. 5, we get

Magnetic moment M = 0.6 J/T

Torque acting on the solenoid

τ = MB sin q = 0.6 × 0.25 sin 30°

$$= 0.6×0.25×\frac{1}{2}$$

= 0.075 N-m

Thus, the magnitude of torque on the solenoid is 0.075 N-m.

**7. A bar magnet of magnetic moment 1.5 J/T 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)?**

**Sol.** Given, magnetic moment of magnet

M = 1.5 J/T

Uniform magnetic field B = 0.22T

(a) (i) Angle θ_{1} = 0° (∵ The magnet lies aligned in the direction of field)

and θ_{2} = 90° (∵ The magnet is aligned normal to the field direction)

Work done in rotating the magnet from angle θ_{1} to θ_{2}

W = – MB (cos θ_{2} – cos θ_{1})

= – 1.5 × 0.22 (cos 90° – cos 0°)

= 0.33 J

(ii) θ1 = 0° and θ2 = 180° (i.e., Magnet is to be aligned opposite to the direction of field)

Work done = – MB (cos θ2 – cos θ1)

= – 1.5 × 0.22 (cos 180° – cos 0°)

= 0.66J

(b) Using the formula of torque

t = MB sin θ

(i) θ = 90° (when magnetic moment M normal to the field B)

t = 1.5 × 0.22 sin 90° = 0.33 N-m

(ii) θ = 180° (when magnetic moment opposite to the field B)

t = 1.5 × 0.22 sin 180° = 0

**8. A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10 ^{–4} m^{2}, carrying a current of 4.0A, 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 are the force and torque on the solenoid, if a uniform horizontal magnetic field of 7.5 × 10–2 T is set up at the angle of 30° with the axis of the solenoid?**

**Sol.** (a) Magnetic moment

M = nIA = 2000 × 4 × 1.6 × 10^{–4}

= 1.28 J/T

(b) The net force on the solenoid is zero, because two equal and opposite forces acting, but their lines of action are parallel so they form a couple thus a torque is applied on it.

Torque on the solenoid

t = MB sin q (here q = 30°)

= 1.28 × 7.5 × 10^{–2} sin 30°

$$= 1.28 × 7.5 × 10^{\normalsize–2}×\frac{1}{2}$$

= 4.8 × 10^{–2} N-m

**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 × 10–2T. 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. What is the moment of inertia of the coil about its axis of rotation?**

**Sol.** Given, number of turns, n = 16

Radius, r = 10 cm = 0.1 m

Current, I = 0.75A

Magnetic field, B = 5.0 × 10^{–2}T

Frequency, f = 2/s

Magnetic moment of the coil,

M = nIA = 16 × 0.75 × π(0.1)^{2}

= 16 × 0.75 × 3.14 × 0.1 × 0.1

= 0.377 J/T

Frequency of oscillation of the coil

$$f=\frac{1}{2\pi}\sqrt{\frac{\text{M×B}}{\text{I}}}$$

where I = Moment of inertia of the coil.

Squaring on both the sides, w

e get

$$f^{2}=\frac{1}{4\pi^{2}}.\frac{\text{MB}}{\text{I}}\\\Rarr\space \text{I}=\frac{\text{MB}}{4\pi^{2}f^{2}}\\=\frac{0.377×5×10^{\normalsize-2}}{4×3.14×3.14×2×2}$$

= 1.2 × 10^{–4} kg-m^{2}

Thus, the moment of inertia of the coil is 1.2 × 10^{–4} kg-m^{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° 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.Sol. **Given, angle of dip δ= 22°

Horizontal component of the earth’s magnetic field H = 0.35 G

Let the magnitude of the earth’s magnetic field at the place is R.

Using the formula,

H = R cos δ

$$\text{or \space R =}\frac{\text{H}}{\text{H}\delta}=\frac{0.35}{\text{cos}22\degree}=\frac{0.35}{0.9272}$$

= 0.38 G

Thus, the value of the earth’s magnetic field at that place is 0.38 G.

**11. At a certain location in Africa, a compass points 12° 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° 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 field at the location.**

**Sol.** Here, angle of declination θ = 12° west

Angle of dip δ = 60°

Horizontal component of earth’s magnetic field

H = 0.16 G

Let the magnitude of earth’s magnetic field is R.

We know that,

H = R cos δ

$$\text{or}\space\text{R}=\frac{\text{H}}{\text{cos}\delta}=\frac{0.16}{\text{cos}\space60\degree}=\frac{0.16×2}{1}$$

= 0.32G = 0.32 × 10^{–4}T

The earth’s magnetic field lies in a vertical plane 12° west of geographical meridian at an angle 60° above the horizontal.

**12. A short bar magnet has a magnetic moment of 0.48 J/T. 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.**

**Sol.**

(a) Magnetic field at point P when the point lies on the axial line.

$$\text{B}=\frac{\mu_{0}}{4\pi}.\frac{2\text{M}}{d^{3}}=\frac{10^{\normalsize-7}×2×0.48}{(0.1)^{3}}$$

= 0.96 × 10^{–4}T

The direction of magnetic field is along the direction of magnetic moment. Thus, the direction of magnetic field is from S to N pole of the magnet.

(b)

∴ Magnetic field at point P (on equatorial line)

$$\text{B}=\frac{\mu_{0}}{4\pi}.\frac{2\text{M}}{d^{3}}\\=10^{\normalsize-7}×\frac{0.48}{(0.1)^{3}}$$

= 0.48 × 10^{–4}T.

The direction of magnetic field B on equatorial line is opposite to the direction of magnetic moment M. Thus, the direction of magnetic field is from N to S pole

**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 0.36G 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.)**

**Sol.** The earth’s magnetic field where the angle of dip is zero, is the horizontal component of earth’s magnetic field.

i.e., H = 0.36 G

The magnetic field on axial line.

$$\text{B}_{1}=\frac{\mu_{0}}{4\pi}.\frac{2m}{d^{3}}$$

This magnetic field is equal to the horizontal component of earth’s magnetic field.

i.e., B_{1} = H …(i)

On the equatorial line of magnetic field due to the magnet

$$\text{B}_{2}=\frac{\mu_{0}}{4\pi}.\frac{m}{d^{3}}=\frac{\text{B}_{1}}{2}=\frac{\text{H}}{2}\qquad...(\text{ii})$$

The total magnetic field on equatorial line at this point.

$$\text{B = B}_{2} + \text{B}_{1}=\frac{\text{H}}{2}+\text{H}\\=\frac{3}{2}\text{H}=\frac{3}{2}×0.36=0.54\text{G}$$

The direction of magnetic field is in the direction of earth’s field.

**14. If the bar magnet in Q. 13 is turned around by 180°, where will the new null points be located?**

**Sol.** When the bar magnet is turned by 180°, then the null points are obtained on the equatorial line.

So, magnetic field on the equatorial line at distance d’ is

$$\text{B'}=\frac{\mu_{0}}{4\pi}.\frac{m}{d^{3}}$$

This magnetic field is equal to the horizontal component of earth’s magnetic field

B’ = H …(i)

From the Q. 13,

$$\text{Magnetic field B}_1 =\frac{\mu_{0}}{4\pi}.\frac{2m}{d^{3}}=\text{H}\qquad\text{...(ii)}$$

Solving equations (i) and (ii), we get

$$\frac{\mu_{0}}{4\pi}.\frac{m}{d^{3}}=\frac{\mu_{0}}{4\pi}.\frac{2m}{d^{3}}\\\text{or}\qquad\frac{1}{d^{3}}=\frac{2}{d^{3}}\\\text{or}\qquad\text{d}^{3}=\frac{d^{3}}{2}=\frac{(14)^{3}}{2}\\\text{or}\space d=\frac{14}{(2)^{1/3}}=11.1\space\text{cm}$$

Hence, the null points are located on the equatorial line at a distance of 11.1 cm.

**15. A short bar magnet of magnetic moment 5.25 × 10 ^{–2} J/T 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° 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 0.42G. Ignore the length of the magnet in comparison to the distances involved.**

**Sol.** Given, magnetic moment m = 5.25 × 10^{–2} J/T

Let the resultant magnetic field be B_{net}. It makes an angle of 45° with B_{e}.

Here B_{e} = 0.42 G = 0.42 × 10^{–4}T

(a) At normal bisector: The magnetic field at point P, due to a short magnet

$$\text{B}=\frac{\mu_{0}}{4\pi}.\frac{m}{r^{3}}\qquad\text{...(i)}$$

The direction of B is along PA, i.e., along N pole to S pole.

Using vector analysis,

$$\text{B}=\frac{\mu_{0}}{4\pi}.\frac{m}{r^{3}}\qquad\text{...(i)}$$

The direction of B is along PA, i.e., along N pole to S pole.

Using vector analysis,

$$\text{tan 45°}=\frac{\text{B sin 90\degree}}{\text{B cos 90\degree} + \text{B}_{e}}\\\text{I}=\frac{\text{B}}{\text{B}_{e}}\\\text{or B = B}_{e}\\\text{0.42 × 10}^{\normalsize–4} =\frac{\mu_{0}}{4\pi}.\frac{m}{r^{3}}\\\text{0.42×10}^{\normalsize -4}=\frac{10^{\normalsize-7}×5.25×10^{\normalsize-2}}{r^{3}}\\\text{r}^{3}=\frac{5.25×10^{\normalsize-9}}{0.42×10^{\normalsize-4}}=12.5×10^{\normalsize-5}$$

r = 0.05 m

or r = 5 cm

**(b) When point lies on axial line:** Let the resultant magnetic field B_{net} makes an angle 45° from earth’s magnetic field B_{e}. The magnetic field on the axial line of the magnet at a distance of r from the centre of magnet

$$\text{B'}=\frac{\mu_{0}}{4\pi}.\frac{2m}{r^{3}}$$

Direction of magnetic field is from S to N., Using vector analysis, we get

$$\text{tan 45\degree}=\frac{\text{B'\text{sin 90\degree}}}{\text{B'}\text{cos} 90\degree + \text{B}_{e}}$$

$$\text{I}=\frac{\text{B'}}{\text{B'}_{e}}\\\text{or B}_{e} = \text{B'}\\0.42×10^{\normalsize-4}=\frac{\mu_{0}}{4\pi}×\frac{2m}{r^{3}}\\\text{or}\space0.42×10^{\normalsize-4}=\frac{10^{\normalsize-7}×2×5.25×10^{\normalsize-2}}{r^{3}}\\\text{r}^{3}=\frac{10^{\normalsize-9}×2×5.25}{0.42×10^{\normalsize-4}}=2.5×10^{\normalsize-5}$$

r = 0.063 or 6.3 cm