Class 12 - Physics - Magnetism and Matter

**Question1.**

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 × 10^{22} 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?

Answer:

(a) The three independent quantities conventionally used for specifying earth’s magnetic field are:

(i) Magnetic declination,

(ii) Angle of dip, and

(iii)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°) 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

Magnetic field strength, B = (μ_{0}M)/ (4 π (r)^{ 3})

Where, μ_{0 }= Permeability of free space = 4π × 10^{−7} Tm A^{−1}

Therefore, B = (4π × 10^{−7} x8 x 10^{22})/ (4π × (6.4 x 10^{6})^{3})

=0.3 G

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 magnetised mineral deposit is an example of a local N-S pole.

**Question2.**

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.]

Answer:

- 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.
- 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.
- 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.
- 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.

**Question 3.**

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

Answer:

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, θ = 30°

Torque is related to magnetic moment (M) as:

T = MB sin θ

Therefore, M = (T)/ (sin θ)

= (4.5 x 10-2)/ (0.25 x sin 30^{0})

=0.36JT^{-1}

Hence, the magnetic moment of the magnet is 0.36 J T^{−1}

** Question 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?

Answer:

Given:

Moment of the bar magnet, M = 0.32 J T^{−1}

External 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 θ, between the bar magnet and the magnetic field is 0°.

Potential energy of the system = -MB cos θ

=-(0.32 x 0.15 cos 0^{0})

= - 4.8 x 10^{-2} J

- The bar magnet is oriented 180° to the magnetic field. Hence, it is in unstable equilibrium. θ = 180°

Potential energy = − MB cos θ

=-(0.32 x 0.15 cos 180^{0})

= - 4.8 x 10^{-2} J

**Question 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?

Answer:

Given:

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 = n I A

= (800 × 3 × 2.5 × 10^{−4})

= 0.6 J T^{−1}

**Question 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?

Answer:

Given:

Magnetic field strength, B = 0.25 T

Magnetic moment, M = 0.6 T^{−1}

The angle θ, between the axis of the solenoid and the direction of the applied field is 30°.

Therefore, the torque acting on the solenoid is given as:

τ = MB cos θ

=0.6 x 0.25 sin 30^{0}

=7.5 x 10^{-2} J

**Question 7.**

A bar magnet of magnetic moment 1.5 J 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 field direction?

(b) What is the torque on the magnet in cases (i) and (ii)?

Answer:

Given:

Magnetic moment, M = 1.5 J T^{−1}

Magnetic field strength, B = 0.22 T

- Initial angle between the axis and the magnetic field, θ
_{1}= 0°

Final angle between the axis and the magnetic field, θ_{2} = 90°

The work required to make the magnetic moment normal to the direction of magnetic field is given as:

W = -MB (cos θ_{2} – cos θ_{1})

= - 1.5 x 0.22 (cos 90^{0} – cos 0^{0})

=-0.33(0 -1)

=0.33J

Initial angle between the axis and the magnetic field, θ_{1} = 0°

Final angle between the axis and the magnetic field, θ_{2} = 180°

The work required to make the magnetic moment opposite to the direction of magnetic field is given as:

W = -MB (cos θ_{2} – cos θ_{1})

= - 1.5 x 0.22 (cos 180^{0} – cos 0^{0})

=-0.33(-1 -1)

=0.66J

- For case (i) : θ
_{2}= θ_{1 }= 90^{0}

Therefore torque τ = MB sin θ

= 1.5 x 0.22 sin 90^{0}

=0.33 J

- For case (ii) : θ
_{2}= θ_{1 }= 180^{0}

Therefore torque τ = MB sin θ

= MB sin 180^{0}

=0J

**Question 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?

Answer:

Given:

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 = 4 A

- The magnetic moment along the axis of the solenoid is calculated as:

M = n A I

= (2000 × 1.6 × 10^{−4} × 4)

= 1.28 Am^{2}

- Magnetic field, B = 7.5 × 10
^{−2}T

Angle between the magnetic field and the axis of the solenoid, θ = 30°

τ = MB sin θ

= (1.28 x 7.5 x 10^{-2} sin 30^{0})

= 4.8 x 10^{-2} Nm

Since the magnetic field is uniform, the force on the solenoid is zero. The torque on the solenoid is 4.8 x 10^{-2} Nm

**Question 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?

Answer:

Given:

Number of turns in the circular coil, n=16

Radius of the coil, r=10cm=0.1m

Current in the coil, I=0.75A

Magnetic field strength, B=5×10^{−2 }T

Cross-section of the coil, A = πr^{2} = π × (0.1)^{2} m^{2}

Frequency of oscillations of the coil, v = 2.0 s^{−1}

f=2s

M= (n I A) = (n I π (r)^{ 2})

=16×0.75×π (0.1)^{2}

=0.377J/T

Frequency is given as f = (1)/ (2π) √ (M×BI) where I = moment of inertia

Or f^{2} = (1/4π^{2}) (MBI)

∴I= (1/4π^{2}) ((MB)/ (f^{2}))

= (0.377×5×10^{−2})/ (4 × (3.14)^{2}×2×2)

=1.2×10^{−4}kgm^{2}

Hence the normal of inertia of the coil 1.2×10^{−4}kgm^{2}

**Question 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.

Answer:

Given:

Horizontal component of earth’s magnetic field, B_{H} = 0.35 G

Angle made by the needle with the horizontal plane = Angle of dip = δ = 22^{0}

Earth’s magnetic field strength = B

We can relate B and B_{H} as:

B_{H} = B cos θ

Therefore, B = (B_{H})/ (cos δ)

= (0.35)/ (cos 22^{0})

=0.377G

Hence, the strength of earth’s magnetic field at the given location is 0.377 G.

**Question 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.

Answer:

Given:

Angle of declination, θ = 12°

Angle of dip, δ = 60^{0}

Horizontal component of earth’s magnetic field, B_{H }= 0.16 G

Earth’s magnetic field at the given location = B

We can relate B and B_{H} as:

B_{H} = B cos δ

Therefore, B = (B_{H})/ (cos)

= (0.16)/ (cos 60^{0})

=0.32G

Earth’s magnetic field lies in the vertical plane, 12° west of the geographic meridian, making an angle of 60° (upward) with the horizontal direction. Its magnitude is 0.32 G.

**Question 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.

Answer:

Given:

Magnetic moment of the bar magnet, M = 0.48 J T^{−1}

(a) Distance, d = 10 cm = 0.1 m

The magnetic field at distance d, from the centre of the magnet on the axis is given by the relation:

B = (μ_{0}2M)/ (4 π (d)^{ 3})

Where, μ_{0 }= Permeability of free space = 4π × 10^{−7} Tm A^{−1}

Therefore, B = (4π × 10^{−7} x 2 x 0.48)/ (4π x (0.1)^{3}))

=0.96 x 10^{-4} T = 0.96 G

The magnetic field is along the S − N direction.

(b) The magnetic field at a distance of 10 cm (i.e., d = 0.1 m) on the equatorial line of the magnet is given as:

B = (μ_{0}M)/ (4 π (d)^{ 3})

= (4π × 10^{−7} x 0.48)/ (4π x (0.1)^{3})) =0.48 G

The magnetic field is along the N − S direction.

**Question 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.)

Answer:

Earth’s magnetic field at the given place, H = 0.36 G

AS the null points lie on the axis of the magnet,

Therefore, B_{axial} = (μ_{0}2M)/ (4 π (d)^{ 3}) = H

Where,

μ_{0}= Permeability of free space

M = Magnetic moment

The magnetic field at the same distance d, on the equatorial line of the magnet is given as:

B_{eq }= (μ_{0}M)/ (4 π (d)^{ 3}) = (H/2)

Total magnetic field, B = (B_{axial }+ B_{eq})

= (H + (H/2)) =0.36 + 0.18 =0.54 G

Hence, the magnetic field is 0.54 G in the direction of earth’s magnetic field.

Question 14.

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

Answer:

The magnetic field on the axis of the magnet at a distance d_{1} = 14 cm, can be written as:

B_{1} = (μ_{0}2M)/ (4 π (d_{1})^{3}) … (1)

Where,

M = Magnetic moment

μ_{0 }= Permeability of free space

H = Horizontal component of the magnetic field at d_{1}

If the bar magnet is turned through 180°, then the neutral point will lie on the equatorial line.

Hence, the magnetic field at a distance d_{2}, on the equatorial line of the magnet can be written as:

B_{1} = (μ_{0}M)/ (4 π (d_{2})^{3}) … (2)

Equating equations (1) and (2), we get:

(2)/ ((d_{1})^{3}) = (1)/ ((d_{2})^{3})

(d_{2}/d_{1})^{3} = (1/2)

Therefore, d_{2} = d_{1} x (1/2)^{1/3}

= 14 x 0.794

=11.1 cm

The new null points will be located 11.1 cm on the normal bisector.

**Question 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

- 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.

Answer:

Given:

Magnetic moment of the bar magnet, M = 5.25 × 10^{−2} J T^{−1}

Magnitude of earth’s magnetic field at a place, H = 0.42 G = 0.42 × 10^{−4} T

- Magnetic field at the equatorial point P at a distance R from the centre of the magnetic dipole on the normal bisector is given by:

B_{eq} = (μ_{0}M)/ (4 πR^{3})

Where μ_{0 }= Permeability of free space = 4π × 10^{−7} Tm A^{−1}

When the resultant field is inclined at 45° with earth’s field, B = H

Therefore, H = (μ_{0}M)/ (4 πR^{3}) = 0.42 × 10^{−4}

R^{3 }= (μ_{0}M)/ (0.42 × 10^{−4 }x 4 π)

= (4 π x 10^{-7} x 5.25 x 10^{-2})/ (4 π x 0.42 x 10^{-4})

=12.5 x 10^{-5}

Therefore, R =0.05m = 5cm

- The magnetic field at a distance R’ from the centre of the axis is given as:

B_{’} = (μ_{0}2M)/ (4 π(R’^{3}))

The resultant field is inclined at 45^{0} with earth’s field.

Therefore, B’ =H

H = (μ_{0}2M)/ (4 π(R’^{3}))

(R’^{3}) = (μ_{0}2M)/ (4 π x H)

= (4 π x 10^{-7} x 2 x 5.25 x 10^{-2})/ (4π x 0.42 x 10^{-4})

=25 x 10^{-}^{5}

R’ = (25 x 10^{-5})^{1/3}

Therefore R’ = 0.063 m = 6.3cm

.