Magnetism and Matter

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Exercise 5.1Bar Magnet, Earth's Magnetism and Magnetic Materials

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 pole? (e) The earth's field, it is claimed, roughly approximates the field due to a dipole of magnetic moment 8 × 10²² A·m² 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 opposite senses. How is such a thing possible at all?

Solution

(a) Three independent quantities to specify earth's magnetic field: 1. Magnetic declination (D): angle between geographic north and magnetic north 2. Angle of dip (I): angle made by earth's field with the horizontal 3. Horizontal component (H) of earth's magnetic field (b) Britain is closer to the magnetic north pole than southern India. The dip angle increases as we move from the magnetic equator to the magnetic poles. So the dip angle at Britain would be greater than 18°. (c) Melbourne (Australia) is in the southern hemisphere. The magnetic field lines of earth enter the southern hemisphere from below and go toward the magnetic north pole. So at Melbourne, the field lines would come out of the ground. (d) At the geomagnetic north pole, the magnetic field lines are directed vertically downward (toward earth's centre). A compass free to move in the vertical plane would point vertically downward. (e) Earth's dipole moment M = 8 × 10²² A·m². The magnetic field at the equator: B = μ₀M/(4πR³) where R = 6.4 × 10⁶ m = (10⁻⁷ × 8 × 10²²) / (6.4 × 10⁶)³ = 8 × 10¹⁵ / (2.62 × 10²⁰) ≈ 3 × 10⁻⁵ T = 0.3 G This matches the observed value of earth's field (≈ 0.3–0.6 G). ✓ (f) Local magnetic poles arise due to deposits of magnetic minerals (magnetite, Fe₃O₄) in the earth's crust. These can create local field variations with poles oriented differently from the main field.

Answer the following: (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) What are the dimensions of the susceptibility χ? (d) A strong magnet is brought near a neutral piece of iron. Is the piece attracted or repelled?

Solution

(a) Yes, earth's magnetic field changes with time. • Short-term variations: daily fluctuations due to ionospheric currents (hours) • Long-term variations: secular variation over centuries and millennia • The field reverses polarity on geological time scales (hundreds of thousands to millions of years) Appreciable change occurs over centuries to millions of years. (b) The earth's core temperature (above 1000°C) exceeds the Curie temperature of iron (~770°C). Above the Curie temperature, iron loses its ferromagnetic properties. So the iron in earth's core cannot be permanently magnetised. The earth's magnetism is believed to originate from electric currents flowing in the liquid outer core (geodynamo theory). (c) Susceptibility χ = M/H (magnetisation/magnetic field intensity). Both M and H have units A/m. Therefore χ is dimensionless. (d) The neutral piece of iron is attracted to the strong magnet. The magnet induces magnetisation in the iron (near end acquires opposite polarity to the nearer pole of the magnet). The induced magnetisation creates an attractive force, since the closer (opposite) pole is always nearer. This is why magnets attract unmagnetised ferromagnetic materials.

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?

Solution

B = 0.25 T, θ = 30°, τ = 4.5 × 10⁻² N·m Using: τ = mB sinθ m = τ/(B sinθ) = 4.5 × 10⁻² / (0.25 × sin30°) = 4.5 × 10⁻² / (0.25 × 0.5) = 4.5 × 10⁻² / 0.125 = 0.36 J/T = 0.36 A·m² Magnetic moment = 0.36 J/T

A short bar magnet of magnetic moment m = 0.32 J/T is placed in a uniform external magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable and (b) unstable equilibrium? What is the potential energy of the magnet in each case?

Solution

m = 0.32 J/T, B = 0.15 T (a) Stable equilibrium: The magnet aligns with the field (θ = 0°). This is the minimum potential energy state. Potential energy = –mB cosθ = –mB cos0° = –mB = –0.32 × 0.15 = –4.8 × 10⁻² J = –0.048 J (b) Unstable equilibrium: The magnet is anti-parallel to the field (θ = 180°). Any slight disturbance and it flips. Potential energy = –mB cos180° = mB = 0.32 × 0.15 = 4.8 × 10⁻² J = +0.048 J

A closely wound solenoid of 800 turns and area of cross section 2.5 × 10⁻⁴ m² carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?

Solution

A current-carrying solenoid acts like a bar magnet because: • It has two poles (north and south) at its two ends • The magnetic field inside and outside resembles that of a bar magnet • It experiences torque in a uniform magnetic field just like a bar magnet • Its magnetic moment is along its axis Magnetic moment: M = NIA = 800 × 3.0 × 2.5 × 10⁻⁴ = 800 × 7.5 × 10⁻⁴ = 0.6 A·m² The magnetic moment of the solenoid is 0.6 A·m², directed along the axis of the solenoid (using right-hand rule for the direction).

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 to turn the magnet so as to align it with the field in the opposite direction? (b) What is the torque on the magnet in the initial and final positions?

Solution

m = 1.5 J/T, B = 0.22 T Initial position: θᵢ = 0° (aligned with field) Final position: θf = 180° (opposite to field) (a) Work done in rotating from θᵢ to θf: W = mB(cosθᵢ – cosθf) = mB(cos0° – cos180°) = mB(1 – (–1)) = 2mB = 2 × 1.5 × 0.22 = 0.66 J (b) Torque: τ = mB sinθ At initial position (θ = 0°): τ = 1.5 × 0.22 × sin0° = 0 At final position (θ = 180°): τ = 1.5 × 0.22 × sin180° = 0 The torque is zero in both the initial and final positions (both are equilibrium positions).

A cylindrical bar magnet is kept along the axis of a circular coil. On rotating the bar magnet about its own axis, will there be a change in the flux through the coil, and will an emf be induced in the coil?

Solution

No, there will be no change in flux through the coil, and no emf will be induced. Reason: A cylindrical bar magnet has axial symmetry — it is symmetric about its own axis. When it rotates about its own axis: • The magnetic field pattern remains identical at every instant • The flux through the coil (which depends only on the external field configuration) does not change • By Faraday's law: EMF = –dΦ/dt = 0 since dΦ/dt = 0 Therefore, no emf is induced in the coil when the cylindrical bar magnet rotates about its own axis.

Two identical bar magnets are kept with their north poles pointing toward each other at distance d apart. A neutral point is found at a distance d/2 from each magnet. If the magnets are moved to distance 2d apart, find the position of the neutral point.

Solution

When two identical bar magnets face each other (N poles toward each other), the neutral points lie on the perpendicular bisector of the line joining them, or on the line joining them between the magnets. For magnets with N poles facing each other, the neutral point is on the line joining them, between the magnets where fields cancel. At distance d apart, neutral point is at d/2 from each. The field on the axial line: B ∝ m/r³ (for short magnets, approximately) When magnets are moved to 2d apart: the neutral point by symmetry will still be at the midpoint = d apart from each magnet (at the center). By symmetry of the configuration (both magnets identical and facing each other), the neutral point is always at the midpoint between the two magnets. New neutral point position: d from each magnet, i.e., at the centre between the two magnets.

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