The circular coil C1 has a radius of R, a length of L, and a number of turns per unit length of n1 = L/2R. The square C2 has a side, a perimeter, and a number of turns per unit length of n2 = L/4a....
Consider the plane S formed by the dipole axis and the axis of the earth. Let P be a point on the magnetic equator and in S. Let Q be the point of intersection of the geographical and magnetic equators. Obtain the declination and dip angle at P and Q.
The declination is zero, P is in the plane, S is in the north, and P is in the plane. The declination for point P is 0 since it is in the plane S created by the dipole axis and the earth's axis....
Assume the dipole model for earth’s magnetic field B which is given by Bv = vertical component of magnetic field = μ0/4π 2m cos θ/r3, BH = horizontal component of magnetic field = μ0/4π 2m sin θm/r3, θ = 90o latitude as measured from magnetic equator. Find loci of points for which i) |B| is minimum ii) dip angle is zero, and iii) dip angle is ±45o.
a) |B| is minimum at the magnetic equator. b) Angle of dip is zero when θ = π/2 c) When dip angle is ±45o θ = tan-1 is the locus.
What are the dimensions of χ, the magnetic susceptibility? Consider an H-atom. Guess an expression for χ, up to a constant by constructing a quantity of dimensions of χ, out of parameters of the atom: e, m, v, R and μ0. Here, m is the electronic mass, v is electronic velocity, R is Bohr radius. Estimate the number so obtained and compare with the value of | χ| equivalent to 10-5 for many solid materials.
χm = I/H = intensity of magnetisation/magnetising force χ is dimensionless as I and H has the same units χ = 10-4
Verify the Ampere’s law for the magnetic field of a point dipole of dipole moment Take C as the closed curve running clockwise along i) the z-axis from z = a > 0 to z = R; ii) along the quarter circle of radius R and centre at the origin, in the first quadrant of x-z plane; iii) along the x-axis from x = R to x = a and iv) along the quarter circle of radius a and centre at the origin in the first quadrant of the x-z plane.
Magnetic field = 0M/4(1/a2-1/R2) along the z-axis b) On the circular arc, the magnetic field at point A is = 0m/4R2. c) (d) The magnetic moment is 0
Use i) the Ampere’s law for H and ii) continuity of lines of B, to conclude that inside a bar magnet a) lines of H run from the N pole to S pole, while b) lines of B must run from the S pole to N pole.
The amperian loop is denoted by the letter C. We can find the angle between the two points by solving the above equation. cos is negative because it is greater than 90o.
A bar magnet of magnetic moment m and moment of inertia I is cut into two equal pieces, perpendicular to length. Let T be the period of oscillations of the original magnet about an axis through the midpoint, perpendicular to the length, in a magnetic field B. What would be the similar period T’ for each piece?
T stands for the time period. The moment of inertia is me. The magnet's mass is m. B stands for magnetic field. T = 2I/MB M' = M/2 magnetic dipole moment T' = T/2 is the time period.
Suppose we want to verify the analogy between electrostatic and magnetostatic by an explicit experiment. Consider the motion of i) electric dipole p in an electrostatic field E and ii) magnetic dipole m in a magnetic field B. Write down a set of conditions on E, B, p, m so that the two movements are verified to be identical.
pE sin θ = μB sin θ pE = μB E = cB pcB = μB p = μ/c
Verify the Gauss’s law for magnetic field of a point dipole of dipole moment m at the origin for the surface which is a sphere of radius R.
P is the point at a distance r from O and OP, then the magnetic field is given as: dS is the elementary area of the surface P, then dS = r2 (r2 sin θ d θr) Solving the above we...
A ball of superconducting material is dipped in liquid nitrogen and placed near a bar magnet. i) In which direction will it move? ii) What will be the direction of it’s magnetic moment?
I The superconducting material will repel the bar magnet. ii) The magnetic moment will be directed from left to right.
From molecular view point, discuss the temperature dependence of susceptibility for diamagnetism, paramagnetism, and ferromagnetism.
The temperature has little effect on the temperature dependence of susceptibility for a diamagnetism. The temperature affects the temperature dependence of susceptibility for paramagnetism and...
Explain quantitatively the order of magnitude difference between the diamagnetic susceptibility of N2 and Cu.
nitrogen Density = 28 g/ 22400 cc copper Density = 8 g/ 22400 cc Ratio = 16 × 10-4 Diamagnetic susceptibility = density of nitrogen/density of copper = 1.6 × 10-4
A proton has spin and magnetic moment just like an electron. Why then its effect is neglected in magnetism of materials?
A comparison of a proton's and an electron's spinning is made by comparing their magnetic dipole moment, which is given as μp = eh/4πmp μe = eh/4πme μp/μe = me/mp = 1/1837 >> 1 μp << μe...
Let the magnetic field on the earth be modelled by that of a point magnetic dipole at the centre of the earth. The angle of dip at a point on the geographical equator a) is always zero b) can be zero at specific points c) can be positive or negative d) is bounded
b) can be zero at specific points c) can be positive or negative d) is bounded
Essential difference between electrostatic shielding by a conducting shell and magneto static shielding is due to a) electrostatic field lines can end on charges and conductors have free charges b) lines of B can also end but conductors cannot end them c) lines of B cannot end on any material and perfect shielding is not possible d) shells of high permeability materials can be used to divert lines of B from the interior region
a) electrostatic field lines can end on charges and conductors have free charges c) lines of B cannot end on any material and perfect shielding is not possible d) shells of high permeability...
A long solenoid has 1000 turns per meter and carries a current of 1 A. It has a soft iron core of μr = 1000. The core is heated beyond the Curie temperature Tc a) the H field in the solenoid is unchanged but the B field decreases drastically b) the H and B fields in the solenoid are nearly unchanged c) the magnetisation in the core reverses direction d) the magnetisation in the core diminishes by a factor of about 108
a) the H field in the solenoid is unchanged but the B field decreases drastically d) the magnetisation in the core diminishes by a factor of about 108
The primary origin(s) of magnetism lies in a) atomic currents b) Pauli exclusion principle c) polar nature of molecules d) intrinsic spin of electron
a) atomic currents d) intrinsic spin of electron
S is the surface of a lump of magnetic material a) lines of B are necessarily continuous across S b) some lines of B must be discontinuous across S c) lines of H are necessarily continuous across S d) lines of H cannot all be continuous across S
a) lines of B are necessarily continuous across S d) lines of H cannot all be continuous across S
A paramagnetic sample shows a net magnetisation of 8 Am-1 when placed in an external magnetic field of 0.6T at a temperature of 4K. When the same sample is placed in an external magnetic field of 0.2T at a temperature of 16K, the magnetisation will be a) 32/3 Am-1 b) 2/3 Am-1 c) 6 Am-1 d) 2.4 Am-1
b) 2/3 Am-1
Consider the two idealized systems: i) a parallel plate capacitor with large plates and small separation and ii) a long solenoid of length L >> R, the radius of the 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’s law for electrostatic fields b) case (ii) contradicts Gauss’s law for magnetic fields c) case (i) agrees withd) case (ii) contradicts
b) In the case of magnetic fields, instance (ii) violates Gauss's law.
In a permanent magnet at room temperature a) magnetic moment of each molecule is zero b) the individual molecules have a non-zero magnetic moment which is all perfectly aligned c) domains are partially aligned d) domains are all perfectly aligned
d) all of the domains are exactly matched
A toroid of n turns, mean radius R and cross-sectional radius a carries current I. It is placed on a horizontal table taken as an x-y plane. Its magnetic moment m a) is non-zero and points in the z-direction by symmetry b) points along the axis of the toroid c) is zero, otherwise, there would be a field falling as 1/r3 at large distances outside the toroid d) is pointing radially outwards
c) is zero; otherwise, a field dropping as 1/r3 at great distances outside the toroid would exist.
Answer the following about magnetism –
(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 a distant past? (e)...
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...
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?
Answer- Mean radius of the Rowland ring is given = 15 cm Number of turns = 3500 Relative permeability of the core is given by μr = 800 Magnetizing current is given by I = 1.2 A Magnetic field...
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 (me
= 9.11 × 10–31 kg). [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.]
Answer – Energy of the electron beam is given by E = 18 keV Or, E = 18 x 103 eV = 18 x 103 x 1.6 x 10-19 J Magnetic field is given by B = 0.04 G Mass of the electron is given by...
A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60°, 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° with this field, what is the magnitude of the other field?
Answer- Magnitude of one of the magnetic field is given by B1 = 1.2 × 10–2 T Suppose that the magnitude of the other field is B2 And the angle between the field is given, θ = 60° We...
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.
(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° in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.
Answer: Number of turns is given = 30 Radius of the coil is given = 12 cm Current in the coil is given = 0.35 A Angle of dip, given by δ = 450 (a) Horizontal component of earth’s magnetic field...
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°. The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable?
Answer – First it is important to decide the direction which would best represent the given situation. We are given that BH = B cos δ = 0.39 × cos 35o G Therefore, BH =...
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 current-carrying the cable is equal and opposite to the horizontal component of earth’s magnetic field.)
Answer – Current in the wire is given by 2.5 A The earth’s magnetic field at a location is given by R= 0.33 G = 0.33 x 10-4 T Angle of dip is zero is given by δ = 0 Horizontal...
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.
Answer – (a) When a substance is placed in an external magnetic field, the domain aligns in the direction of the magnetic field. The process of alignment consumes some energy. The substance retains...
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 magnetisation
of a ferromagnet?
Answer – (a) At a lower temperature, thermal motion is reduced, and the tendency to disturb the alignment of the dipoles is reduced. (b) The induced dipole moment is always in the opposite direction...
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° with earth’s field on (a) its normal bisector and (b) its axis. The 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 : Magnetic moment of the bar magnet is given by M = 5.25 × 10–2 J T–1 Magnitude of earth’s magnetic field at a place is given by H = 0.42 G = 0.42×10-4 T The magnetic field on the...
If the bar magnet in exercise 5.13 is turned around by 180°, where will the new null points be located?
Answer – The magnetic field at a distance d1 = 14 cm on the axis of the magnet can be written as – ${{B}_{1}}=\frac{{{\mu }_{0}}2M}{4\pi {{d}_{1}}^{3}}=H$ Where, M is the magnetic moment...
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 is given by H = 0.36 G The magnetic field at distance d from the axis of the magnet can be calculated using the following formula –...
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.
Magnetic moment of the bar magnet is given by M = 0.48 J T–1 Distance, d = 10 cm = 0.1 m The magnetic field at distance d from the magnet's centre on the axis can be calculated using the following...
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’s field at the location.
Answer – We are given the angle of declination, θ = 12° Angle of dip, given by δ = 60° Horizontal component of earth’s magnetic field is given by BH = 0.16 G Earth’s magnetic field at the given...
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.
Answer – Horizontal component of earth’s magnetic field is given by BH = 0.35 G Angle made by the needle with the horizontal plane is known as the Angle of dip, given by δ=22° Earth’s magnetic field...
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 – Number of turns in the circular coil is given by N = 16 Radius of the coil is given by r = 10 cm = 0.1 m Cross-section of the coil, A is given by – $A=\pi {{r}^{2}}=\pi {{(0.1)}^{2}}$...
A closely wound solenoid of 2000 turns and area of cross-section 1.6×10-4 m2, carrying 4.0 A current, is suspended through its centre, thereby 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 the horizontal magnetic field of 7.5 × 10–2 T is set up at an angle...
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...
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 the torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?
Answer – Magnetic field strength is given by B = 0.25 T Magnetic moment is by M = 0.6 T-1 The angle θ between the direction of the applied field...
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?
Answer – It is given that the number of turns in the solenoid, n = 800 Area of cross section is given by, A = 2.5×10-4 m2 And the current in the solenoid, I = 3.0 A Because a magnetic field develops...
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 – Moment of the bar magnet is given by M = 0.32 JT–1 External magnetic field, is given by B = 0.15 T The magnetic field is aligned with the bar magnet. This system is said to be in...
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 4.5×10-2J. What is the magnitude of the magnetic moment of the magnet?
Answer – We are given the magnetic field strength, B = 0.25 T Torque on the bar magnet is given by, T = 4.5×10-2J Θ = 30° is the angle between the external magnetic field and the bar...
Answer the following:
(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...
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...