Magnetism And Matter

The three independent quantities used to specify the earth’s magnetic field are the following.

1) Magnetic declination: It is the angle between the geographic north and the magnetic north at a place.

Note: If the compass deflects right of geographic north, then declination is positive and if deflection is towards the left of geographic north then it is a negative declination

2) Angle of Dip: It is the angle between the horizontal plane and the magnetic axis, as observed in the compass.

3) The horizontal component of the earth’s magnetic field

The angle of dip will be greater than 18° in Britain. The angle of dip increases from equator towards pole. Britain is farther from equator as compared to India.

Note: The angle of dip increases from equator towards pole. At equator angle of dip is zero while at poles it is 90 degrees. The magnetic field lines are perpendicular at poles.

The magnetic field lines would come out of the ground at Melbourne in Australia. The magnetic field lines emanate from magnetic north pole (near Geographic south) and merge at magnetic South Pole (near geographic north). Australia is near the magnetic north pole so, magnetic field lines will seem to be emerging.

The compass will get aligned in vertical direction (up-down) if is held vertically at the North Pole. This is due to the fact that magnetic field is perpendicular at the Poles and the magnetic needle of the compass aligns to it.

Note: There will be no predictable motion in horizontal plane due to very weak horizontal component of magnetic field.

Given:

Magnetic moment, m = 8 × 10²² JT^-1

Radius of earth, r = 6.4 × 10⁶m

Our task is to check the order of m, we shall attempt to do so by finding the magnetic field of earth using the available information. If we are close to the standard measured value then the order of m is correct.

We know that magnetic field strength,

…(1)

Where, �₀ = permeability of free space

�₀ = 4π × 10^-7 TmA^-1

m = magnetic moment

r = radius

Putting values in equation (1) we get,

⇒ B = 0.3 × 10^-4T

⇒ B = 0.3 G

Which is a close value to the strength of earth’s magnetic field. Hence the order of magnetic moment is correct.

There may be several local north and south poles, such a thing is possible due to deposits of minerals which are magnetic in nature.

The earth’s magnetic field changes with time due to many reasons, changes is the speed of convection current in core is one of them. These changes are very slow and will take hundreds of years to show substantial effects.

The earth’s core contains iron in molten state which is not ferromagnetic in nature hence geologists do not regard this as a source of earth’s magnetism.

Fact: After Curie temperature (T_c) materials lose permanent magnetism. This is due to inability is magnetic domains to align properly. After this temperature magnetic materials attain paramagnetic nature.

The charged currents here refer to the convection currents in the core of earth, which is responsible for earth’s magnetic nature. This current is sustained partly by the Coriolis force due to rotation of earth, which rotates the molten core.

Earth changed its magnetic orientation several times in history. The magnetic orientation of the earth were weekly recorded in the rocks during solidification.

Earth’s atmosphere has a layer called Ionosphere, which mainly constitutes of electrons and other charges particles. A magnetic field is associated with these charged particles when they are in motion. So we can say that Ionosphere is one of the agencies which can distort the shape of earth’s dipole at greater altitudes.

The magnetic field has a tendency to affect the path of charged particles in circular motion. Since Interstellar space is Vast hence a minor change in path due to magnetic field can become substantial deviation from the destination.

Given:

Angle between axis of bar magnet and external magnetic field, θ = 30°

Strength of external magnetic field, B = 0.25 T

Torque experienced by bar magnet, Τ = 4.5 × 10^-2J

We know that, when a Bar magnet is placed in a uniform magnetic field it experiences a Net torque given by,

T = M × B …(1)

T = M × B × Sinθ

Where,

M = magnetic moment

T = Torque on bar magnet

θ = angle between magnetic field and magnetic moment

From equation (1) we can write,

M = 0.36 JT^-1

Hence, the magnitude of moment of the Bar magnet is 0.36JT^-1.

Given:

Magnetic moment of magnet, M = 0.32JT^-1

Strength of external magnetic field, B = 0.15T

(a) When the magnetic moment is aligned (0°) with the magnetic field then we consider this as stable equilibrium, if the magnet is rotated, then it will have the tendency to come back in this position.

θ = 0°

We know that,

U = - M.B

U = -M × B × Cos θ …(1)

Where, U = potential energy

M = magnetic moment

B = magnetic field

θ = Angle between filed and magnetic moment

By putting the given values in the equation (1), we have,

U = -0.32 × 0.15 × Cos 0°

U = -0.048 J

Potential energy of the system in stable equilibrium is -0.048 J.

(b) Whereas, in case of unstable equilibrium the moment is at 180° with the magnetic field. If the magnet is rotated then it will never come back in the initial position.

θ = 180°

Now, by putting values in equation (1), we get,

U = -0.32 × 0.15 × Cos 180°

U = 0.048 J

Potential energy of the system in unstable equilibrium is 0.048J.

Note: The direction of the magnetic moment is from south pole to north pole inside the magnet.

Given:

Number of turns, N = 800

Area of cross-section, A = 2.5 × 10^-4 m²

Current, I = 3A

The solenoid acts like a bar magnet because the magnetic axis of the solenoid is along its length, which is analogous to the magnetic axis of Bar magnet.

We know that,

Magnetic moment, M = N × I × A …(1)

Now putting the values in equation 1, we get

M = 800 × 3A × 2.5 × 10^-4m²

M = 0.6 JT^-1

Magnetic field strength, B = 0.25 T

Magnetic moment, M = 0.6 JT⁻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 = MBsinθ
= 0.6 JT^-1 × 0.25 T × sin 30^o

= 0.6 JT^-1 × 0.25 T × (1/2)

= 0.075 J
= 7.5 × 10^-2 J

The magnitude of torque is 7.5 × 10^-2 J.

(a) Magnetic moment, M = 1.5 J T⁻¹

Magnetic field strength, B = 0.22 T

(i) Initial angle between the axis and the magnetic field, θ₁ = 0°

Final angle between the axis and the magnetic field, θ₂ = 90°

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

W = -MB(cosθ₂ - cosθ₁)

= -1.5 J T⁻¹ x0.22 T(cos 90^o-cos 0^o)

= -0.33 Jx(0-1)

= 0.33 J

(ii) Initial angle between the axis and the magnetic field, θ₁ = 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 θ₂ - cos θ₁)

= -1.5 J T⁻¹ x.22 T(cos 180^o-cos 0^o)

= -0.33 Jx(-1-1)

= 0.66 J

(b) For case i) θ = θ₂ = 90^o

τ = MBsinθ = 1.5 J T⁻¹ x 0.22 T x 1 = .33 J

(Remember the torque and the work applied by the torque are not the same thing)

For case ii) θ = θ₂ = 180^o

τ = MBsinθ = 1.5 J T⁻¹ x.22 Tx0 = 0 J

Number of turns on the solenoid, n = 2000

Area of cross-section of the solenoid, A = 1.6 × 10^-4 m²

Current in the solenoid, I = 4 A

(a) The magnetic moment along the axis of the solenoid is calculated as:

M = n.A.I = 2000 × 1.6 × 10^-4 m² x 4A

= 1.28 Am²

(b) Magnetic field, B = 7.5 × 10^-2 T

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

τ = MBsinθ = 1.28 JT^-1 × 7.5x10^-2 T × sin30^o

= 4.8 × 10^-2 N-m

= 4.8x10^-2 J

But the Force is zero as the Magnetic field is uniform.

Number of turns in the circular coil, N = 16

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

Cross-section of the coil, A = πr² = π × (0.1)² m²

Current in the coil, I = 0.75 A

Magnetic field strength, B = 5.0 × 10⁻² T

Frequency of oscillations of the coil, f = 2.0 s⁻¹

Magnetic moment, M = NIA = NI × πr²

= 16 × 0.75 A × π × (0.1)² m²

= 0.377 J T⁻¹

Now the frequency is given by,

I = Moment of Inertia of the coil (Remember this equation)

So,

Or,

⇒ I = 1.19 x 10^-4 kgm²

The moment of inertia of the coil about its axis of rotation is 1.19 × 10^-4 kgm²

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°

Earth’s magnetic field strength = B

We know, B_H = B cosδ

B = B_H/cos δ = 0.35/(cos 22^o) = 0.377 G

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

Angle of declination, θ = 12°

Angle of dip, δ = 60°

Horizontal component of earth’s magnetic field, B_H = 0.16 G

Earth’s magnetic field at the given location = B

We know that,

B_H = B cosδ

B = B_H/cos δ = 0.16 G/(cos 60^o) = 0.32 G

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.

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:

μ₀ = permeability of the free space = 4 x 10^-7

⇒ B₁ = 0.96 x 10^-4 T

∴ B₁ = 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:

= 0.48 G

The magnetic field is along the N − S direction.

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

The magnetic field at a distance d, on the axis of the magnet is given as:

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

as:

(NOTE: Refer the pictures of the previous question)

= B₁/2

Hence the total magnetic field is B₁ + B_{2 =} B₁ + B₁/2 = (0.36 + 0.18) G = 0.54 G

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

The magnetic field on the axis of the magnet at a distance d₁ = 14 cm, can be written as:

…………(1)

Where,

M = Magnetic moment

= Permeability of free space

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

…………(2)

Equating (1) and (2) we get,

Or,

Or,

Or, d₂ = 14x 0.794 = 11.1cm

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

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

(a) The magnetic field at a distance R from the centre of the magnet on the normal bisector is given by the relation:

When the resultant field is inclined at 45° with earth’s field, B = H (Note: This is the key point of the question, try to sort out this kind of data from the question when you are reading it, so you need not read it again.)

= H = 0.42 × 10⁻⁴ T

R³ =

⇒

⇒ R³ = 12.5 x 10^-5 m³

⇒ R = 0.05 m = 5 cm

(b) The magnetic field at a distance R₁ from the centre of the magnet on its axis is given as:

The resultant field is inclined at 45° with earth’s field.

So, B₁ = H

R₁³ =

⇒ R₁³ = 25 x10^-5 m³

⇒ R_{1 =} 0.063m = 6.3cm

At high temperatures, due to the random thermal motion of molecules in a paramagnetic sample, the alignments of dipoles get disrupted. On reducing the temperature random thermal motion reduces, thus disruption is reduced. Hence, a paramagnetic sample displays greater magnetization when cooled.

The magnetism in a diamagnetic substance is due to induced dipole moment (opposite to the magnetizing field). Hence, the internal motion of the atoms, which depends on the temperature, does not affect the diamagnetism of a material.

A toroid using bismuth core has a magnetic field slightly greater than a toroid whose core is empty because bismuth is a diamagnetic substance.

The permeability of ferromagnetic materials is inversely dependent on the applied magnetic field, i.e. it is lower for a higher field and vice versa.

The magnetic field lines are always nearly normal to the surface of ferromagnetic materials at every point because permeability of ferromagnetic material is always greater than one.

Yes, the maximum possible magnetization of a paramagnetic sample can be of the same order of magnitude as the magnetization of a ferromagnet but this requires very strong magnetic fields.

The relation between B (i.e. the external magnetic field) and H (i.e. magnetic intensity) in ferromagnetic materials is shown in the following curve.

From the above curve it is clear that magnetization persists even when the external field is removed, this depicts the irreversibility of magnetization in a ferromagnet.

After going through repeated cycles of magnetization, a carbon steel piece dissipates greater heat energy than a soft iron piece, because the dissipated heat energy is directly proportional to the area of a hysteresis loop and the carbon steel piece has a greater hysteresis curve area than a soft iron piece.

The record of hysteresis loop cycles during magnetization (i.e. value of magnetization) corresponds to bits of information. Therefore, the material displaying hysteresis loops like a ferromagnet can be used for storing information.

Ferromagnetic material like Ceramic is used for coating magnetic tapes in cassette players and for building memory stores in modern computers.

The certain region of space can be surrounded by soft iron coil to shield it from magnetic field.

Current in the cable (I) = 2.5 A

Earth’s magnetic field at the location (H) = 0.33 G = 0.33 × 10^-4T

Angle of dip (δ) = 0

Let the line of neutral points be at a distance “r” meters from the horizontal cable.

The magnetic field at the neutral point due to current carrying cable,

,

where μ₀ = Permeability of free space = 4π × 10^-7 TmA¹

Horizontal component of earth’s magnetic field (H’) = H × cosδ

At neutral points, magnetic field due to a current-carrying cable is equal and opposite to the horizontal component of earth’s magnetic field.

∴ H’ = H_N

⇒ H × cosδ =

⇒ 0.33 × 10^-4 T × cos0^o =

⇒ r =

⇒ r = 15.15 × 10^-3m

⇒ r = 1.515cm

Current in the wires (I) = 1.0 A (east to west)

The earth’s magnetic field at the place (H) = 0.39G = 0.39 × 10^-4T

The angle of dip (δ) = 35°

Magnetic declination (θ) ∼ 0^o

Number of wires in the cable (n) = 4

The horizontal component of earth’s magnetic field (H’) = H × cosδ

Magnetic fields due to current carrying cables (B) =

where μ₀ = Permeability of free space = 4π × 10^-7TmA¹

The resultant magnetic fields at point 4.0 cm below the cables

R = 4cm = 0.04m

Resultant horizontal magnetic fields (H_h) = H’- B

⇒ H_h = H’- B

⇒ H_h =

⇒ H � �_h = (0.39 × 10^-4T × cos35^o)-

⇒ H_h = 0.39 × 10^-4T × 0.819 - 4 × 2 × 25 × 10^-7T

⇒ H_h = 0.319 × 10^-4T-0.2 × 10^-4T

⇒ H_h = 0.119 × 10^-4

Or H_h ∼ 0.12 × 10^-4 T or 0.12 Gauss

Resultant vertical magnetic field (H � �_{v �}) = vertical component of earth’s magnetic field = H × sinδ

= 0.39 × 10^-4T × sin35^o

= 0.39 × 10^-4T × 0.573

= 0.22 × 10^-4T

∴ Resultant magnetic field =

= 0.25 × 10^-4T

(a) Number of turns in the coil (n) = 30

Radius of coil (r) = 12cm = 0.12m

Current in the coil (I) = 0.35A

Angle of dip (δ) = 45^o

Magnetic fields due to current carrying coils (B) =

where, μ₀ = Permeability of free space = 4π × 10^-7TmA¹

B =

= 5.49 × 10^-5 T

∴ Horizontal component of the earth’s magnetic field = B sinδ

= 5.49 × 10^-5 T × sin45^o

= 5.49 × 10^-5 T × 0.707

= 3.88 × 10^-5 T

(b) When 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, then the needle will reverse its original direction (i.e. the needle will point from East to West).

Magnitude of one of the magnetic fields (B₁) = 1.2 × 10^–2 T

The angle between the magnetic field directions (θ) = 60°

The angle between the dipole and the magnetic field B₁ is θ₁ = 15°

Let magnitude of second of the magnetic fields = B₂

∴ The angle between the dipole and the magnetic field B₂ is θ₂ = θ-θ₁ = 45°

Let magnetic moment of the dipole be M

Thus, at rotational equilibrium,

Torque due to field B₁ = Torque due to field B₂

⇒ M × B₁sinθ₁ = M × B₂ sinθ₂

⇒

⇒ B₂ = 9 × 10^-3 T

Thus, the magnitude of the other magnetic field = 4.39 × 10^-3T

Energy of the electron beam(E) = 18keV = 18000eV = 18000 × 1.6 × 10^-19 = 2.88 × 10^-15J

Mass of electron (m_e) = 9.11 × 10^–31 kg

Horizontal magnetic field (B) = 0.04 G = 0.04 × 10^-4T

Distance of deflection from the beam (r) = 30 cm = 0.3m

Let velocity of electrons be v.

∴ Energy of the beam = Kinetic energy of electrons

⇒ E = 1/2 m_ev²

⇒

⇒

∴ v = 7.95 × 10⁷m/s

∵ The electron beam deflects along a circular path in a magnetic field.

Let the radius of the circular path be r m.

The force due to magnetic field on the electron = qvBsinθ = e × v × B × sin 90°

Here, q = charge on the electron,

θ = angle between velocity and magnetic field.

∵ The force due to magnetic field on electron = centripetal force on electron

⇒

⇒

⇒

⇒ r = 11.3 m

Let the deflection of electron beam in upward and downward directions be x = r(1-cosΦ),

where Φ = angle of deflection.

∵

⇒ = 0.026

⇒ Φ = sin^-1(0.026) = 1.521°

Thus, Deflection(x) = r(1- cosΦ) = 11.3 × (1- cos(1.521)°)

= 11.3 × (1- 0.90)

= 11.3 × 0.01

= 1.1310^-3m

Number of atomic dipoles (n) = 2 × 10²⁴

Initially, Dipole moment of each atomic dipole (M) = 1.5 × 10^-23JT^-1

Magnetic field (B₁) = 0.64 T and Temperature (θ₁) = 4.2K

∵ Degree of magnetic saturation = 15%

∴ Effective Dipole moment (M₁) = 15% of Total dipole moment

⇒ M₁ = × (no. of atomic dipole × dipole moment of each)

⇒ M₁ = × 2 × 10²⁴ × 1.5 × 10^-23JT^-1 = 4.5 JT^-1

When, Magnetic field (B₂) = 0.98 T and Temperature (θ₂) = 2.8 K

Let the dipole moment be M_2.

According to the Curie’s Law, magnetization of a paramagnetic material is directly proportional to applied magnetic field and inversely proportional to temperature.

∴ Ratio of magnetic dipole moments (i.e. M₁:M₂) = (B₁θ₂/B₂θ₁)

⇒ ⇒ M₂ = M₁ × ⇒ M₂ = 4.5 JT^-1 × = 10.336 JT^-1

Thus, total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K = 10.336 JT^-1

Radius of ring (r) = 15cm = 0.15m

Number of turns in the ring (n) = 3500

Relative permeability of the ferromagnetic core (μ_r) = 800

Current in the Rowland ring (I) = 1.2A

Magnetic Field due to circular coil (B) =

Where, μ₀ = Permeability of free space = 4π × 10^-7TmA^-1

∴ B = = 4.48T

We know,

It follows from the definition of μ₁ and l.

μ_{1 =} i × A =

l = mvr = m

Where, r is the radius of the circular path traced by the electron around the nucleus, with the mass of electron m and charge (-e) and time period T.

Divide the above equations-

⇒

⇒

Clearly μ₁ and l will be antiparallel.

On the other hand, is obtained on the basis of quantum mechanics.