NEET Physics: Magnetism, Electromagnetic Induction and AC · Chapter 2

Magnetism and Matter

What to remember

A bar magnet is a magnetic dipole with moment m (unit A m²). In a uniform field it feels torque τ = m × B (magnitude mB sinθ) and has potential energy U = −m·B, lowest when m is along B.
Magnetic field lines are closed loops that never cross. Isolated magnetic poles do not exist, so the net flux through any closed surface is zero: ∮ B · dA = 0.
B = μ₀(H + M), M = χH, μr = 1 + χ. Diamagnetic: χ small and negative. Paramagnetic: χ small and positive. Ferromagnetic: χ large and positive.
Soft iron (low retentivity, low coercivity) is used for electromagnet cores. Steel and alnico (high retentivity, high coercivity) are used for permanent magnets.

Bar magnet, field lines and the equivalent solenoid

A bar magnet has a north pole and a south pole. A freely suspended magnet comes to rest along the north–south direction. Like poles repel and unlike poles attract. If a bar magnet is cut into two pieces, each piece is a complete magnet with two poles. A single north or south pole cannot be separated.

Magnetic field lines are drawn to picture the field. Their properties:

Outside the magnet they run from the north pole to the south pole. Inside the magnet they run from south to north. So they form continuous closed loops.
The tangent to a line at a point gives the direction of the field there.
Lines never intersect, because the field has one direction at each point.
The closer the lines, the stronger the field.
The number of lines through a surface is a measure of the magnetic flux.

Bar magnet as an equivalent solenoid. The field of a bar magnet looks like the field of a current-carrying solenoid of finite length. A solenoid with N turns, current I and area A has magnetic moment

m = N I A

The moment is a vector along the axis, directed from the south pole to the north pole of the magnet. The magnet behaves as if it were made of many tiny current loops. If a magnet is cut perpendicular to its length into k equal pieces, each piece has moment m/k, because N becomes N/k.

Field of a short bar magnet at a distance r from its centre (r much larger than the length):

On the axis: B = (μ₀/4π) (2m/r³), in the direction of m.
On the equatorial line: B = (μ₀/4π) (m/r³), in the direction opposite to m.

So the axial field is twice the equatorial field at the same distance. Both fall as 1/r³. Here μ₀/4π = 10⁻⁷ T m A⁻¹.

Dipole in a uniform magnetic field

A magnetic dipole of moment m placed in a uniform field B at angle θ to the field has:

Net force: zero (the forces on the two poles are equal and opposite).
Torque: τ = m × B, magnitude τ = mB sinθ. It tries to turn the dipole into line with the field.
Potential energy: U = −m·B = −mB cosθ.

Angle θ	Torque	Potential energy	Equilibrium
0° (m along B)	0	−mB (minimum)	Stable
90°	mB (maximum)	0	Not an equilibrium
180° (m opposite to B)	0	+mB (maximum)	Unstable

Work done by an external agent to turn the dipole from θ₁ to θ₂ is W = mB (cosθ₁ − cosθ₂). From 0° to 180° it is 2mB.

A small magnet pivoted in a uniform field and displaced slightly oscillates with period T = 2π √(I/(mB)), where I is its moment of inertia about the axis of rotation. The period is larger for a weaker field, and T is proportional to 1/√B.

Magnetism and Gauss's law

For electric charges, the flux through a closed surface equals the enclosed charge divided by ε₀. Magnetic poles always occur in pairs, and isolated magnetic monopoles have not been found. Every field line that enters a closed surface also leaves it. So

∮ B · dA = 0

for any closed surface. This is Gauss's law for magnetism. It means that the lines of B have no beginning and no end. The unit of magnetic flux is the weber (Wb = T m²).

Magnetisation and magnetic intensity

In a material, the atoms act as tiny dipoles. The following quantities describe the response of a material to an applied field.

Quantity	Symbol	Meaning	Unit
Magnetisation	M	Net magnetic moment per unit volume	A m⁻¹
Magnetic intensity	H	Field due to external currents (the field that is "applied")	A m⁻¹
Magnetic field in the material	B	Total field	T
Magnetic susceptibility	χ	M/H, measures how easily the material is magnetised	No unit
Relative permeability	μr	B/B₀ = μ/μ₀	No unit
Permeability	μ	μ₀ μr	T m A⁻¹

Relations:

B = μ₀ (H + M)
M = χ H
μr = 1 + χ
B = μ₀ μr H = μ H

For a solenoid with n turns per metre, carrying current I, the intensity in the core is H = nI. With a core of relative permeability μr, the field is B = μ₀ μr n I.

Magnetic properties of materials

Diamagnetic materials. The induced magnetisation is opposite to the applied field. The susceptibility is small and negative (−1 ≤ χ < 0) and the relative permeability is slightly less than 1. A diamagnetic body is weakly repelled by a magnet. In a non-uniform field it tends to move from the stronger part to the weaker part of the field. The effect does not depend on temperature. Diamagnetism is present in all materials but is visible only when the other effects are absent. Examples: bismuth, copper, lead, silicon, nitrogen (at STP), water, sodium chloride.

Paramagnetic materials. The atoms have permanent magnetic moments, which are randomly oriented. A field partly lines them up, so the magnetisation is along the field. The susceptibility is small and positive and μr is slightly more than 1. A paramagnetic body is weakly attracted. In a non-uniform field it tends to move from the weaker to the stronger part. Examples: aluminium, sodium, calcium, oxygen (at STP), copper chloride.

Curie's law: for a paramagnetic substance, χ = C/T, where C is the Curie constant and T is the absolute temperature. The susceptibility falls as the temperature rises, since thermal motion disturbs the alignment.

Ferromagnetic materials. These show very strong magnetisation in the direction of the field. The susceptibility is large and positive (χ ≫ 1, often thousands) and μr is very large. Atomic moments line up in small regions called domains, and an applied field makes the domains grow and turn along the field. A ferromagnet is strongly attracted and moves from the weaker to the stronger part of a non-uniform field. Examples: iron, cobalt, nickel, gadolinium, alnico.

Above a certain temperature, called the Curie temperature, a ferromagnet becomes paramagnetic. For iron, this is 1043 K. Above it, the susceptibility follows χ = C/(T − T_c), where T_c is the Curie temperature.

Property	Diamagnetic	Paramagnetic	Ferromagnetic
Susceptibility χ	Small, negative	Small, positive	Large, positive
Relative permeability μr	Slightly less than 1	Slightly more than 1	Much greater than 1
Effect of a magnet	Weakly repelled	Weakly attracted	Strongly attracted
Motion in non-uniform field	Stronger → weaker	Weaker → stronger	Weaker → stronger
Atomic dipoles	None permanent	Permanent, random	Permanent, in domains
Temperature effect	Almost none	χ ∝ 1/T	Becomes paramagnetic above T_c
Examples	Bismuth, copper, water	Aluminium, sodium, oxygen	Iron, cobalt, nickel

Hysteresis, permanent magnets and electromagnets

When a ferromagnetic specimen is taken through a cycle of magnetisation by changing H, the graph of B against H is a closed loop, called the hysteresis loop. The value of B for a given H depends on the earlier history of the specimen. Features of the loop, described in words:

Starting from an unmagnetised sample, B rises with H along a curve that finally flattens (saturation).
When H is brought back to zero, B does not fall to zero. The remaining value of B is the retentivity (remanence).
To bring B to zero, a field must be applied in the reverse direction. The reverse value of H needed is the coercivity.
The area of the loop is the energy lost as heat per unit volume in one cycle of magnetisation.

Use	Material	Retentivity	Coercivity	Loop area	Reason
Permanent magnet	Steel, alnico, cobalt steel	High	High	Large	Must stay magnetised and not be demagnetised easily
Electromagnet core	Soft iron	Low	Low	Small	Must magnetise and demagnetise easily with little energy loss

An electromagnet has a soft iron core inside a solenoid. When a current flows, the core becomes strongly magnetised, and when the current is switched off, it loses most of its magnetism. Electromagnets are used in electric bells, loudspeakers and cranes. Soft iron is also the core material for devices that carry alternating current, because the small loop area gives small heating loss.

#### Worked examples

Example 1. A bar magnet of moment 2 A m² is placed at 30° to a uniform field of 0.5 T. Find the torque and the potential energy.

τ = mB sinθ = 2 × 0.5 × 0.5 = 0.5 N m.

U = −mB cosθ = −2 × 0.5 × 0.866 ≈ −0.87 J.

Example 2. A bar magnet is equivalent to a solenoid of 100 turns carrying 2 A, with cross-section 4 cm². Find its moment, and its axial field at 0.1 m (take μ₀/4π = 10⁻⁷ T m A⁻¹).

m = N I A = 100 × 2 × 4 × 10⁻⁴ = 0.08 A m².

B = 10⁻⁷ × 2 × 0.08/(0.1)³ = 1.6 × 10⁻⁷ × 10³ = 1.6 × 10⁻⁴ T.

Example 3. A paramagnetic sample has χ = 6 × 10⁻⁵ at 200 K. Find χ at 300 K.

By Curie's law, χ is proportional to 1/T. χ = 6 × 10⁻⁵ × 200/300 = 4 × 10⁻⁵.

Example 4. A material has H = 100 A m⁻¹ and M = 1900 A m⁻¹. Find B and μr.

B = μ₀ (H + M) = 4π × 10⁻⁷ × 2000 = 8π × 10⁻⁴ T. χ = M/H = 19, so μr = 1 + χ = 20.

#### Formula summary

Quantity	Formula	Units
Magnetic moment of a solenoid or loop	m = NIA	A m²
Torque on dipole	τ = mB sinθ	N m
Potential energy of dipole	U = −mB cosθ	J
Work to turn the dipole from θ₁ to θ₂	W = mB (cosθ₁ − cosθ₂)	J
Period of oscillation	T = 2π√(I/mB)	s
Axial field of short magnet	B = μ₀ 2m/(4π r³)	T
Equatorial field of short magnet	B = μ₀ m/(4π r³)	T
Gauss's law	∮ B · dA = 0	Wb
Magnetic field in material	B = μ₀ (H + M)	T
Susceptibility	χ = M/H; μr = 1 + χ	No unit
Curie's law	χ = C/T

Constants: μ₀ = 4π × 10⁻⁷ T m A⁻¹.

NEET traps

1. The net force on a dipole in a uniform field is zero, but the torque is not.
2. The potential energy is minimum (−mB) when m is along B, and maximum (+mB) when it is opposite. The torque is zero in both positions, but only the first is stable.
3. The torque is maximum at θ = 90°, while the potential energy is zero at 90°.
4. The axial field is twice the equatorial field, and the equatorial field is directed opposite to m.
5. The field lines run from S to N inside a magnet. Outside, they run from N to S.
6. Diamagnetism is present in every material. It is not the same as "no magnetism", and its χ is negative.
7. A diamagnetic body moves towards the weaker field. A paramagnetic or ferromagnetic body moves towards the stronger field.
8. Curie's law χ = C/T applies to paramagnetic materials. Do not apply it to diamagnetic ones, whose χ is almost independent of temperature.
9. Soft iron is used in electromagnets (low retentivity, low coercivity). Steel is used for permanent magnets (high retentivity, high coercivity). Do not swap them.
10. Retentivity is the value of B left at H = 0. Coercivity is the reverse H needed to make B zero. They are different quantities.

One-liners

1. A bar magnet always has two poles of equal strength.
2. Magnetic field lines are closed loops.
3. Field lines never cross each other.
4. Magnetic moment of a bar magnet points from south to north.
5. A bar magnet is equivalent to a solenoid, with m = NIA.
6. The net magnetic flux through a closed surface is zero.
7. The unit of magnetic moment is A m².
8. A dipole in a uniform field has zero net force.
9. The potential energy of a dipole is −mB cosθ.
10. B = μ₀ (H + M).
11. χ = M/H, and μr = 1 + χ.
12. Bismuth and copper are diamagnetic.
13. Aluminium and sodium are paramagnetic.
14. Iron, cobalt and nickel are ferromagnetic.
15. Above the Curie temperature a ferromagnet behaves as a paramagnet.

Practice questions

The torque on a magnetic dipole of moment m placed at an angle θ to a uniform magnetic field B has magnitude
1. mB cosθ
2. m sinθ / B
3. mB
4. mB sinθ
Answer
D. mB sinθ
τ = m × B, so τ = mB sinθ.
A bar magnet of magnetic moment 2 A m² is placed in a uniform magnetic field of 0.5 T at an angle of 30° to the field. The torque on it is
1. 1.0 N m
2. 0.5 N m
3. 0.25 N m
4. 0.87 N m
Answer
B. 0.5 N m
τ = mB sinθ = 2 × 0.5 × sin30° = 0.5 N m. Check: with cos30° the answer would be 0.87, which is the wrong function.
The net magnetic flux through any closed surface is
1. equal to the number of poles inside divided by μ₀
2. positive and equal to μ₀ times the enclosed current
3. zero
4. negative
Answer
C. zero
Magnetic poles do not occur singly, so ∮ B · dA = 0 for any closed surface.
A magnetic dipole of moment 3 A m² is placed in a uniform field of 0.2 T with its moment at 60° to the field. Its potential energy is
1. −0.3 J
2. −0.52 J
3. −0.6 J
4. +0.3 J
Answer
A. −0.3 J
U = −mB cosθ = −3 × 0.2 × 0.5 = −0.3 J.
A magnet of moment 0.4 A m² lies along a uniform field of 0.5 T. The work needed to turn it through 180° is
1. 0.2 J
2. 0.4 J
3. 0.1 J
4. 0.8 J
Answer
B. 0.4 J
W = U(180°) − U(0°) = mB − (−mB) = 2mB = 2 × 0.4 × 0.5 = 0.4 J.
A paramagnetic substance of susceptibility 2 × 10⁻⁵ is placed in a magnetic intensity of 5 × 10³ A m⁻¹. The magnetisation is
1. 10 A m⁻¹
2. 1.0 A m⁻¹
3. 0.01 A m⁻¹
4. 0.1 A m⁻¹
Answer
D. 0.1 A m⁻¹
M = χH = 2 × 10⁻⁵ × 5 × 10³ = 0.1 A m⁻¹.
A ferromagnetic material has a susceptibility of 999. Its relative permeability is
1. 1/999
2. 999
3. 1000
4. 998
Answer
C. 1000
μr = 1 + χ = 1000.
The susceptibility of a paramagnetic sample is 1.2 × 10⁻⁵ at 300 K. Its susceptibility at 400 K is
1. 9 × 10⁻⁶
2. 1.2 × 10⁻⁵
3. 1.6 × 10⁻⁵
4. 6 × 10⁻⁶
Answer
A. 9 × 10⁻⁶
By Curie's law χ ∝ 1/T. χ = 1.2 × 10⁻⁵ × 300/400 = 9 × 10⁻⁶.
A short bar magnet has a magnetic moment of 1.2 A m². Using μ₀/4π = 10⁻⁷ T m A⁻¹, the magnetic field on its axis at a distance of 0.1 m from its centre is
1. 2.4 × 10⁻⁵ T
2. 1.2 × 10⁻⁴ T
3. 2.4 × 10⁻⁴ T
4. 4.8 × 10⁻⁴ T
Answer
C. 2.4 × 10⁻⁴ T
B_axial = (μ₀/4π)(2m/r³) = 10⁻⁷ × 2.4/10⁻³ = 2.4 × 10⁻⁴ T. The equatorial value would be 1.2 × 10⁻⁴ T.
A bar magnet is equivalent to a solenoid of 100 turns carrying 2 A, with a cross-sectional area of 4 cm². Its magnetic moment is
1. 0.08 A m²
2. 0.02 A m²
3. 0.8 A m²
4. 8 A m²
Answer
A. 0.08 A m²
m = NIA = 100 × 2 × 4 × 10⁻⁴ = 0.08 A m².
In a material the magnetic intensity is 100 A m⁻¹ and the magnetisation is 1900 A m⁻¹. Using μ₀ = 4π × 10⁻⁷ T m A⁻¹, the magnetic field B in the material is
1. 4π × 10⁻⁵ T
2. 8π × 10⁻³ T
3. 7.6π × 10⁻⁴ T
4. 8π × 10⁻⁴ T
Answer
D. 8π × 10⁻⁴ T
B = μ₀(H + M) = 4π × 10⁻⁷ × 2000 = 8π × 10⁻⁴ T.
Which of the following is a diamagnetic substance?
1. Oxygen at STP
2. Bismuth
3. Aluminium
4. Iron
Answer
B. Bismuth
Bismuth, copper, lead, silicon and water are diamagnetic. Aluminium and oxygen are paramagnetic and iron is ferromagnetic.
Above the Curie temperature, a ferromagnetic substance behaves as
1. a paramagnetic substance
2. a diamagnetic substance
3. a stronger ferromagnet
4. a non-magnetic substance with χ = −1
Answer
A. a paramagnetic substance
The domains break up due to thermal motion, and the substance becomes paramagnetic.
When the magnetising field of a ferromagnet is reduced to zero, the magnetic field that remains in it is called its
1. susceptibility
2. retentivity
3. coercivity
4. permeability
Answer
B. retentivity
The value of B at H = 0 is the retentivity (remanence).
A small bar magnet of moment of inertia 2.5 × 10⁻⁶ kg m² and magnetic moment 0.5 A m² oscillates in a uniform field of 0.2 T. The period of small oscillations is
1. 5π × 10⁻³ s
2. 5π × 10⁻⁵ s
3. π × 10⁻² s
4. 2π × 10⁻² s
Answer
C. π × 10⁻² s
T = 2π√(I/mB) = 2π√(2.5 × 10⁻⁶/0.1) = 2π × 5 × 10⁻³ = π × 10⁻² s.
A bar magnet of magnetic moment 6 A m² is cut into three equal pieces by planes perpendicular to its length. The magnetic moment of each piece is
1. 6 A m²
2. 18 A m²
3. 1 A m²
4. 2 A m²
Answer
D. 2 A m²
Each piece has N/3 turns in the equivalent-solenoid model, with the same I and A, so m becomes 6/3 = 2 A m².
Given below are two statements: Statement I: Magnetic field lines form closed loops. Statement II: Magnetic field lines of a magnet can intersect each other.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
Field lines are closed and never intersect.
Given below are two statements: Statement I: Isolated magnetic monopoles have not been found. Statement II: The net magnetic flux through any closed surface is zero.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
Both are true and are linked: poles occur in pairs.
Given below are two statements: Statement I: A diamagnetic substance in a non-uniform field moves towards the stronger part of the field. Statement II: A paramagnetic substance in a non-uniform field moves towards the stronger part of the field.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
A diamagnetic substance moves towards the weaker part; a paramagnetic one moves towards the stronger part.
Given below are two statements: Statement I: The susceptibility of a diamagnetic substance is negative. Statement II: The susceptibility of a paramagnetic substance is independent of temperature.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
Paramagnetic susceptibility follows Curie's law, χ ∝ 1/T.
Given below are two statements: Statement I: A ferromagnetic substance becomes paramagnetic above its Curie temperature. Statement II: Above the Curie temperature the susceptibility varies as 1/(T − T_c).
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
Both are correct (Curie–Weiss form).
Which of the following statements are correct? 1. Soft iron has low coercivity. 2. Soft iron is suitable for the core of an electromagnet. 3. Steel is used for permanent magnets because its coercivity is low.
1. 2 and 3 only
2. 1 and 2 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
B. 1 and 2 only
Steel is used for permanent magnets because its coercivity and retentivity are high, so 3 is wrong.
Which of the following statements are correct? 1. The relative permeability of a diamagnetic substance is greater than 1. 2. The relative permeability of a paramagnetic substance is slightly greater than 1. 3. The susceptibility of a ferromagnetic substance is large and positive.
1. 1 and 2 only
2. 1 and 3 only
3. 2 and 3 only
4. 1, 2 and 3
Answer
C. 2 and 3 only
For a diamagnetic substance χ is negative, so μr = 1 + χ is slightly less than 1. Statement 1 is wrong.
Given below are two statements: Statement I: At the same distance, the field of a short bar magnet on its equatorial line is half the field on its axis. Statement II: The field on the equatorial line is directed along the magnetic moment of the magnet.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
B_equatorial = (μ₀/4π)(m/r³) = ½ B_axial, but it points opposite to m.
Given below are two statements: Statement I: The potential energy of a magnetic dipole in a uniform field is maximum when its moment is along the field. Statement II: A magnetic dipole is in stable equilibrium when its moment is along the field.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
U = −mB cosθ is minimum, not maximum, at θ = 0°, so Statement I is wrong.
Given below are two statements: Statement I: Outside a bar magnet the field lines run from the south pole to the north pole. Statement II: Inside a bar magnet the field lines run from the north pole to the south pole.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are incorrect
Outside they run from N to S and inside from S to N, so both statements are incorrect.
Assertion (A): The net magnetic flux through a closed surface is zero. Reason (R): Isolated magnetic poles do not exist.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
Without isolated poles, every line entering a closed surface also leaves it.
Assertion (A): Soft iron is used as the core of an electromagnet. Reason (R): Soft iron has low retentivity and low coercivity.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
The core must magnetise and lose its magnetism easily, which low retentivity and low coercivity give.
Assertion (A): Permanent magnets are made of steel or alnico. Reason (R): Alnico is a ferromagnetic alloy.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
Both are true, but the reason for the choice is high retentivity and high coercivity, not that alnico is an alloy.
Assertion (A): Diamagnetism is present in all materials. Reason (R): Diamagnetic substances have permanent atomic magnetic dipoles.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
C. A is true but R is false
Diamagnetic substances have no permanent atomic dipoles, so R is false.
Assertion (A): A ferromagnetic substance remains ferromagnetic at all temperatures. Reason (R): Above the Curie temperature a ferromagnetic substance becomes paramagnetic.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
D. A is false but R is true
A is false and R is true.
Match List I with List II. List I: a. Diamagnetic substance b. Paramagnetic substance c. Ferromagnetic substance d. Ferromagnet above its Curie temperature. List II: i. Small positive susceptibility varying as 1/T ii. Small negative susceptibility iii. Large positive susceptibility iv. Susceptibility varying as 1/(T − T_c)
1. a-iii, b-i, c-ii, d-iv
2. a-ii, b-i, c-iv, d-iii
3. a-i, b-ii, c-iii, d-iv
4. a-ii, b-i, c-iii, d-iv
Answer
D. a-ii, b-i, c-iii, d-iv
Diamagnetic: small negative χ. Paramagnetic: Curie's law. Ferromagnetic: large χ. Above T_c: χ = C/(T − T_c).
Match List I with List II. List I: a. Torque on a dipole b. Potential energy of a dipole c. Magnetisation d. Axial field of a short magnet. List II: i. Magnetic moment per unit volume ii. m × B iii. −m·B iv. (μ₀/4π)(2m/r³)
1. a-ii, b-iii, c-iv, d-i
2. a-ii, b-iii, c-i, d-iv
3. a-iii, b-ii, c-i, d-iv
4. a-i, b-iii, c-ii, d-iv
Answer
B. a-ii, b-iii, c-i, d-iv
Direct matching of each quantity to its expression.
Match List I with List II. List I: a. Magnetic moment b. Magnetic susceptibility c. Magnetic field d. Magnetic flux. List II: i. weber ii. no unit iii. A m² iv. tesla
1. a-iii, b-ii, c-i, d-iv
2. a-iv, b-ii, c-iii, d-i
3. a-ii, b-iii, c-iv, d-i
4. a-iii, b-ii, c-iv, d-i
Answer
D. a-iii, b-ii, c-iv, d-i
m in A m², χ is a ratio, B in tesla, flux in weber.
Match List I with List II. List I: a. Retentivity b. Coercivity c. Area of hysteresis loop d. Soft iron. List II: i. Energy lost per unit volume per cycle ii. B remaining at H = 0 iii. Reverse H needed to make B zero iv. Core of electromagnets
1. a-ii, b-iii, c-i, d-iv
2. a-iii, b-ii, c-i, d-iv
3. a-ii, b-iii, c-iv, d-i
4. a-i, b-iii, c-ii, d-iv
Answer
A. a-ii, b-iii, c-i, d-iv
Definitions of the loop terms.
Match List I with List II. A magnet of moment m = 2 A m² is in a uniform field B = 0.5 T. List I (torque): a. θ = 90° b. θ = 30° c. θ = 0° d. m = 4 A m², B = 1 T, θ = 30°. List II: i. 0.5 N m ii. 0 iii. 2.0 N m iv. 1.0 N m
1. a-iv, b-ii, c-i, d-iii
2. a-i, b-iv, c-ii, d-iii
3. a-iv, b-i, c-ii, d-iii
4. a-iv, b-i, c-iii, d-ii
Answer
C. a-iv, b-i, c-ii, d-iii
τ = mB sinθ: (a) 1.0; (b) 0.5; (c) 0; (d) 4 × 1 × 0.5 = 2.0 N m.
Match List I with List II. A magnet of moment 2 A m² is in a uniform field of 0.5 T. List I (potential energy): a. θ = 0° b. θ = 180° c. θ = 90° d. θ = 60°. List II: i. 0 ii. −0.5 J iii. +1.0 J iv. −1.0 J
1. a-iv, b-iii, c-i, d-ii
2. a-i, b-iii, c-iv, d-ii
3. a-iii, b-iv, c-i, d-ii
4. a-iv, b-iii, c-ii, d-i
Answer
A. a-iv, b-iii, c-i, d-ii
U = −mB cosθ with mB = 1 J: −1.0, +1.0, 0, −0.5 J.
A magnet of moment 5 A m² is at 90° to a uniform field of 0.4 T and is released. The work done by the field in turning the magnet into line with the field is
1. 0.5 J
2. 1.0 J
3. 4.0 J
4. 2.0 J
Answer
D. 2.0 J
U(90°) = 0 and U(0°) = −mB = −2 J. Work done by the field = 0 − (−2) = 2.0 J.
A small magnet in a uniform field of 0.25 T at 30° to the field experiences a torque of 0.1 N m. Its magnetic moment is
1. 0.46 A m²
2. 0.8 A m²
3. 1.6 A m²
4. 0.4 A m²
Answer
B. 0.8 A m²
m = τ/(B sinθ) = 0.1/(0.25 × 0.5) = 0.8 A m².
A rod placed in a strong non-uniform magnetic field moves slowly towards the region of weaker field. Its susceptibility is about −10⁻⁵. The rod is
1. paramagnetic
2. a perfect magnet
3. diamagnetic
4. ferromagnetic
Answer
C. diamagnetic
Negative χ and movement to the weaker part of the field show diamagnetic behaviour.
A solenoid has 400 turns per metre and carries 0.5 A. It has a core of relative permeability 500. Using μ₀ = 4π × 10⁻⁷ T m A⁻¹, the magnetic field in the core is
1. 4π × 10⁻³ T
2. 4π × 10⁻² T
3. 8π × 10⁻⁵ T
4. 8π × 10⁻² T
Answer
B. 4π × 10⁻² T
B = μ₀μr nI = 4π × 10⁻⁷ × 500 × 400 × 0.5 = 4π × 10⁻² T. Without the core it would be 8π × 10⁻⁵ T.
The susceptibility of a paramagnetic sample is 6 × 10⁻⁵ at 200 K. At what temperature is it 2 × 10⁻⁵?
1. 600 K
2. 400 K
3. 300 K
4. 67 K
Answer
A. 600 K
χ ∝ 1/T, so T = 200 × 6 × 10⁻⁵/(2 × 10⁻⁵) = 600 K.
The core of a transformer is magnetised and demagnetised many times each second. The best choice of core material, to keep heating small, is
1. copper, because it is diamagnetic
2. steel, because its retentivity is high
3. alnico, because its coercivity is high
4. soft iron, because its hysteresis loop has a small area
Answer
D. soft iron, because its hysteresis loop has a small area
The energy lost per cycle is proportional to the loop area, which is small for soft iron.
A small magnet oscillates in a uniform field with period T. If the field is made four times as strong, the new period is
1. 2T
2. T/2
3. 4T
4. T/4
Answer
B. T/2
T = 2π√(I/mB) ∝ 1/√B. A fourfold B halves the period.

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1.The camel's hump is a store of-
2.The chemical composition of the ruby gemstone is:
3.The covalent bonding behaviour of the carbon compounds corresponds to which of the following characteristics of these compounds?
4.The conversion of liquid state into solid state is known as-
5.The basic modification with a “prefix” or “suffix” in the nomenclature is the indicator of:
6.The formula unit mass of a substance is the sum of the atomic masses of all atoms in a formula unit of a/an:
7.The gas usually filled in the electric bulb is
8.The gas released on the reaction of metal carbonate with hydrochloric acid turns lime water milky. The released gas is:
9.The DNA in the cell nucleus is the information source for the making of which of the following substances?
10.The drying of milk of lime (white washing) is due to the action of-

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