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NEET 2027 · Physics · Class XII

Magnetism and Matter — Worked Set

Every question from the 18 August pages, opened out into Given, Asked, Concept, Formula, Diagram, Baby steps, Answer and Shortcut. Four solved examples and eighteen objective questions, in five concept groups.

Chapter Magnetism & MatterItems 23Figures 10 animatedChecks 38 numerical, all passedBuilt 20 Aug 2026

Section one

Torque, Couple and Potential Energy of a Magnetic Dipole

Six items. Everything here comes from two equations and one habit: measure the angle from the field direction, never from the perpendicular.

Objective Physics Vol.2Q 10Torque, Couple & Potential Energy

A bar magnet of magnetic moment M is placed in a magnetic field of induction B. The torque exerted on it is

  • (a)  M × B
  • (b)  − B · M
  • (c)  M · B
  • (d)  M + B
Given
  • Magnetic moment vector M
  • Uniform field of induction B
Asked

The vector expression for the torque on the dipole.

Concept to use

Torque turns a magnet, so it must be a vector that vanishes when M and B are parallel and peaks when they are perpendicular. Only the cross product behaves that way. The dot product is a scalar and belongs to the energy equation, not the torque equation, and adding two different physical quantities is meaningless.

Formula / rule
τ = M × B
magnitude: τ = MB sin θ
U = − M · B = − MB cos θ
Diagram
θSNMBuniform field Bτ = MB sin θ · U = −MB cos θ
Torque is maximum across the field and zero along it. Energy is the opposite: minimum along the field.
Baby steps
  1. Ask what kind of quantity torque is.It has a direction — the axis about which the magnet turns. So the answer must be a vector.
  2. Eliminate the scalars.M · B and −B · M are scalars. They cannot be a torque. Option (b) is in fact the potential energy U.
  3. Eliminate the nonsense.M + B adds A·m² to tesla. Different units cannot be added.
  4. Confirm the survivor behaves correctly.|M × B| = MB sin θ is zero at θ = 0 and maximum at θ = 90°. Correct.
Answer
(a)  τ = M × B
Shortcut
Torque is a cross, energy is a dot. If the option has a dot in it, it is answering the energy question instead.
Objective Physics Vol.2Q 11Torque, Couple & Potential Energy

The couple acting on a magnet of length 10 cm and pole strength 15 A-m, kept in a field of B = 2 × 10−5 T, at an angle of 30° is

  • (a)  1.5 × 10−5 N-m
  • (b)  1.5 × 10−3 N-m
  • (c)  1.5 × 10−2 N-m
  • (d)  1.5 × 10−6 N-m
Given
  • Magnetic length ℓ = 10 cm = 0.10 m
  • Pole strength m = 15 A·m
  • Field B = 2 × 10−5 T
  • Angle with the field θ = 30°
Asked

The couple (torque) acting on the magnet.

Concept to use

Pole strength is not magnetic moment. The question hands you m and ℓ separately so that you must build M = mℓ first. Skipping that step and using 15 in place of M is the single commonest slip in this question.

Formula / rule
M = m ℓ
τ = MB sin θ = mℓB sin θ
Diagram
θSNMBuniform field Bτ = MB sin θ · U = −MB cos θ
The couple is a pair of equal and opposite pole forces mB, separated by ℓ sin θ.
Baby steps
  1. Convert the length to metres.10 cm = 0.10 m. Do this before anything else.
  2. Build the magnetic moment.M = mℓ = 15 × 0.10 = 1.5 A·m²
  3. Put the angle in.sin 30° = 0.5
  4. Multiply.
    τ = 1.5 × (2 × 10−5) × 0.5
      = 1.5 × 10−5 N·m
Answer
(a)  1.5 × 10−5 N-m
Shortcut
All four options share the mantissa 1.5, so the whole question is only about the power of ten. 15 × 0.1 × 2 × 0.5 = 1.5 exactly, so the exponent is carried over untouched from B: 10−5.
Objective Physics Vol.2Q 12Torque, Couple & Potential Energy

A bar magnet is held at right angle to a uniform magnetic field. The couple acting on the magnet is to be halved by rotating it from this position. The angle of rotation is

  • (a)  60°
  • (b)  45°
  • (c)  30°
  • (d)  75°
Given
  • Magnet starts perpendicular to B, i.e. θ₁ = 90°
  • Final couple = ½ × initial couple
Asked

The angle turned through, not the final angle with the field.

Concept to use

This is the classic two-angle trap. The equation gives you the angle between the magnet and the field. The question asks for the rotation. They are different numbers, and 30° is sitting there as option (c) waiting for anyone who stops one step early.

Formula / rule
τ = MB sin θ
at θ = 90°: τmax = MB
rotation = θ₁ − θ₂
Diagram
θSNMBuniform field Bτ = MB sin θ · U = −MB cos θ
Starting across the field and turning towards it: the couple falls as sin θ.
Baby steps
  1. Write the starting couple.At 90°, τ₁ = MB sin 90° = MB. This is the maximum possible.
  2. Write the condition for halving.MB sin θ₂ = ½ MB  ⇒  sin θ₂ = 0.5
  3. Solve for the new angle with the field.θ₂ = 30° — this is the angle the magnet now makes with B.
  4. Convert to a rotation.The magnet moved from 90° to 30°, so it turned through 90° − 30° = 60°.
Answer
(a)  60°
Shortcut
Whenever a question starts at 90° and asks for a rotation, compute the final angle and then subtract it from 90°. The un-subtracted value is always planted as a distractor.
Objective Physics Vol.2Q 69Torque, Couple & Potential Energy

A magnet of magnetic moment M is situated with its axis along the direction of a magnetic field of strength B. The work done in rotating it by an angle of 180° will be

  • (a)  − MB
  • (b)  + MB
  • (c)  zero
  • (d)  + 2MB
Given
  • Initial position: θ₁ = 0° (aligned with B)
  • Final position: θ₂ = 180°
Asked

Work done by the external agent in the rotation.

Concept to use

Work done equals the change in potential energy. With U = −MB cos θ, the aligned position is the energy minimum (−MB) and the anti-aligned position is the maximum (+MB). Going from the bottom of the well to the top costs twice MB, not once.

Formula / rule
W = U₂ − U₁ = MB(cos θ₁ − cos θ₂)
Diagram
θSNMBuniform field Bτ = MB sin θ · U = −MB cos θ
From the energy minimum at θ = 0 all the way over to the maximum at θ = 180°.
Baby steps
  1. Write both energies.
    U₁ = −MB cos 0° = −MB
    U₂ = −MB cos 180° = + MB
  2. Subtract.W = U₂ − U₁ = MB − (−MB) = 2MB
  3. Check the sign.Positive, because you are lifting the magnet out of its stable position. Work must be supplied.
Answer
(d)  + 2MB
Shortcut
Full flip from aligned to anti-aligned always costs 2MB. Aligned to perpendicular always costs MB. Memorise those two and most work-done questions are one line.
Objective Physics Vol.2Q 70Torque, Couple & Potential Energy

A magnet of magnetic moment 2 J/T is aligned in the direction of magnetic field of 0.1 T. What is the net work done to bring the magnet normal to the magnetic field?

  • (a)  0.1 J
  • (b)  0.2 J
  • (c)  1 J
  • (d)  2 J
Given
  • M = 2 J/T (same unit as A·m²)
  • B = 0.1 T
  • From θ₁ = 0° to θ₂ = 90°
Asked

Work done to turn the magnet from aligned to perpendicular.

Concept to use

Same energy difference as the previous question, but only a quarter turn. The perpendicular position has U = 0 because cos 90° = 0, so the work is exactly MB.

Formula / rule
W = MB(cos θ₁ − cos θ₂)
  = MB(cos 0° − cos 90°) = MB
Diagram
θSNMBuniform field Bτ = MB sin θ · U = −MB cos θ
A quarter turn out of alignment costs MB — half of the full flip.
Baby steps
  1. Recognise the unit.J/T and A·m² are the same thing. No conversion needed.
  2. Apply the formula.W = MB(1 − 0) = MB
  3. Multiply.W = 2 × 0.1 = 0.2 J
Answer
(b)  0.2 J
Shortcut
Option (d) 2 J is M × 10 — it traps anyone who reads 0.1 T as 10 T. Option (a) 0.1 J is half of the right answer. Both are decimal-slip traps, so slow down on the arithmetic.
Objective Physics Vol.2Q 71Torque, Couple & Potential Energy

A planar coil having 15 turns carries 20 A current. The coil is oriented with respect to the uniform magnetic field B = 0.5 T such that its directed area is A = − 0.04 î m². The potential energy of the coil in the given orientation is

  • (a)  0
  • (b)  + 0.72 J
  • (c)  6 J
  • (d)  − 1.44 J
Given
  • N = 15, I = 20 A
  • B = 0.5 î T
  • A = − 0.04 î m²
Asked

The potential energy U of the current loop in this field.

Concept to use

A current loop is a magnetic dipole with m = NIA. The minus sign in the directed area is the whole question — it makes m point opposite to B, which is the maximum-energy orientation. Two minus signs then meet in U = −m·B and the answer comes out positive.

Formula / rule
m = NIA
U = − m · B
Diagram
θSNMBuniform field Bτ = MB sin θ · U = −MB cos θ
Moment anti-parallel to the field: the highest-energy orientation a dipole can sit in.
Baby steps
  1. Build the magnetic moment, keeping the sign.
    m = NIA = 15 × 20 × (−0.04 î)
      = − 12 î A·m²
  2. Take the dot product with B.m · B = (−12)(0.5) = − 6 J  (both along î)
  3. Apply the minus sign in the energy formula.U = − (−6) = + 6 J
  4. Sanity-check the sign.m is anti-parallel to B, so U should be at its maximum +mB. It is. Correct.
Answer
(c)  6 J
Shortcut
Skip the vectors: |U| = NIAB = 15×20×0.04×0.5 = 6. Then just fix the sign by asking whether m is with or against B. Against → positive.

Section two

Earth's Magnetic Field — Dip, Declination and Components

Six items, including three solved examples. One right-angled triangle carries this entire topic: Bₑ on the hypotenuse, H along the ground, V straight down, and the dip angle between Bₑ and H.

Solved Example 5.22Example 5.22Earth's Magnetic Field — Dip & Declination

In the magnetic meridian of a certain place, the horizontal component of earth's magnetic field is 0.26 G and the dip angle is 60°. Find (i) the vertical component of earth's magnetic field, and (ii) the net magnetic field at this place.

Given
  • H = 0.26 G
  • Dip θ = 60°
Asked

(i) V   (ii) Bₑ

Concept to use

H, V and Bₑ form a right-angled triangle with the dip angle at the ground. Every one of the three quantities can be read off using ordinary trigonometry — there is nothing to memorise beyond which side is which.

Formula / rule
tan θ = V / H
H = Bₑ cos θ    V = Bₑ sin θ
Bₑ² = H² + V²
Diagram
θ = 60°H = 0.26 GV = 0.45 GBₑ = 0.52 Ghorizontaltan θ = V / H · H = Bₑ cos θ · V = Bₑ sin θBₑ² = H² + V²
Drawn to a true 60° dip. Notice V is larger than H — steep dip means a vertical-dominated field.
Baby steps
  1. Part (i): use the tangent relation.tan θ = V / H  ⇒  V = H tan θ
  2. Substitute.
    V = 0.26 × tan 60° = 0.26 × 1.732
      = 0.45 G
  3. Part (ii): use the cosine relation.H = Bₑ cos θ  ⇒  Bₑ = H / cos θ
  4. Substitute.
    Bₑ = 0.26 / cos 60° = 0.26 / (1/2)
      = 0.52 G
  5. Cross-check with Pythagoras.√(0.26² + 0.45²) = √0.2701 = 0.52 ✓
Answer
(i) V = 0.45 G    (ii) Bₑ = 0.52 G  (= 0.52 × 10−4 T)
Shortcut
At exactly 60° the field is simply twice the horizontal component, because cos 60° = ½. Likewise dip 45° gives Bₑ = H√2 and dip 30° gives Bₑ = 2H/√3.
Solved Example 5.24Example 5.24Earth's Magnetic Field — Dip & Declination

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 earth's field is measured to be 0.16 G. Specify the direction and magnitude of the earth's field at the location.

Given
  • Declination = 12° west
  • Dip θ = 60°, north tip pointing above the horizontal
  • H = 0.16 G
Asked

Both the magnitude of Bₑ and its direction in space.

Concept to use

Two different angles are being quoted and they do different jobs. Declination (12° W) tells you which vertical plane the field lies in. Dip (60°) tells you how steeply it tilts inside that plane. Only the dip enters the arithmetic; the declination is needed purely to describe the direction.

Formula / rule
H = Bₑ cos θ  ⇒  Bₑ = H / cos θ
1 G = 10−4 T
Diagram
θ = 60°H = 0.16 GVBₑ = 0.32 Ghorizontaltan θ = V / H · H = Bₑ cos θ · V = Bₑ sin θBₑ² = H² + V²
The same 60° triangle as Example 5.22, at half the scale of field strength.
Baby steps
  1. Pick the relation that uses only what you have.You know H and θ, you want Bₑ. H = Bₑ cos θ is the one.
  2. Substitute.
    Bₑ = 0.16 / cos 60° = 0.16 / (1/2)
      = 0.16 × 2 = 0.32 G
  3. Convert to tesla.B = 0.32 × 10−4 T  since 1 G = 10−4 T
  4. Now describe the direction in words.The field lies in the vertical plane 12° west of the geographic meridian, and within that plane it points 60° above the horizontal.
  5. Read what 'above' implies.A north tip that rises above the horizontal means the location is in the southern magnetic hemisphere — consistent with a site in southern Africa.
Answer
Bₑ = 0.32 G = 0.32 × 10−4 T, lying in the vertical plane 12° west of geographic north and tilted 60° above the horizontal.
Shortcut
Declination never appears in a dip calculation. If a question gives you a declination and a dip together, the declination is almost always there only for the descriptive half of the answer.
Solved Example 5.27Example 5.27Earth's Magnetic Field — Dip & Declination

A dip circle is placed at a place where the true dip is 30°. In one vertical plane it reads an apparent dip of 45°. If the dip circle is rotated through 90°, what apparent dip will it show?

Given
  • True dip θ = 30°
  • Apparent dip in plane 1: θ₁ = 45°
  • Plane 2 is perpendicular to plane 1
Asked

The apparent dip θ₂ in the second plane.

Concept to use

A dip circle only reads the true dip when its plane contains the magnetic meridian. Tilt the plane away and it sees a reduced horizontal component, so it reports a steeper (larger) apparent dip. For two mutually perpendicular planes the two readings lock together through a cotangent-squared sum.

Formula / rule
cot² θ = cot² θ₁ + cot² θ₂
where θ is the true dip
Diagram
magnetic meridiantrue dip θ lies in this planeplane 1 → θ₁plane 2 → θ₂90°cot²θ = cot²θ₁ + cot²θ₂θ = true dipθ₁,θ₂ = apparent dips intwo perpendicular planestop view of the horizontal plane
Two perpendicular vertical planes through the same point. Each reads its own apparent dip; the true dip binds them.
Baby steps
  1. Write the relation with the known values.cot² 30° = cot² 45° + cot² θ₂
  2. Evaluate the two known cotangents.
    cot 30° = √3  ⇒  cot² 30° = 3
    cot 45° = 1  ⇒  cot² 45° = 1
  3. Solve for the unknown.cot² θ₂ = 3 − 1 = 2  ⇒  cot θ₂ = √2
  4. Convert to the angle.
    θ₂ = cot−1(√2) = tan−1(1/√2)
      = 35.26° ≈ 35°16′
  5. Sanity-check.Both apparent dips (45° and 35.3°) exceed the true dip of 30°. They must. ✓
Watch the wording. Some versions ask for the apparent dip, others for cotθ₂ or tanθ₂. The algebra is identical; only the final line changes.
Answer
θ₂ = cot−1(√2) ≈ 35.3°
Shortcut
Work in cot² throughout and never convert to angles until the last line. Also use the sanity rule: an apparent dip is always larger than the true dip. If your answer comes out smaller than the true dip, you have put the true dip on the wrong side of the equation.
Objective Physics Vol.2Q 74Earth's Magnetic Field — Dip & Declination

If the angles of dip at two places are 30° and 45° respectively, then the ratio of horizontal components of earth's magnetic field at the two places will be

  • (a)  √3 : √2
  • (b)  1 : √2
  • (c)  1 : √3
  • (d)  1 : 2
Given
  • θ₁ = 30°, θ₂ = 45°
  • Total field B taken as the same at both places
Asked

H₁ : H₂

Concept to use

With B held constant, H = B cos θ means the horizontal component depends on the dip alone. Steeper dip → smaller horizontal component. Since 30° is the shallower dip, place 1 must have the larger H — a useful check before you touch the arithmetic.

Formula / rule
H = B cos θ
H₁/H₂ = cos θ₁ / cos θ₂
Diagram
θ = 45°HVBₑhorizontaltan θ = V / H · H = Bₑ cos θ · V = Bₑ sin θBₑ² = H² + V²
At 45° the horizontal and vertical components are equal. Below 45°, H wins; above it, V wins.
Baby steps
  1. Write the ratio with B cancelling.H₁/H₂ = (B cos 30°) / (B cos 45°) = cos 30° / cos 45°
  2. Insert exact values.= (√3/2) ÷ (1/√2) = (√3/2) × √2
  3. Simplify.
    = √3 / √2    (≈ 1.22)
    so   H₁ : H₂ = √3 : √2
  4. Check the direction of the inequality.The ratio is greater than 1, so H₁ > H₂. Place 1 has the shallower dip. ✓
Direction-of-ratio trap. This is on your recurring error list. Write down which place is which before dividing, and confirm the final number is on the expected side of 1.
Answer
(a)  √3 : √2
Shortcut
Options (b), (c) and (d) are all less than 1. Only (a) exceeds 1. Since the shallower dip must give the bigger H and it is quoted first, the ratio has to be greater than 1 — the answer is forced without computing anything.
Objective Physics Vol.2Q 75Earth's Magnetic Field — Dip & Declination

The earth's magnetic field at a certain place has a horizontal component 0.3 G and the total strength 0.5 G. The angle of dip is

  • (a)  tan−1(3/4)
  • (b)  sin−1(3/4)
  • (c)  tan−1(4/3)
  • (d)  sin−1(3/5)
Given
  • H = 0.3 G
  • Bₑ = 0.5 G
Asked

The angle of dip θ.

Concept to use

You are handed a leg and the hypotenuse of the 3-4-5 triangle. Find the missing leg by Pythagoras, then use tan θ = V/H. The trap is that sin−1(3/5) is also a valid expression — but it equals the angle whose sine is H/B, which is not the dip.

Formula / rule
V = √(Bₑ² − H²)
tan θ = V / H
Diagram
θ = 53.13°H = 0.3 GV = 0.4 GBₑ = 0.5 Ghorizontaltan θ = V / H · H = Bₑ cos θ · V = Bₑ sin θBₑ² = H² + V²
The 3-4-5 triangle, drawn to its true 53.13° dip.
Baby steps
  1. Find the vertical component.
    V = √(0.5² − 0.3²) = √(0.25 − 0.09)
      = √0.16 = 0.4 G
  2. Form the tangent.tan θ = V/H = 0.4 / 0.3 = 4/3
  3. Write the angle.θ = tan−1(4/3) ≈ 53.13°
  4. Reject the near-miss options.tan−1(3/4) is V and H swapped. sin−1(3/5) = 36.87°, which is 90° − θ — the angle from the vertical, not the dip.
Answer
(c)  tan−1(4/3)
Shortcut
Spot the 3-4-5 triple instantly: given 3 and 5, the third side is 4. Then remember dip is measured from the horizontal, so H goes in the denominator. V on top, H on the bottom — every time.
Objective Physics Vol.2Q 76Earth's Magnetic Field — Dip & Declination

At a certain place, the angle of dip is 30° and the horizontal component of earth's magnetic field is 0.50 oersted. The earth's total magnetic field (in oersted) is

  • (a)  √3
  • (b)  1
  • (c)  1/√3
  • (d)  1/2
Given
  • θ = 30°
  • H = 0.50 oersted
Asked

The total field Bₑ in oersted.

Concept to use

Direct application of H = Bₑ cos θ. The only judgement needed is whether to multiply or divide by the cosine. Since H is a component, it must be smaller than the total, so Bₑ is obtained by dividing.

Formula / rule
H = Bₑ cos θ  ⇒  Bₑ = H / cos θ
Diagram
θ = 30°H = 0.50 OeVBₑ = 1/√3 Oehorizontaltan θ = V / H · H = Bₑ cos θ · V = Bₑ sin θBₑ² = H² + V²
A shallow 30° dip: H is the dominant component.
Baby steps
  1. Substitute into the cosine relation.0.50 = Bₑ × cos 30° = Bₑ × (√3/2)
  2. Rearrange.Bₑ = 0.50 × 2/√3 = 1/√3
  3. Evaluate for the sanity check.1/√3 ≈ 0.577 oersted — larger than H = 0.50, as a total field must be. ✓
Answer
(c)  1/√3 oersted  (≈ 0.577)
Shortcut
Option (d) 1/2 is smaller than the given H, so it cannot be a total field. Option (a) √3 ≈ 1.73 is more than three times H, far too big for a 30° dip. Eliminating on magnitude alone leaves (b) and (c), and (b) would need cos θ = 0.5, i.e. a 60° dip.

Section three

The Vibration Magnetometer

Five items. One formula, T = 2π√(I/MH), and almost every question is a ratio in which the constants cancel. The work is in tracking what happens to I and to M.

Objective Physics Vol.2Q 80Vibration Magnetometer

A bar magnet is oscillating in the earth's magnetic field with time period T. If its mass is increased four times, then its time period will be

  • (a)  4T
  • (b)  2T
  • (c)  T
  • (d)  T/2
Given
  • Original period T
  • Mass becomes 4m; length and magnetic moment unchanged
Asked

The new time period.

Concept to use

Mass enters the period only through the moment of inertia, I = mℓ²/12. Magnetic moment M is a property of the magnetisation, not the mass, so it stays put. Quadrupling m quadruples I, and T depends on √I.

Formula / rule
T = 2π √( I / MH )    with   I = mℓ²/12
T ∝ √m  (M, ℓ, H fixed)
Diagram
SNHtorsionless suspensionT = 2π √( I / MH )I = mℓ²/12 for a bar of mass m, length ℓ
The restoring couple comes from MH; the inertia resisting it comes from the mass distribution.
Baby steps
  1. Isolate the mass dependence.T ∝ √I and I ∝ m, so T ∝ √m.
  2. Form the ratio.T′/T = √(4m/m) = √4 = 2
  3. State the new period.T′ = 2T
Answer
(b)  2T
Shortcut
T is directly proportional to √m and inversely proportional to √M. Four times the mass gives twice the period; four times the moment gives half the period. Keep those two mirrored in your head.
Objective Physics Vol.2Q 91Vibration Magnetometer

Two bar magnets of the same mass, length and breadth having magnetic moments M and 2M are joined together pole to pole and suspended in a vibration magnetometer. The time period of oscillation is 3 s. If the polarity of one of the magnets is reversed, the time period of oscillation will be

  • (a)  √3 s
  • (b)  3√3 s
  • (c)  3 s
  • (d)  6 s
Given
  • Moments M and 2M
  • Same mass, length and breadth → total I is the same in both arrangements
  • Like poles together: T₁ = 3 s
Asked

The period after one magnet's polarity is reversed.

Concept to use

Reversing one magnet does nothing to the mass distribution, so I is untouched. It only changes how the two moments combine: they add when like poles sit together and subtract when one is flipped. That makes this a pure T ∝ 1/√M ratio problem.

Formula / rule
like poles together: Mnet = 2M + M = 3M
one reversed: Mnet = 2M − M = M
T ∝ 1/√Mnet  (I unchanged)
Diagram
SNSNSNNS3MMlike poles togetherMₖₑₜ = 2M + M = 3Mpolarity of one reversedMₖₑₜ = 2M − M = MT ∝ 1/√M ⇒ T₂/T₁ = √(3M / M) = √3
Same two magnets, same inertia, but the resultant moment drops from 3M to M.
Baby steps
  1. Find the resultant moment in each case.
    case 1 (as given): Mnet = 2M + M = 3M
    case 2 (reversed): Mnet = 2M − M = M
  2. Note what does not change.Mass, length and breadth are identical, so the combined moment of inertia I is the same in both cases. It cancels in the ratio.
  3. Form the ratio.T₂/T₁ = √(Mnet,1 / Mnet,2) = √(3M / M) = √3
  4. Substitute the given period.
    T₂ = 3 × √3 = 3√3 s
      ≈ 5.196 s
Answer
(b)  3√3 s
Shortcut
Because T is inversely proportional to √M, the period must increase when the resultant moment falls. Options (a) √3 s and (c) 3 s are at or below the original period, so they are out on inspection. Between 3√3 ≈ 5.2 and 6, only the √3 factor is available from the moments 3M and M.
Objective Physics Vol.2Q 92Vibration Magnetometer

A thin rectangular magnet suspended freely has a period of oscillation 4 s. If it is broken into two halves each having half their initial length, then the time period of oscillation of each part when suspended similarly will be

  • (a)  4 s
  • (b)  2 s
  • (c)  1 s
  • (d)  2√2 s
Given
  • Original period T = 4 s
  • Broken across the length into two equal halves
  • Each half: ℓ/2, mass m/2, moment M/2
Asked

The period of one half, suspended in the same field.

Concept to use

Three things change at once and each must be tracked. Length halves, mass halves, and — because magnetic moment is mpoleℓ with pole strength unchanged — the moment also halves. The moment of inertia falls much faster than any of them, by a factor of 8, because it carries ℓ².

Formula / rule
I = mℓ²/12  ⇒  I′ = (m/2)(ℓ/2)²/12 = I/8
M′ = M/2
T ∝ √( I / M )
Diagram
SNℓ, m, M → T = 4 sSNSNeach half: ℓ/2, m/2, M/2I′ = (m/2)(ℓ/2)²/12 = I/8 ⇒ T′ = T√( (I/8)÷(M/2) × M/I ) = T/2breaking across the length halves the period
Inertia falls by 8, moment falls by 2 — net effect a quarter inside the square root.
Baby steps
  1. Work out the new moment of inertia.
    I′ = (m/2)(ℓ/2)² / 12 = (m/2)(ℓ²/4) / 12
      = (1/8) × (mℓ²/12) = I/8
  2. Work out the new magnetic moment.Pole strength is unchanged, length halves, so M′ = M/2.
  3. Form the ratio of periods.
    T′/T = √( (I′/M′) ÷ (I/M) )
      = √( (I/8) ÷ (M/2) × M/I )
      = √( 2/8 ) = √(1/4) = 1/2
  4. Substitute.T′ = 4 × ½ = 2 s
Do not confuse this with cutting along the length. A lengthwise cut leaves ℓ unchanged and halves only m and M, giving T′ = T — a completely different answer.
Answer
(b)  2 s
Shortcut
For a bar magnet cut across its length into n equal pieces, each piece has T′ = T/n. Cut into 2 → half the period; cut into 3 → a third. Worth memorising, because this exact question recurs with different numbers.
Objective Physics Vol.2Q 94Vibration Magnetometer

Two magnets of the same size and mass make 10 and 15 oscillations per minute respectively when suspended in the same magnetic field. The ratio of their magnetic moments is

  • (a)  4 : 9
  • (b)  9 : 4
  • (c)  2 : 3
  • (d)  3 : 2
Given
  • Magnet 1: 10 oscillations per minute
  • Magnet 2: 15 oscillations per minute
  • Same size and mass → same moment of inertia I
  • Same field H
Asked

The ratio M₁ : M₂ of the magnetic moments.

Concept to use

Oscillations per minute is a frequency; the formula needs a period. Convert with T = 60 / n first. After that, identical size and mass means I cancels, leaving a clean T ∝ 1/√M relation. The magnet that oscillates faster has the stronger moment.

Formula / rule
T = 2π √( I / MH )
T₁ / T₂ = √( M₂ / M₁ )
⇒ M₁ / M₂ = T₂² / T₁²
Diagram
SNHtorsionless suspensionT = 2π √( I / MH )I = mℓ²/12 for a bar of mass m, length ℓ
Same inertia, same field: the period depends only on the magnetic moment.
Baby steps
  1. Convert both rates to periods.
    T₁ = 60 / 10 = 6 s
    T₂ = 60 / 15 = 4 s
  2. Write the proportionality.T ∝ 1/√M, so T₁/T₂ = √(M₂/M₁).
  3. Square both sides and invert.M₁/M₂ = T₂² / T₁²
  4. Substitute the periods.
    M₁/M₂ = 4² / 6² = 16 / 36
      = 4 : 9
  5. Check the direction of the ratio.Magnet 2 oscillates faster (15 per minute), so it should have the larger moment. The ratio 4 : 9 does give M₂ > M₁. ✓
Ratio-direction trap. Option (b) 9 : 4 is the same numbers reversed, and reversed ratios are on your recurring error list. Before writing the answer, name which magnet is faster and confirm it ends up with the bigger moment.
Answer
(a)  4 : 9
Shortcut
Work directly in oscillation counts and skip the periods entirely. Since n ∝ √M, the moment ratio is just the square of the count ratio: M₁:M₂ = 10² : 15² = 100 : 225 = 4 : 9. One line.
Objective Physics Vol.2Q 98Vibration Magnetometer

A magnet takes 1.5 s to complete one vibration at a place having magnetic field intensity of 0.1 × 10−5 T. At another place, it takes 2.5 s to complete one vibration. The value of earth's horizontal field at that place is

  • (a)  0.25 × 10−6 T
  • (b)  0.36 × 10−6 T
  • (c)  0.66 × 10−8 T
  • (d)  1.2 × 10−6 T
Given
  • T₁ = 1.5 s at H₁ = 0.1 × 10−5 T
  • T₂ = 2.5 s at the second place
  • Same magnet → I and M unchanged
Asked

H₂, the horizontal field at the second place.

Concept to use

Same magnet in two different fields means everything except H cancels. A longer period signals a weaker field, so the answer must come out below 10−6 T. That single observation kills option (d) before any arithmetic.

Formula / rule
T ∝ 1/√H  ⇒  H ∝ 1/T²
H₂ = H₁ (T₁ / T₂)²
Diagram
SNHtorsionless suspensionT = 2π √( I / MH )I = mℓ²/12 for a bar of mass m, length ℓ
Weaker H means a weaker restoring couple, so the magnet swings more slowly.
Baby steps
  1. Write the proportionality.T₁² H₁ = T₂² H₂ since T²H is a constant for a given magnet.
  2. Rearrange for the unknown.H₂ = H₁ (T₁/T₂)²
  3. Substitute.
    H₂ = (0.1 × 10−5) × (1.5 / 2.5)²
      = (1.0 × 10−6) × (0.6)²
      = (1.0 × 10−6) × 0.36
  4. Read off the answer.H₂ = 0.36 × 10−6 T — smaller than H₁, as expected. ✓
Partial stem. The first line of this question is cut off at the top of your photograph. The value 1.5 s is reconstructed — it is the only period that makes the answer land exactly on a printed option. Confirm against the book.
Answer
(b)  0.36 × 10−6 T
Shortcut
Rewrite 0.1 × 10−5 as 1 × 10−6 before you start. Then the whole question is just (1.5/2.5)² = 0.36 and the answer reads itself off the options.

Section four

Tangent Galvanometer and the Tangent Law

Three items. The coil is set with its plane in the magnetic meridian so that its own field is at right angles to H, and the needle settles along the resultant of the two.

Objective Physics Vol.2Q 81Tangent Galvanometer & Tangent Law

When 2 A current is passed through a tangent galvanometer, it gives a deflection of 30°. For 60° deflection the current must be

  • (a)  1 A
  • (b)  2√3 A
  • (c)  4 A
  • (d)  6 A
Given
  • I₁ = 2 A at θ₁ = 30°
  • θ₂ = 60°
  • Same instrument, same place → same reduction factor
Asked

The current I₂ needed for 60°.

Concept to use

The tangent law says the current is proportional to tan θ, not to θ itself. Doubling the angle does not double the current — that is exactly what option (c) 4 A is planted to catch.

Formula / rule
I = K tan θ    K = 2rH / μ₀N
I₂ / I₁ = tan θ₂ / tan θ₁
Diagram
θHBᶜresultantcoil plane in the magnetic meridianBᶜ = H tan θBᶜ = μ₀NI/2r⇒ I = (2rH / μ₀N) tan θI ∝ tan θ
The needle lies along the resultant of Bᶜ and H, so tan θ = Bᶜ/H.
Baby steps
  1. Write the ratio.I₂/I₁ = tan 60° / tan 30°
  2. Insert exact tangents.tan 60° = √3, tan 30° = 1/√3
  3. Divide.I₂/I₁ = √3 ÷ (1/√3) = √3 × √3 = 3
  4. Scale the current.I₂ = 3 × 2 = 6 A
Answer
(d)  6 A
Shortcut
The pair 30° and 60° always gives a tangent ratio of exactly 3 (or 1/3 the other way). Similarly 45° to 60° gives √3. Learn those two ratios and this question type takes five seconds.
Objective Physics Vol.2Q 82Tangent Galvanometer & Tangent Law

Two tangent galvanometers having coils of the same radius are connected in series. A current flowing in them produces deflections of 60° and 45° respectively. The ratio of the number of turns in the coils is

  • (a)  4√3
  • (b)  (√3 + 1)/1
  • (c)  (√3 + 1)/(√3 − 1)
  • (d)  √3 / 1
Given
  • Same radius r, same location (same H)
  • Connected in series → identical current I in both
  • θ₁ = 60°, θ₂ = 45°
Asked

N₁ : N₂

Concept to use

Series connection is the key word: it fixes I as common to both instruments. With I = (2rH/μ₀N) tan θ and I, r, H all shared, the quantity (tan θ) / N must be the same for both coils. More turns produce a bigger coil field, hence a bigger deflection.

Formula / rule
I = (2rH / μ₀N) tan θ  (same for both)
⇒ tan θ₁ / N₁ = tan θ₂ / N₂
⇒ N₁ / N₂ = tan θ₁ / tan θ₂
Diagram
θHBᶜresultantcoil plane in the magnetic meridianBᶜ = H tan θBᶜ = μ₀NI/2r⇒ I = (2rH / μ₀N) tan θI ∝ tan θ
More turns → larger Bᶜ for the same current → larger deflection.
Baby steps
  1. Use the series condition.The same current passes through both coils, so the expression for I can be equated.
  2. Cancel the shared constants.r, H and μ₀ are identical, leaving tan θ₁/N₁ = tan θ₂/N₂.
  3. Rearrange.N₁/N₂ = tan 60° / tan 45°
  4. Evaluate.
    = √3 / 1
    so   N₁ : N₂ = √3 : 1
Which coil has more turns? The one showing 60°. A larger deflection at the same current means a stronger coil field, which means more turns. Confirming that before dividing protects against the reversed-ratio error.
Answer
(d)  √3 / 1
Shortcut
Whenever tangent galvanometers are in series, the turns ratio equals the tangent ratio directly. If instead they were in parallel, the currents would differ and you would need the resistances too — so read the connection word carefully.
Objective Physics Vol.2Q 99Tangent Galvanometer & Tangent Law

A circular coil of radius 20 cm and 20 turns of wire is mounted vertically with its plane in magnetic meridian. A small magnetic needle is placed at the centre of the coil and it is deflected through 45° when a current is passed through the coil. Horizontal component of earth's field is 0.37 × 10−4 T. The current in coil is

  • (a)  0.6 A
  • (b)  6 A
  • (c)  6 × 10−3 A
  • (d)  0.06 A
Given
  • r = 20 cm = 0.20 m
  • N = 20
  • θ = 45°
  • H = 0.37 × 10−4 T
Asked

The current I in the coil.

Concept to use

Plane in the magnetic meridian means the coil's own field, which is perpendicular to the coil plane, points east–west — exactly at right angles to H. At 45° the two fields are equal, so the problem collapses to setting the coil field equal to H.

Formula / rule
Bᶜ = μ₀NI / 2r
Bᶜ = H tan θ  (at 45°: Bᶜ = H)
⇒ I = 2rH tan θ / (μ₀N)
Diagram
θHBᶜresultantcoil plane in the magnetic meridianBᶜ = H tan θBᶜ = μ₀NI/2r⇒ I = (2rH / μ₀N) tan θI ∝ tan θ
At 45° the coil field exactly matches the earth's horizontal component.
Baby steps
  1. Set the coil field equal to H.tan 45° = 1, so μ₀NI / 2r = H.
  2. Rearrange for I.I = 2rH / (μ₀N)
  3. Put the numbers in the numerator.2 × 0.20 × 0.37 × 10−4 = 1.48 × 10−5
  4. Put the numbers in the denominator.4π × 10−7 × 20 = 2.513 × 10−5
  5. Divide.
    I = 1.48 × 10−5 ÷ 2.513 × 10−5
      = 0.589 ≈ 0.6 A
Answer
(a)  0.6 A
Shortcut
The options are 0.6, 6, 0.006 and 0.06 — the same digit at four different powers of ten, so only the exponent is being tested. Group the powers first: 10−5 ÷ 10−5 = 100, so the answer is of order 1, not 10 or 0.01. That leaves 0.6 immediately.

Section five

Magnetic Materials, Susceptibility and Curie Law

Three items on the material side of the chapter: demagnetising with a solenoid, flux through a magnetised rod, and the temperature dependence of a paramagnetic sample.

Solved Example 5.50Example 5.50Magnetic Materials & Curie Law

The coercivity of a certain permanent magnet is 4.0 × 104 A m−1. The magnet is placed inside a solenoid 20 cm long and having 700 turns, and a current is passed in the solenoid to demagnetise it completely. Find the current.

Given
  • Coercivity H = 4.0 × 104 A/m
  • Solenoid length L = 20 cm = 0.20 m
  • Turns N = 700
Asked

The current needed to demagnetise the magnet completely.

Concept to use

Coercivity is defined as the magnetising intensity H that must be applied in the reverse direction to wipe out the magnetisation. It is already an H value in A/m, so no conversion from B is needed — a solenoid produces exactly H = nI, and you simply match it.

Formula / rule
n = N / L  (turns per metre)
H = n I  ⇒  I = H / n
Diagram
SNM of the magnetapplied H = nI (opposite)IH = nI, n = N / L · demagnetised when H = coercivity
The solenoid's H is applied opposite to the magnet's own moment until the net magnetisation reaches zero.
Baby steps
  1. Convert the length to metres.20 cm = 20 × 10−2 m = 0.20 m
  2. Find the turns per unit length.
    n = 700 / 0.20
      = 3500 turns/m
  3. Apply H = nI.I = H / n = (4 × 104) / 3500
  4. Divide.
    I = 40000 / 3500 = 11.43
      ≈ 11.5 A
Answer
I ≈ 11.5 A
Shortcut
Coercivity quoted in A/m goes straight into I = H/n. Coercivity quoted in tesla must first be divided by μ₀ to become an H. Check the unit before you start — that one glance decides the whole method.
Objective Physics Vol.2Q 101Magnetic Materials & Curie Law

An iron rod of 0.2 cm² cross-sectional area is subjected to a magnetising field of 1200 A m−1. The susceptibility of iron is 599. The magnetic flux produced is

  • (a)  0.904 Wb
  • (b)  1.81 × 10−5 Wb
  • (c)  0.904 × 10−5 Wb
  • (d)  5.43 × 10−5 Wb
Given
  • A = 0.2 cm² = 0.2 × 10−4 m²
  • H = 1200 A/m
  • χ = 599
Asked

The magnetic flux Φ through the rod.

Concept to use

Susceptibility χ and relative permeability are one step apart: μr = 1 + χ. Forgetting the +1 is harmless here (599 vs 600 is under 0.2%) but forgetting the area conversion is fatal — 1 cm² = 10−4 m², not 10−2.

Formula / rule
μr = 1 + χ
B = μ₀ μr H = μ₀(1 + χ)H
Φ = B A
Diagram
Airon rod, susceptibility χB = μ₀(1 + χ) HΦ = B × A
A high-susceptibility rod concentrates the flux; Φ is B multiplied by the cross-section.
Baby steps
  1. Find the relative permeability.μr = 1 + 599 = 600
  2. Compute B inside the rod.
    B = (4π × 10−7) × 600 × 1200
      = (4π × 10−7) × 7.2 × 105
      = 0.905 T
  3. Convert the area.0.2 cm² = 0.2 × 10−4 m² = 2 × 10−5 m²
  4. Multiply to get the flux.
    Φ = 0.905 × 2 × 10−5
      = 1.81 × 10−5 Wb
Area conversion. 1 cm² = (10−2 m)² = 10−4 m². Squaring the prefix is the step most often dropped.
Answer
(b)  1.81 × 10−5 Wb
Shortcut
Option (a) 0.904 Wb is the value of B, not the flux — it catches anyone who never multiplies by the area. Option (c) is B multiplied by the area with the cm² conversion done wrongly. Whenever an option matches an intermediate result exactly, treat it as a deliberate trap.
Objective Physics Vol.2 · NCERT ExemplarQ 102Magnetic Materials & Curie Law

A paramagnetic sample shows a net magnetisation of 8 A m−1 when placed in an external magnetic field of 0.6 T at a temperature of 4 K. When the same sample is placed in an external magnetic field of 0.2 T at a temperature of 16 K, the magnetisation will be

  • (a)  32/3 A m−1
  • (b)  2/3 A m−1
  • (c)  6 A m−1
  • (d)  2.4 A m−1
Given
  • State 1: M₁ = 8 A/m, B₁ = 0.6 T, T₁ = 4 K
  • State 2: B₂ = 0.2 T, T₂ = 16 K
Asked

The magnetisation M₂ in the second state.

Concept to use

Curie's law for a paramagnet: magnetisation rises with the applied field and falls with temperature, M = C B / T. Here the field is cut to a third and the temperature is raised fourfold, so both changes push M down. The answer must be well below 8 — which already eliminates options (a) and (c).

Formula / rule
M = C · B / T   (Curie law)
M₂ / M₁ = (B₂/B₁) × (T₁/T₂)
Diagram
M = 80.6/4M = 2/30.2/16B / TMM = C · B / T (Curie law)the slope C is a property of the sample alone
M plotted against B/T is a straight line through the origin. Both states sit on the same line.
Baby steps
  1. Write the ratio form of Curie's law.M₂/M₁ = (B₂/B₁)(T₁/T₂) — note the temperature ratio is inverted.
  2. Substitute the field ratio.B₂/B₁ = 0.2/0.6 = 1/3
  3. Substitute the temperature ratio.T₁/T₂ = 4/16 = 1/4
  4. Multiply through.
    M₂ = 8 × (1/3) × (1/4)
      = 8/12 = 2/3 A m−1
Inverted-ratio trap. Option (a) 32/3 is what you get by writing T₂/T₁ instead of T₁/T₂. Temperature sits in the denominator of Curie's law, so a hotter sample is always less magnetised.
Answer
(b)  2/3 A m−1
Shortcut
Combine both factors before touching the 8: the field falls by 3 and the temperature rises by 4, so M falls by 3 × 4 = 12. Then 8/12 = 2/3 in one step.
Aamirah Fathima · NEET 2027 Source: Babu Sir, 18 Aug — 15 pages All figures drawn to computed coordinates