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CBSE Class 12 Physics · 11 questions · 26 marks
The compass needle that guides a ship and the tiny magnet inside a speaker obey the same rules as the current loop you met in the last chapter. Here you will treat a bar magnet as a magnetic dipole, learn why magnetic field lines never begin or end anywhere, and see how the Earth itself behaves like a giant tilted magnet. The last section explains why iron sticks to a magnet, copper is quietly pushed away, and a transformer core must be made of soft iron.
A closed surface of arbitrary shape is drawn around one pole of a strong bar magnet. The net magnetic flux through this surface is:
Answer
Zero — magnetic field lines form continuous closed loops that pass through the body of the magnet, so every line entering the closed surface also leaves it. This is Gauss's law in magnetism, ∮B·dS = 0, and it is the formal statement that isolated magnetic poles have never been found.
At a certain place on the Earth the angle of dip is 45°. The ratio of the vertical component to the horizontal component of the Earth's magnetic field there is:
Answer
1 : 1 — the dip angle satisfies tanδ = B_V/B_H. With δ = 45°, tan45° = 1, so B_V = B_H. The total field is then B = √2 B_H and points at 45° below the horizontal.
For a diamagnetic substance the magnetic susceptibility χ is:
Answer
Small and negative — a diamagnetic material has no permanent atomic moments; an applied field induces moments opposing it, so the magnetisation M points against H and χ = M/H is negative, typically of order -10⁻⁵. Consequently μ_r = 1 + χ is slightly less than 1 and the material is weakly repelled by a magnet.
A bar magnet of magnetic moment 0.60 A m² is held at 30° to a uniform magnetic field of 0.25 T. The torque acting on it is:
Answer
0.075 N m — the torque on a magnetic dipole is τ = mB sinθ = 0.60 × 0.25 × sin30° = 0.60 × 0.25 × 0.5 = 0.075 N m. The torque acts to rotate the magnet into alignment with the field; it would be maximum (0.15 N m) at 90° and zero at 0°.
Assertion (A): A rod of a diamagnetic substance, suspended freely in a non-uniform magnetic field, tends to move from the stronger part of the field towards the weaker part. Reason (R): In a diamagnetic substance the induced magnetisation is directed opposite to the applied magnetic field.
Answer
Both A and R are true and R is the correct explanation of A — the induced moment of a diamagnetic sample opposes the applied field, so the sample behaves like a magnet aligned against B. In a non-uniform field such an anti-aligned dipole is pushed towards the region of weaker field. A paramagnetic rod, whose magnetisation is along the field, does the opposite.
Give two clear points of difference between paramagnetic and ferromagnetic substances, with one example of each.
Answer
1. Strength of magnetisation: a paramagnetic substance is only feebly magnetised in the direction of the applied field, with a small positive susceptibility of the order of 10⁻⁵ to 10⁻³, while a ferromagnetic substance is very strongly magnetised, with susceptibility running into hundreds or thousands. 2. Behaviour on removing the field: a paramagnetic substance loses its magnetisation almost at once and shows no hysteresis, whereas a ferromagnetic substance retains magnetisation (retentivity) and shows a hysteresis loop because of the alignment of magnetic domains. Examples: aluminium is paramagnetic; iron is ferromagnetic.
At a place in central India the horizontal component of the Earth's magnetic field is 0.32 G and the angle of dip is 60°. Calculate the total magnetic field and the vertical component there, expressing both in tesla.
Answer
The horizontal and vertical components are related to the total field B by B_H = B cosδ and B_V = B sinδ. Total field: B = B_H/cosδ = 0.32/cos60° = 0.32/0.5 = 0.64 G. Vertical component: B_V = B sinδ = 0.64 × sin60° = 0.64 × 0.866 = 0.554 G. (Cross-check: B_V = B_H tanδ = 0.32 × 1.732 = 0.554 G.) Converting, 1 G = 10⁻⁴ T, so B = 0.64×10⁻⁴ T = 6.4×10⁻⁵ T and B_V = 0.554×10⁻⁴ T = 5.5×10⁻⁵ T.
Sketch in words the hysteresis loop of a ferromagnetic material and define retentivity and coercivity from it. State which of soft iron and steel is preferred for the core of a transformer, and why.
Answer
The loop is obtained by plotting the magnetic field B in the specimen against the magnetising intensity H as H is taken from zero up to a large positive value, back down through zero to a large negative value and then up again. The returning curve does not retrace the outgoing one but encloses a loop, showing that B lags behind H. Retentivity (remanence) is the value of B that remains in the specimen when H is brought back to zero. Coercivity is the magnitude of the reverse magnetising intensity H needed to reduce that remaining B to zero. Soft iron is preferred for a transformer core. Its loop is narrow, so the energy dissipated per cycle (proportional to the loop area) is small, and its low retentivity and coercivity let it magnetise and demagnetise easily as the alternating current reverses. Steel, with a wide loop and high retentivity, is instead used for permanent magnets.
(a) Derive an expression for the magnetic field due to a short bar magnet at a point on its axial line, treating the magnet as a magnetic dipole of moment m. (b) Show that a small magnet suspended in a uniform field B executes simple harmonic motion for small angular displacements, with time period T = 2π√(I/mB). (c) A magnet of moment of inertia 8.0×10⁻⁶ kg m² and moment 0.50 A m² oscillates in a field of 2.0×10⁻⁵ T. Find its time period.
Answer
(a) Let the magnet have poles of strength q_m at a separation 2l, so that m = q_m × 2l, with the north pole at N and the south pole at S. Take a point P on the axis at distance r from the centre O, on the side of the north pole. Distance of P from N = (r - l); distance of P from S = (r + l). Field at P due to the north pole, directed from N towards P (away from the magnet): B_N = (μ₀/4π)·q_m/(r - l)² Field at P due to the south pole, directed from P towards S (towards the magnet): B_S = (μ₀/4π)·q_m/(r + l)² These are oppositely directed and B_N > B_S, so the resultant points away from the magnet along the axis: B = (μ₀q_m/4π)[1/(r - l)² - 1/(r + l)²] B = (μ₀q_m/4π) × [(r + l)² - (r - l)²]/(r² - l²)² Since (r + l)² - (r - l)² = 4rl, B = (μ₀q_m/4π) × 4rl/(r² - l²)² Using m = 2q_m l: B = (μ₀/4π) × 2mr/(r² - l²)² For a short magnet, r >> l, so l² may be neglected against r²: B(axial) = (μ₀/4π) × 2m/r³, directed along m. (b) Let the magnet be displaced through a small angle θ from the field direction. The restoring torque is τ = -mB sinθ. For small θ, sinθ ≈ θ (in radian), so τ = -mBθ. By the rotational form of Newton's second law, τ = I d²θ/dt², where I is the moment of inertia about the suspension axis: I d²θ/dt² = -mBθ d²θ/dt² = -(mB/I)θ. This is the standard equation of angular SHM, d²θ/dt² = -ω²θ, with ω² = mB/I, so ω = √(mB/I). Hence the time period is T = 2π/ω = 2π√(I/mB). (c) I = 8.0×10⁻⁶ kg m², m = 0.50 A m², B = 2.0×10⁻⁵ T. mB = 0.50 × 2.0×10⁻⁵ = 1.0×10⁻⁵. I/mB = 8.0×10⁻⁶/1.0×10⁻⁵ = 0.80 s². √0.80 = 0.894 s. T = 2 × 3.14 × 0.894 = 5.6 s.
(a) Define magnetisation M, magnetic intensity H and magnetic susceptibility χ, and derive the relation μ_r = 1 + χ. (b) A long solenoid wound with 1000 turns per metre carries a current of 0.50 A and is filled with an iron core of susceptibility 900. Calculate H, M and B inside the core, and comment on the role of the core.
Answer
(a) Magnetisation M of a material is the net magnetic dipole moment developed per unit volume, M = m(net)/V, measured in A/m. It records how strongly the atomic moments inside the specimen have been lined up. Magnetic intensity (or magnetising field) H describes the part of the field due to the free current in the windings alone, H = B₀/μ₀ = nI for a solenoid, also in A/m. It is independent of the material placed inside. Magnetic susceptibility χ measures how readily a material responds, defined by M = χH; it is a pure number. Derivation: Inside a magnetised material the total field is the sum of the field due to the coil current and the field due to the magnetisation of the material: B = μ₀(H + M) Substituting M = χH: B = μ₀(H + χH) = μ₀H(1 + χ) Writing B = μH, where μ is the permeability of the material, μ = μ₀(1 + χ) and the relative permeability is μ_r = μ/μ₀ = 1 + χ. For a vacuum χ = 0 and μ_r = 1; for a diamagnetic material μ_r is just below 1; for a ferromagnetic material μ_r is very large. (b) Magnetic intensity: H = nI = 1000 × 0.50 = 500 A/m. Magnetisation: M = χH = 900 × 500 = 4.5×10⁵ A/m. Total field: B = μ₀(H + M) = 4π×10⁻⁷ × (500 + 4.5×10⁵) H + M = 4.505×10⁵ A/m B = 1.2566×10⁻⁶ × 4.505×10⁵ = 0.566 T. Without the core the field would be only B₀ = μ₀H = 1.2566×10⁻⁶ × 500 = 6.3×10⁻⁴ T. The iron core therefore multiplies the field by about 901 (that is, by μ_r = 1 + χ), because the applied field aligns the magnetic domains of the iron, whose own field far outweighs the field of the winding.
Read the passage and answer the questions that follow: A survey team led by Nandini is mapping a mineral belt. At their base camp the compass needle settles 1.5° west of true north, a dip circle held in the magnetic meridian reads 42°, and a magnetometer gives the horizontal component of the Earth's field as 0.38 G. The team knows that the Earth behaves roughly like a huge bar magnet tilted with respect to its rotation axis, and that dip varies from 0° near the magnetic equator to 90° at the magnetic poles. (a) Name the element of the Earth's magnetic field that the 1.5° reading represents, and explain what it means for a hiker using a compass. (b) Calculate the total magnetic field of the Earth at the base camp, in tesla. (c) Calculate the vertical component of the field there. (d) The team later works at a site where the dip needle rests horizontally. State the value of the vertical component there and explain what this tells them about the location.
Answer
(a) The 1.5° reading is the magnetic declination: the angle between the geographic meridian (true north) and the magnetic meridian at that place. For a hiker it means the compass needle does not point to true north; to walk due north the hiker must correct the compass reading by 1.5° towards the east at this camp. (b) B_H = B cosδ, so B = B_H/cosδ. cos42° = 0.7431. B = 0.38/0.7431 = 0.511 G. Converting, B = 0.511×10⁻⁴ T = 5.1×10⁻⁵ T. (c) B_V = B_H tanδ = 0.38 × tan42° = 0.38 × 0.9004 = 0.342 G. That is B_V = 0.342×10⁻⁴ T = 3.4×10⁻⁵ T. (Equivalently B_V = B sinδ = 0.511 × 0.669 = 0.342 G.) (d) If the dip needle rests horizontally the angle of dip is 0°, so B_V = B sin0° = 0. The entire field is horizontal, B = B_H. This tells the team they are on the magnetic equator, the line on which the Earth's field has no vertical component at all.
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