Physics · Magnetism · chapter 5 of 14

Class 12 Physics Chapter 5: Magnetism and Matter MCQs with answers

10 multiple-choice questions on Magnetism and Matter, each with its answer and the reason, written from the NCERT chapter. For the CBSE board, JEE Main and NEET, which are set from the same NCERT chapter. Open a question's solution to check your working; then take the free test to find your next chapter.

Question 1
A long magnetised steel knitting needle is snapped into many short pieces. Each piece
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Answer: (c) is a magnet with its own north and south pole

Magnetic poles cannot be isolated: breaking a magnet gives smaller magnets, each with a north and a south pole. Isolated poles (monopoles) do not exist.

Question 2
Which statement about the magnetic field lines of a bar magnet is NOT correct?
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Answer: (c) They start at the north pole and end at the south pole.

The lines of a magnet run outside from N to S and continue inside the magnet from S to N, forming closed loops; they do not begin or end anywhere. The other three are properties stated for field lines.

Question 3
The figure shows a closed surface S₁ that encloses the north pole of a bar magnet but not its south pole. The net magnetic flux through S₁ is
SNS₁
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Answer: (b) zero

Gauss's law for magnetism: the net magnetic flux through any closed surface is zero. The lines leaving S₁ outside the magnet are balanced by the lines entering S₁ inside the magnet, which run from S to N.

Question 4
A small magnetic needle of moment 0.80 J T⁻¹ is held with its moment at 150° to a uniform magnetic field of 0.050 T. The torque on it is
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Answer: (a) 0.020 N m

τ = mB sin θ = 0.80 × 0.050 × sin 150° = 0.80 × 0.050 × 0.5 = 0.020 N m. Using the size of cos 150° gives 0.035 N m; leaving out sin θ gives 0.040 N m.

Question 5
A bar magnet of moment 2.0 J T⁻¹ lies in a uniform field of 0.30 T, its moment making 120° with the field. Taking the energy as zero when the moment is perpendicular to the field, its potential energy is
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Answer: (b) 0.30 J

U = −mB cos θ = −2.0 × 0.30 × cos 120° = −0.60 × (−0.5) = +0.30 J. The sign is positive because the moment points more against the field than along it.

Question 6
A bar magnet of moment 0.50 A m² in a uniform field of 0.40 T is turned slowly from 60° to 120° with the field. The work done by the external torque is
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Answer: (a) 0.20 J

W = U(120°) − U(60°) = mB(cos 60° − cos 120°) = 0.50 × 0.40 × (0.5 + 0.5) = 0.20 J. It is not zero: the two angles are not symmetric about the stable position θ = 0°. 0.40 J is 2mB, the work from 0° to 180°.

Question 7
At a point on the equatorial line (normal bisector) of a short bar magnet, far from it, the magnetic field is
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Answer: (d) opposite to the magnetic moment m

BE = −(μ₀/4π) m/r³: the minus sign shows the field on the normal bisector is directed opposite to m.

Question 8
The field of a short bar magnet at a point on its axis at a distance r is B. At a point on its equatorial line at a distance 2r, the field has magnitude
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Answer: (d) B/16

Axial: B = (μ₀/4π)(2m/r³). Equatorial at 2r: (μ₀/4π) m/(2r)³ = (μ₀/4π) m/8r³ = B/16. Forgetting the factor 2 of the axial field gives B/8; counting it twice gives B/32; ignoring the change of distance gives B/2.

Question 9
The field of a short bar magnet at a point on its axis 10 cm from its centre is 4.0 × 10⁻⁴ T. At a point on its equatorial line 20 cm from its centre, the field is
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Answer: (c) 2.5 × 10⁻⁵ T

Halving for the equatorial position and dividing by 2³ = 8 for the doubled distance: B = 4.0 × 10⁻⁴ × (1/2) × (1/8) = 2.5 × 10⁻⁵ T. Dropping the 1/2 gives 5.0 × 10⁻⁵ T.

Question 10
The SI unit of magnetisation M is
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Answer: (d) A m⁻¹

M is the net magnetic moment per unit volume: (A m²)/m³ = A m⁻¹, the same unit as H.

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