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Class 12 Physics Chapter 5: Magnetism and Matter case study question with answers

One case-based question of the kind the board asks: a passage and a figure, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.

Case study · 4 marks
In a school laboratory, Rohit places a short bar magnet of magnetic moment 1.6 A m² at O on a table, its moment pointing along OP, as shown. P is on the axis of the magnet, 40 cm from O; Q is on its equatorial line, 20 cm from O. Both distances are large compared with the size of the magnet, so the short-magnet formulae apply. He then maps the field with a small compass needle, which sets itself along the field at each point. Take μ₀/4π = 10⁻⁷ T m A⁻¹.
SNmPQOP = 40 cmOQ = 20 cmO
Part 1 · 1 mark
The magnitude of the field of the magnet at P is
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Answer: (a) 5.0 × 10⁻⁶ T

BP = (μ₀/4π)(2m/r³) = 10⁻⁷ × 2 × 1.6/(0.40)³ = 5.0 × 10⁻⁶ T. Leaving out the 2 gives 2.5 × 10⁻⁶ T.

Part 2 · 1 mark
At Q, the north end of the compass needle points
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Answer: (a) parallel to OP, but in the opposite direction

On the equatorial line BE = −(μ₀/4π) m/r³: the field is opposite to m, and the needle's north end points along the field.

Part 3 · 2 marks
Find the ratio of the magnitude of the field at Q to that at P.
Answer and marking scheme

BQ = (μ₀/4π) m/r³ = 10⁻⁷ × 1.6/(0.20)³ = 2.0 × 10⁻⁵ T. BP = 5.0 × 10⁻⁶ T. BQ/BP = 4: the factor 2 on the axis is outweighed by the factor 8 from halving the distance.

  • B_Q = μ₀m/4πr³ = 2.0 × 10⁻⁵ T (1)
  • ratio 4 : 1 (1)
Or
A second small needle of moment 0.010 A m² is held at Q with its moment pointing in the direction of OP. Find its potential energy in the magnet's field (zero when perpendicular to the field) and state whether this position is stable.
Answer and marking scheme

At Q the field, 2.0 × 10⁻⁵ T, is opposite to OP, so the needle's moment is anti-parallel to B: U = −m′·B = +m′B = 0.010 × 2.0 × 10⁻⁵ = 2.0 × 10⁻⁷ J. Energy is maximum: an unstable equilibrium.

  • field at Q opposite to OP, 2.0 × 10⁻⁵ T; U = +m′B = 2.0 × 10⁻⁷ J (1)
  • maximum energy, unstable equilibrium (1)
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