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EAPCET Physics · Intermediate second year

Magnetism and Matter

Magnetism and Matter covers bar magnets, the earth’s magnetism and the classification of magnetic materials.

Subject
Physics
Taught in
Intermediate second year
Physics in the paper
40 of 160 questions

Key formulas and results

  • Axial field: B=μ04π⋅2mr3B = \frac{\mu_{0}}{4\pi}\cdot\frac{2m}{r^{3}}; equatorial field is half this, in the opposite direction
  • Torque on a magnet: τ⃗=m⃗×B⃗\vec{\tau} = \vec{m} \times \vec{B}
  • Horizontal component of the earth’s field: BH=Bcos⁡δB_{H} = B\cos\delta
  • Susceptibility: small negative for diamagnetic, small positive for paramagnetic, large positive for ferromagnetic
  • Curie’s law for paramagnets: χ∝1/T\chi \propto 1/T
  • Work to turn a magnet from θ1\theta_{1} to θ2\theta_{2} in a field: W=MB(cos⁡θ1−cos⁡θ2)W = MB(\cos\theta_{1} - \cos\theta_{2})
  • Resultant of two magnetic moments at angle θ\theta: M=M12+M22+2M1M2cos⁡θM = \sqrt{M_{1}^{2} + M_{2}^{2} + 2M_{1}M_{2}\cos\theta}
  • Period of a magnet oscillating in a uniform field: T=2πI/MBT = 2\pi\sqrt{I/MB}
  • Magnetic moment of a current loop: M=NIAM = NIA
  • B=μ0(H+I)B = \mu_{0}(H + I), where II = magnetic moment per unit volume (intensity of magnetisation); χ=I/H\chi = I/H and μr=μ/μ0=1+χ\mu_{r} = \mu/\mu_{0} = 1 + \chi
  • Short bar magnet at a general point (r,θ)(r, \theta): B=μ0M4πr31+3cos⁡2θB = \frac{\mu_{0}M}{4\pi r^{3}}\sqrt{1 + 3\cos^{2}\theta}
  • Magnet in a uniform field: U=−M⃗⋅B⃗U = -\vec{M} \cdot \vec{B}; work to rotate from θ1\theta_{1} to θ2\theta_{2}: W=MB(cos⁡θ1−cos⁡θ2)W = MB(\cos\theta_{1} - \cos\theta_{2})
  • Period of a magnet oscillating in a uniform field: T=2πI/MBT = 2\pi\sqrt{I/MB}; in a vibration magnetometer, B=BHB = B_{H}
  • Tangent galvanometer: I=2RBHμ0Ntan⁡θI = \frac{2RB_{H}}{\mu_{0}N}\tan\theta, so I∝tan⁡θI \propto \tan\theta
  • Angle of dip: tan⁡δ=BV/BH\tan\delta = B_{V}/B_{H}; apparent dip δ′\delta' in a plane at angle β\beta to the magnetic meridian: tan⁡δ′=tan⁡δcos⁡β\tan\delta' = \frac{\tan\delta}{\cos\beta}

What this chapter covers

  • Bar magnet as a dipole
  • Magnetic field on the axis and equator
  • Earth’s magnetism
  • Diamagnetic, paramagnetic and ferromagnetic materials
  • Curie’s law

How it is usually asked

  • Classify a material by susceptibility
  • Angle of dip
  • Magnet in a uniform field
  • Axial and equatorial fields

Understanding Magnetism and Matter

A bar magnet behaves like a magnetic dipole with moment m, just as a small current loop does. Its field on the axis is twice its field on the equatorial line at the same distance, and the two point in opposite directions. In a uniform field a magnet feels a torque m × B that tries to align it with the field, but no net force.

The earth behaves roughly like a giant bar magnet. At any place its field is described by the declination, the angle of dip δ and the horizontal component B_H = B cos δ, with the vertical component B sin δ. The dip is zero at the magnetic equator and 90° at the magnetic poles.

Materials respond to a magnetic field in three ways. Diamagnetic materials such as bismuth and copper are weakly repelled and have a small negative susceptibility. Paramagnetic materials such as aluminium are weakly attracted, with susceptibility falling as 1/T (Curie’s law). Ferromagnetic materials such as iron are strongly attracted, but become paramagnetic above their Curie temperature. Superconductors expel magnetic fields and act as perfect diamagnets.

Worked example

EAPCET-style question

At a place, the horizontal component of the earth’s magnetic field is 3×10−5 T3 \times 10^{-5}\ \text{T} and the angle of dip is 60∘60^\circ. The total magnetic field of the earth there is

  1. (1)6×10−5 T6 \times 10^{-5}\ \text{T}
  2. (2)1.5×10−5 T1.5 \times 10^{-5}\ \text{T}
  3. (3)3.46×10−5 T3.46 \times 10^{-5}\ \text{T}
  4. (4)5.2×10−5 T5.2 \times 10^{-5}\ \text{T}

Solution

  1. BH=Bcos⁡δB_{H} = B\cos\delta, so B=BH/cos⁡δB = B_{H}/\cos\delta.
  2. cos⁡60∘=0.5\cos 60^\circ = 0.5.
  3. B=3×10−50.5=6×10−5 TB = \frac{3 \times 10^{-5}}{0.5} = 6 \times 10^{-5}\ \text{T}.
  4. The value 5.2×10−5 T5.2 \times 10^{-5}\ \text{T} is the vertical component BHtan⁡δB_{H}\tan\delta, not the total field.

Answer: 6×10−5 T6 \times 10^{-5}\ \text{T}

Shortcuts and results to remember

  • H and I (magnetisation) have the same SI unit, A/m, so the susceptibility χ=I/H\chi = I/H has no unit.
  • Cutting a bar magnet into nn equal parts along its length gives each part moment M/nM/n; bending a magnet into a semicircle reduces its moment to 2M/π2M/\pi.
  • A magnet in a uniform field feels zero net force; θ=0∘\theta = 0^\circ is stable (U=−MBU = -MB), θ=180∘\theta = 180^\circ is unstable (U=+MBU = +MB), θ=90∘\theta = 90^\circ gives the maximum torque MBMB.
  • On the axial line the field is 2μ0M4πr3\frac{2\mu_{0}M}{4\pi r^{3}} along MM; on the equatorial line it is μ0M4πr3\frac{\mu_{0}M}{4\pi r^{3}} opposite to MM.

More practice questions

Practice question 1

A short magnet of moment 3M3M is placed at the origin pointing along +x+x, and another short magnet of moment 6M6M is placed at (a,a)(a, a), also pointing along +x+x. The resultant magnetic field at (a,0)(a, 0) is

  1. (1)Zero
  2. (2)6μ0M4πa3\frac{6\mu_{0}M}{4\pi a^{3}}
  3. (3)12μ0M4πa3\frac{12\mu_{0}M}{4\pi a^{3}}
  4. (4)3μ0M4πa3\frac{3\mu_{0}M}{4\pi a^{3}}

Solution

  1. (a,0)(a, 0) lies on the axial line of the 3M3M magnet at distance aa: B1=μ04π⋅2×3Ma3=6μ0M4πa3B_{1} = \frac{\mu_{0}}{4\pi}\cdot\frac{2 \times 3M}{a^{3}} = \frac{6\mu_{0}M}{4\pi a^{3}} along +x+x
  2. (a,0)(a, 0) lies on the equatorial line of the 6M6M magnet at distance aa: B2=μ04π⋅6Ma3=6μ0M4πa3B_{2} = \frac{\mu_{0}}{4\pi}\cdot\frac{6M}{a^{3}} = \frac{6\mu_{0}M}{4\pi a^{3}}, opposite to its moment, i.e. along −x-x
  3. The two fields are equal and opposite, so the resultant is zero

Answer: Zero

Practice question 2

A short magnet of moment 6M6M at the origin points along +x+x, and a short magnet of moment 8M8M at (a,a)(a, a) points along +y+y. The magnitude of the resultant field at (0,a)(0, a) is

  1. (1)2μ0M4πa3\frac{2\mu_{0}M}{4\pi a^{3}}
  2. (2)10μ0M4πa3\frac{10\mu_{0}M}{4\pi a^{3}}
  3. (3)14μ0M4πa3\frac{14\mu_{0}M}{4\pi a^{3}}
  4. (4)20μ0M4πa3\frac{20\mu_{0}M}{4\pi a^{3}}

Solution

  1. (0,a)(0, a) is on the equatorial line of the 6M6M magnet at distance aa: B1=6μ0M4πa3B_{1} = \frac{6\mu_{0}M}{4\pi a^{3}} along −x-x
  2. (0,a)(0, a) is at distance aa from (a,a)(a, a) along −x-x, perpendicular to the 8M8M moment, so it is on that magnet’s equatorial line: B2=8μ0M4πa3B_{2} = \frac{8\mu_{0}M}{4\pi a^{3}} along −y-y
  3. The two fields are perpendicular: B=62+82 μ0M4πa3B = \sqrt{6^{2} + 8^{2}}\,\frac{\mu_{0}M}{4\pi a^{3}}
  4. B=10μ0M4πa3B = \frac{10\mu_{0}M}{4\pi a^{3}}

Answer: 10μ0M4πa3\frac{10\mu_{0}M}{4\pi a^{3}}

Common mistakes to avoid

  • Using B sin δ for the horizontal component — the horizontal component is B cos δ and the vertical component is B sin δ.
  • Forgetting that the equatorial field of a magnet is half the axial field and opposite in direction.
  • Applying Curie’s law to diamagnetic materials — diamagnetic susceptibility is negative and almost independent of temperature.
  • Placing the angle of dip at 90° on the equator — it is 0° at the magnetic equator and 90° at the magnetic poles.

How to prepare it

Learn one example of each material type — questions often ask you to pick the odd one out.

Once the formulas feel familiar, test them the way the exam does: practise Magnetism and Matter questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Physics chapters

See every chapter in the EAPCET Physics syllabus.

Practise Magnetism and Matter chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Magnetism and Matter part of the EAPCET Physics syllabus?
Yes. Magnetism and Matter is taught in Intermediate second year Physics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Magnetism and Matter for EAPCET?
Learn one example of each material type — questions often ask you to pick the odd one out.
How many Physics questions are there in EAPCET?
40 of the 160 questions in the Engineering paper are Physics, spread across 30 chapters.
What mistakes do students make in Magnetism and Matter?
Using B sin δ for the horizontal component — the horizontal component is B cos δ and the vertical component is B sin δ. Forgetting that the equatorial field of a magnet is half the axial field and opposite in direction. Applying Curie’s law to diamagnetic materials — diamagnetic susceptibility is negative and almost independent of temperature. Placing the angle of dip at 90° on the equator — it is 0° at the magnetic equator and 90° at the magnetic poles.