Magnetic Field and Earth's Magnetism
ICSE · Class 12 · Physics
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Quick Quiz: Magnetic Field and Earth's Magnetism
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A short bar magnet has a magnetic moment of 0.6 J/T. The magnetic field at a point on its axial line at a distance of 20 cm from its centre is:
A current-carrying circular coil of 50 turns has a radius of 5 cm and carries a current of 2 A. What is the magnetic dipole moment of the coil?
An electron revolves in a circular orbit of radius 0.53 Å with speed 2.2 × 10⁶ m/s around a hydrogen nucleus. What is the magnetic dipole moment associated with this orbital motion?
A bar magnet of magnetic moment 2.5 J/T is placed in a uniform magnetic field of 0.4 T. The magnet makes an angle of 30° with the field direction. What is the torque acting on the magnet?
Sample Questions
A bar magnet with magnetic moment 1.5 J/T is initially aligned with a uniform magnetic field B = 0.3 T. The work done in rotating the magnet from its initial position (θ₁ = 0°) to θ₂ = 180° is:
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0.9 J
Step 1: Work done W = mB(cos θ₁ − cos θ₂). Step 2: θ₁ = 0°, cos 0° = 1; θ₂ = 180°, cos 180° = −1. Step 3: W = 1.5 × 0.3 × (1 − (−1)) = 0.45 × 2 = 0.9 J. Step 4: This equals 2mB, the energy difference between stable and unstable equilibrium. Wrong option 0.45 J comes from using (cos 0° − cos 90°) = 1 − 0 = 1, giving W = mB = 0.45 J (this is work for rotation to 90° only). Wrong option 1.8 J comes from incorrectly doubling 2mB. Wrong option 0 J is incorrect — zero torque at 180° does not mean zero work done to reach there.
The ratio of the magnetic field at an axial point to the field at an equatorial point at the same distance from a short bar magnet is:
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2 : 1
Step 1: For a short bar magnet, the axial field is B_axial = (μ₀/4π)(2m/r³). Step 2: The equatorial field is B_equatorial = (μ₀/4π)(m/r³). Step 3: Ratio = B_axial / B_equatorial = [(μ₀/4π)(2m/r³)] / [(μ₀/4π)(m/r³)] = 2/1 = 2 : 1. Step 4: This is a fundamental result — the axial field is always twice the equatorial field at the same distance for a short dipole. This mirrors the electric dipole result. Wrong option 1:2 confuses axial with equatorial. Wrong option 4:1 comes from incorrectly squaring the factor. This result is also expressed as B_end-on = −2 × B_broad-side-on in vector form.
At a place where the angle of dip is θ, the horizontal and vertical components of Earth's field are equal. What is the value of θ?
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45°
Step 1: The relationship between components is: B_H = B_E cos θ and B_V = B_E sin θ. Step 2: Given condition: B_H = B_V. So B_E cos θ = B_E sin θ. Step 3: Dividing both sides: 1 = tan θ, therefore tan θ = 1. Step 4: θ = tan⁻¹(1) = 45°. At 45° dip, both horizontal and vertical components are equal and each equals B_E/√2. Wrong option 30°: tan 30° = 1/√3 ≠ 1. Wrong option 60°: tan 60° = √3 ≠ 1. Wrong option 90°: at poles, B_H = 0, not equal to B_V.
The horizontal component of Earth's magnetic field at a place is 0.3 G and the total field is 0.6 G. What is the angle of dip at that place?
Show answer
60°
Step 1: Use the relation B_H = B_E cos θ, so cos θ = B_H / B_E. Step 2: cos θ = 0.3 / 0.6 = 0.5. Step 3: θ = cos⁻¹(0.5) = 60°. Step 4: Verification: B_V = B_E sin 60° = 0.6 × (√3/2) = 0.52 G. Check: √(B_H² + B_V²) = √(0.09 + 0.27) = √0.36 = 0.6 G ✓. Wrong option 30°: would give cos 30° = √3/2 ≈ 0.866, meaning B_H = 0.6 × 0.866 = 0.52 G ≠ 0.3 G. Wrong option 45°: would give cos 45° = 0.707, B_H = 0.424 G ≠ 0.3 G.
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