Magnetic Field and Earth's Magnetism — Practice Quiz

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.

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.

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.

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.