Concept of a magnetic field — practice questions

1. Identify the given values: Force, F = 1.5 N Current, I = 3.0 A Length, L = 0.25 m Angle, θ = 90° (perpendicular)
2. Recall the formula for magnetic force: $F = BIL \sin(\theta)$
3. Rearrange for B: $B = \frac{F}{IL \sin(\theta)}$
4. Substitute the values into the formula: $B = \frac{1.5 \text{ N}}{(3.0 \text{ A})(0.25 \text{ m}) \sin(90°)}$
5. Calculate the result (since sin(90°) = 1): $B = \frac{1.5}{0.75}$ $B = 2.0 \text{ T}$
6. State the answer with units: The magnetic flux density is 2.0 Tesla.

1. Identify given values and convert units: Magnetic Flux Density, B = 1.2 T Coil dimensions: 10 cm = 0.10 m, 5 cm = 0.05 m Rotation angle = 30°
2. Calculate the area of the coil: Area, A = length × width = 0.10 m × 0.05 m = 0.0050 m²

Part 1: Initial Flux 3. The field is perpendicular to the coil's plane, so the angle θ between the field B and the normal to the area A is 0°. 4. Use the magnetic flux formula: Φ = BA cos(θ) 5. Substitute values: Φ_initial = (1.2 T) × (0.0050 m²) × cos(0°) Φ_initial = 0.0060 × 1 Φ_initial = 6.0 × 10⁻³ Wb

Part 2: New Flux after Rotation 6. The coil is rotated by 30°. The angle θ between the normal and the field lines is now 30°. 7. Use the magnetic flux formula again: Φ_new = BA cos(θ) 8. Substitute new values: Φ_new = (1.2 T) × (0.0050 m²) × cos(30°) Φ_new = 0.0060 × 0.866 Φ_new = 0.005196 Wb 9. State the answer to 2 significant figures: The new magnetic flux is 5.2 × 10⁻³ Wb (or 5.2 mWb).