Eddy Currents and Applications - UNSOLVED PRACTICE SET

Q7. What are eddy currents? Why are they called "eddy" currents?

Q8. Explain how laminating a transformer core reduces eddy current losses. Why must the laminations be insulated from each other?

Q9. A copper plate is swung like a pendulum between the poles of a strong magnet. Describe what happens and explain why.

Q10. Give two applications where eddy currents are useful and two where they are harmful.

Q11. Why is the core of an induction furnace made of a conducting material rather than an insulating material?

Q12. A magnet falls through a vertical aluminum pipe. Compare its motion with falling through a PVC pipe of the same dimensions. Explain the difference.

Q13. Explain the formation of eddy currents in a conducting plate when a magnetic field changes near it. Use Lenz's Law to explain the direction of eddy currents. Why do eddy currents cause heating?

Q14. Describe the working of an induction furnace with a diagram. How does it use eddy currents to melt metals? Why is it preferred over conventional furnaces for certain applications?

Q15. A thick copper plate of area 100 cm² and thickness 2 mm is placed perpendicular to a magnetic field that changes from 0 to 0.5 T in 0.1 s. The resistivity of copper is 1.7 × 10⁻⁸ Ω m.

(a) Calculate the induced EMF in the plate.

(b) Estimate the resistance of the eddy current paths.

(c) Calculate the approximate power dissipated as heat due to eddy currents.

(d) If the plate is laminated into 20 thin sheets insulated from each other, how does the power loss change?

Q16. In a school physics demonstration, a strong magnet is dropped through a copper tube (length 1.2 m, inner diameter 2.5 cm, wall thickness 2 mm) and a PVC tube of identical dimensions.

(a) Predict and explain the difference in fall times.

(b) Calculate the approximate resistance of the copper tube wall for circumferential current paths.

(c) If the magnet's field is 0.8 T at the tube wall, estimate the induced EMF as the magnet falls at 0.5 m/s.

(d) Estimate the power dissipated as eddy currents and compare with the magnet's loss of gravitational potential energy.

(e) The copper tube is cut lengthwise into a C-shape. Predict what happens now and explain using Lenz's Law.

[Given: ρ_copper = 1.7 × 10⁻⁸ Ω m]

Q17. An induction cooktop uses a coil with 30 turns carrying alternating current at 25 kHz, producing a peak magnetic field of 0.2 T at the pan surface. A steel pan bottom has area 300 cm², thickness 3 mm, and resistivity 10⁻⁷ Ω m.

(a) Calculate the rate of change of magnetic field (dB/dt) assuming sinusoidal variation.

(b) Estimate the induced EMF in the pan bottom.

(c) Calculate the resistance of the pan bottom for eddy current paths.

(d) Estimate the power dissipated as heat.

(e) If the pan is replaced with an aluminum one (ρ = 2.8 × 10⁻⁸ Ω m), will the heating increase or decrease? Explain why copper-bottomed pans don't work well on induction cooktops despite copper having lower resistivity.

Q18. A transformer core has volume 5000 cm³ and is made of iron with resistivity 10⁻⁷ Ω m. The core operates at 50 Hz with peak magnetic field 1.2 T.

(a) Calculate the approximate eddy current power loss in a solid core.

(b) The core is laminated into sheets of thickness 0.5 mm insulated from each other. Calculate the reduction in eddy current loss.

(c) Calculate the power loss in the laminated core.

(d) If the transformer handles 10 kW, what percentage of power is lost to eddy currents in (a) and (c)?

(e) Explain why transformer cores are laminated parallel to the magnetic field direction, not perpendicular to it.