Question

Lenz's law in supercoductors: What happens if a superconducting ring approaches a bar magnet?

Assume that the alignment of a bar magnet passes through the center of a superconducting ring. Can we say that if we accelerate the ring toward a pole of the bar magnet applying a force F, the induced emf in the ring produces an opposing force exactly equal to F? What is the difference between using a copper ring and a superconductive one in this experiment regarding Lenz's law?

Answer 1

Lenz's law states that an induced current opposes the change in magnetic flux. When a superconducting ring approaches a bar magnet, the induced emf in the ring produces an opposing force equal to the applied force. Superconductive rings have zero resistance, making them more efficient in conserving energy compared to copper rings.

Step-by-step explanation:

Lenz's law states that an induced current will flow in a direction that opposes the change in magnetic flux that caused the current. When a superconducting ring approaches a bar magnet, applying a force F, the induced emf in the ring will produce an opposing force exactly equal to F, as per Lenz's law.

The difference between using a copper ring and a superconducting one in this experiment is that the superconductive ring will have zero resistance, allowing for the induced current to flow without any energy loss, while the copper ring will experience resistance, leading to some energy being lost as heat. This makes the superconductive ring more efficient in terms of conserving energy.