EasyPhysicsHigh School (11–12)Electricity & Magnetism › Electromagnetic InductionConceptualA-LevelAPIB
A current enters a coil at the end P and leaves it at the end Q…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
A current enters a coil at the end P and leaves it at the end Q. (a) The current is growing. Which end of the coil is at the higher potential? Explain. (b) The current is falling. Which end is higher now?
Need a reminder? The key ideaShow
The flux linkage of a coil is . The self-induced emf is , and it opposes the change of the current. For a long solenoid, . The energy stored is , with .
The worked solution
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. A growing current. The emf opposes the growth. It must push current from Q towards P inside the coil, against the current. A source that pushes current from Q to P has its positive end at P. So P is at the higher potential.
Step 2. A falling current. The emf now opposes the fall. It pushes current from P towards Q, along the current. Its positive end is at Q. So Q is at the higher potential.
Answer: (a) P is higher. (b) Q is higher.
Watch out. The sign depends on how the current changes, not on the direction of the current. When the current is steady, both ends are at the same potential if the wire has no resistance.
Discussion (0)
Log in to join the discussion.
Loading…