Скачать или смотреть HSC Physics 2017 answers to short answer

Here are the solutions to the short answer section of the HSC Physics paper for 2017.

ERRATA:
I make an error in Q27b in determining the polarity at P

Here is the correct explanation:
Firstly, the flux is decreasing from 10-12s
Therefore an emf must be generated that opposes this, so it needs to produce an increasing flux in that are.
This means that the loop generates a clockwise current.
There are now two way you can look at this as to why P is positive .
1. We always connect a voltmeter so that the current ENTERS the voltmeter at the positive terminal , or red terminal. This means P is the positive terminal
2. If conventional current is clockwise then electron flow is anti-clockwise.
With the voltmeter being a high resistance you could argue there will be an increase negative charge at Q as opposed to P. Again, P is positive

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