2021 A/L Physics Past Paper · Question 50 (English Medium)

Answer: (5) Graph (5)

What the question asks

We want: the graph of the induced current $I$ against the time $t$, as the triangle moves right through the field.

The field points into the paper. The loop goes in point first, then comes out of the other side of the field point first.

The plan: follow the loop in three parts. Going in, being inside, and coming out.

Relevant theory

The magnitude of the electromotive force induced in a conducting coil or a circuit at any instant is directly proportional to the rate of change of the magnetic flux or the flux linkage number intersecting it at that instant.

The direction of the induced electromotive force is such that its current flows in a direction opposing the change of flux causing it.

Applying it to this question

Read the figure. The loop is an equilateral triangle with a point at the front. The field is a block of crosses ahead of it.

A current flows only while the number of field lines through the loop is changing. That is Faraday's law.

A moving edge that cuts across the field makes an induced voltage:

$$E = Blv$$

In words: a longer cutting edge, a stronger field or a faster loop gives a bigger voltage. Here $l$ is the length of the edge that is cutting across the boundary of the field.

Reasoning and calculation

Part 1, going in. Push a triangle-shaped net into a rain shower, point first. At first only the tip is wet. Then more and more of the net is wet.

The length of loop crossing the edge of the field grows steadily from nothing to the full side. So $E = Blv$ grows steadily and the current rises along a slanted line.

The flux into the paper is growing. The current opposes it, so it runs anticlockwise. Call this positive.

When the whole loop has just gone in, the cutting length drops away at once, so the current drops back to zero in one jump.

Part 2, inside the field. The loop is wholly in the field. The flux does not change, so there is no current. The graph lies along zero.

Part 3, coming out. The point goes out first. Again the cutting length grows steadily from nothing to the full side, so the current again rises along a slanted line.

But now the flux is falling. The current opposes that, so it runs clockwise. It is negative.

When the whole loop is out, the current drops to zero in one jump.

Reading the options

The graph must have: a rising line to a peak and then a drop to zero, a flat stretch at zero, then a line going down to a negative peak and then back to zero.

Only graph (5) does this. It is option (5).

Why the other options fail

There is no steady current while the loop is wholly inside, and the current must turn the other way on the way out.

The current grows along a slanted line, not in flat steps, and it is zero while the loop is wholly inside.

On the way out the flux falls, so the current turns the other way. Both lines cannot be positive.

The flat blocks are for a square loop. A triangle comes in point first, so the current rises along a slanted line.

Checks

Does it make sense?

The same triangle goes in and comes out. So the rising line going in and the rising line coming out are the same size. One is above zero and one is below. Graph (5) shows exactly that.

$2 * v * t * k$

Conclusion

Going in, the current rises along a slanted line and drops to zero. Inside, it is zero. Coming out, it rises along a slanted line the other way and drops to zero.

This is graph (5). The answer is (5).

Ask three questions. Is the flux changing? How fast? Which way?

A flat stretch on the graph means a steady current. For a triangle going in point first the current rises along a slanted line.

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