2023 A/L Physics Past Paper · Question 47 (English Medium)

Answer: (1) $2.5\,\text{mV}$, anticlockwise

What the question asks

We want: the largest e.m.f. and the direction of the current as the loop enters.

Set the directions first. We look at the page as drawn. The crosses mean the field $B$ points into the page. The loop moves to the right with $v$, its tip first.

What each letter means:

$B$ ($\text{T}$): the flux density of the field, $0.1\,\text{T}$.

$v$ ($\text{m s}^{-1}$): the speed of the loop, $0.5\,\text{m s}^{-1}$.

$a$ ($\text{m}$): the side of the triangle, $0.05\,\text{m}$.

$l$ ($\text{m}$): the length of the loop that lies along the edge of the field. It is the length that cuts the field.

$E$ ($\text{V}$): the induced e.m.f.

Relevant theory

We know that the electromotive force induced in a straight conductor of length $l$ moving in speed $v$ in a direction perpendicular to a magnetic field is given by $E = Blv$

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

The first quote gives the e.m.f. of a straight conductor of length $l$ moving with speed $v$ across a field $B$: $E = Blv$.

The second quote, Lenz's law, says that the induced current flows in the direction that opposes the change of flux causing it.

For example, push a wide broom across a floor covered in dust. The wider the part of the broom that touches new dust, the more dust it sweeps each second.

A loop entering a field is like that: the longer the part of the loop crossing the edge of the field, the more flux it sweeps in each second.

Applying it to this question

Apply Lenz's law.

As the loop moves in, more of its area is inside the field. So the flux into the page through the loop grows.

The induced current opposes this growth. It makes its own field out of the page inside the loop.

A current that makes a field out of the page, seen from the front, goes round anticlockwise.

Reasoning and calculation

Find the size.

The tip enters first. The part of the loop along the edge of the field is a straight line across the triangle. It starts at zero and gets longer as the triangle moves in.

It is longest just as the back side of the triangle reaches the edge. Then it is the back side itself, $l = a = 0.05\,\text{m}$.

So the e.m.f. $E = Blv$ is largest at that moment.

$0.05$

$l = a$

$2.5 \times 10^{-3}$

$E = Blv = 0.1 \times 0.05 \times 0.5$

$2.5$

$E = 2.5 \times 10^{-3} \times 1000$

State the direction: the current goes round the loop anticlockwise, as we see the page.

Check it a second way, with Fleming's right hand rule. The part of the loop in the field moves to the right, across a field into the page. The rule gives the current upward in that part.

Up on the right-hand part of the loop, then left along the top and down the left: that is anticlockwise. The two ways agree.

Reading the options

Now match it with the five options.

We have $2.5\,\text{mV}$, anticlockwise. Option (1) is $2.5\,\text{mV}$, anticlockwise. It matches.

Option (2) has the right size but turns the other way. Options (3), (4) and (5) have a smaller size.

Why the other options fail

The size is right, but a clockwise current would add to the growing flux into the page, against Lenz's law.

$0.5\,\text{mV}$ needs a cut length of $0.01\,\text{m}$, but the cut length reaches the full side, $0.05\,\text{m}$.

Both the size and the direction are wrong: the cut length reaches $0.05\,\text{m}$, and the current turns anticlockwise.

$0.25\,\text{mV}$ needs a cut length of $0.005\,\text{m}$, and the current is anticlockwise, not clockwise.

Checks

$B * a * v * 1000$

Check the sign of the change. While the loop enters, the flux grows, so the current is anticlockwise.

When the loop later leaves the field, the flux falls, and the current turns clockwise. The question asks about entering, so anticlockwise is the one we want.

Conclusion

$E = 2.5\,\text{mV}$

The largest e.m.f. is $E = Bav = 0.1 \times 0.05 \times 0.5 = 2.5\,\text{mV}$, and the current goes anticlockwise as the loop enters.

The answer is (1).

The quotes give $E = Blv$ for a moving conductor and Lenz's law. That the cut length of the triangle grows to the full side $a$, and the check with Fleming's right hand rule, are our reasoning from them and the figure.

Teacher's note

For a loop entering a field, find the longest part of the loop along the edge of the field. Use $E = Blv$ with that length. For the direction, ask: is the flux growing or falling? The current opposes that change.

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