2023 A/L Physics Past Paper · Question 23 (English Medium)

Answer: (2) will be anti-clockwise in $A$ and clockwise in $B$.

What the question asks

What we know:

The wire is straight and carries a current $I$ upwards on the page.

Loop $A$ is on the left of the wire and loop $B$ is on the right. Both lie in the same plane as the wire.

The current $I$ is getting smaller.

We want: the direction of the induced current in $A$ and in $B$, clockwise or anti-clockwise.

What each letter means:

$I$ ($\text{A}$): the current in the straight wire.

$B$ ($\text{T}$): the magnetic flux density made by the wire.

$r$ ($\text{m}$): the distance from the wire.

$\mu_0$ ($\text{H m}^{-1}$): the magnetic permeability of free space.

We say a direction is "into the page" or "out of the page", and we look at the loops from the front.

Relevant theory

When a right handed corkscrew is passed in the direction of an electric current passing through a conductor, the sense of rotation of the screw represents the direction of the magnetic field formed around the conductor.

Then the magnetic flux density at P due to the current passing through the whole conducting wire can be derived by using mathematical integration for Biot Savart's expression which leads to the expression for flux density at P as

$B = \left(\dfrac{\mu_0}{4\pi}\right)\dfrac{2I}{r}$

$B = \dfrac{\mu_0 I}{2\pi r}$

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.

The first quote gives the way of the field round a wire. Turn a right handed corkscrew along the current, and its turning shows the field.

The second quote says the field gets weaker as $I$ gets smaller. The third quote says a changing flux induces an electromotive force. The fourth quote says the induced current flows so that it opposes the change.

For example, think of a room kept warm by a heater. If the heater is turned down, the room starts to cool. You close the window to keep the warmth.

The loop does the same. When the flux through it falls, the loop sends a current that makes flux in the same way, to bring back what is lost.

Applying it to this question

$B = \dfrac{\mu_0 I}{2\pi r}$

$\phi \propto B \propto I$

Reasoning and calculation

Loop $A$ is on the left of the wire.

The current goes up. Turn the corkscrew up along the wire. On the left of the wire the field comes out of the page.

The field through $A$ is out of the page, and it gets weaker.

Lenz says the loop makes flux in the same way, out of the page, to bring back what is lost. A field out of the page inside a loop needs a current that goes anti-clockwise.

So in $A$ the induced current is anti-clockwise.

Loop $B$ is on the right of the wire.

On the right of the wire the field goes into the page.

The field through $B$ is into the page, and it gets weaker.

Lenz says the loop makes flux in the same way, into the page, to bring back what is lost. A field into the page inside a loop needs a current that goes clockwise.

So in $B$ the induced current is clockwise.

Reading the options

Now match it with the five options.

(1) Clockwise in $A$ is wrong. We found anti-clockwise.

(2) Anti-clockwise in $A$ and clockwise in $B$. This matches.

(3) This swaps the two loops. It does not match.

(4) Anti-clockwise in $B$ is wrong. We found clockwise.

(5) A changing flux gives a current. So zero does not match.

Why the other options fail

In $A$ the current has to be anti-clockwise, to bring back the field that comes out of the page.

This swaps the two loops. The field comes out of the page at $A$ and goes into the page at $B$.

In $B$ the current has to be clockwise, to bring back the field that goes into the page.

The field of the wire gets weaker, so the flux through each loop changes, and a current is induced.

Checks

Do the two answers make sense?

The two loops are on opposite sides of the wire, so the field of the wire goes the opposite way through them. So the two induced currents go in opposite ways: anti-clockwise in $A$, clockwise in $B$.

Both are in the way that opposes the fall. That is the same as saying that both loops try to keep the flux they had.

Conclusion

The induced current is anti-clockwise in $A$ and clockwise in $B$.

The answer is (2).

First find which way the field of the wire goes through each loop, with the corkscrew rule.

Then ask whether that flux is growing or falling. The loop sends a current that opposes the change: if the flux falls, it makes flux the same way.

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