2022 A/L Physics Past Paper · Question 39 (English Medium)

Answer: (1) $\dfrac{1}{4}\omega r^2 B$

Given information

What we know:

A semicircular wire loop of radius $r$ has its straight edge through the centre $O$. The resistance of the loop is $2\ \Omega$.

The loop turns about $O$, in its own plane, at the steady angular velocity $\omega$.

The magnetic field $B$ points into the page. In the figure the crosses are only on the left of the straight edge, so the curved half starts outside the field.

The question has no numbers except $2\ \Omega$. So we keep the letters and check with sample values $B = 0.5\ \text{T}$, $r = 0.2\ \text{m}$ and $\omega = 10\ \text{rad s}^{-1}$.

What the question asks

We want: the size of the current in the loop.

What each letter means:

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

$r$ ($\text{m}$): the radius of the semicircle.

$\omega$ ($\text{rad s}^{-1}$): the angular velocity of the loop.

$\Phi$ ($\text{Wb}$): the magnetic flux through the loop.

$\varepsilon$ ($\text{V}$): the induced emf.

$R$ ($\Omega$): the resistance of the loop, $2\ \Omega$.

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

Relevant theory

What is happening:

The loop turns. The edge of the loop swings into the region where the field is, and the curved half follows.

The field is the same everywhere in its region, so the flux through the loop is $B$ times the area of the loop that is inside the field.

Turning by an angle $\theta$ brings a sector of the loop of angle $\theta$ into the field. Its area is $\dfrac{1}{2}r^2\theta$.

As the loop keeps turning, the flux through it keeps changing. A changing flux induces an emf, and the loop is a closed circuit, so a current flows.

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 current flowing in a conductor is directly proportional to the potential difference between its ends provided temperature and its other physical factors remain constant

The area inside the field is $\dfrac{1}{2}r^2\theta$ and $\theta = \omega t$. So the flux is:

$$\Phi = B \times \dfrac{1}{2}r^2\omega t$$

The flux grows steadily with time. Its rate of change is:

$$\dfrac{d\Phi}{dt} = \dfrac{1}{2}B\omega r^2$$

In words: the emf equals the rate at which the flux changes, which is $\dfrac{1}{2}B\omega r^2$.

The current is the emf divided by the resistance.

For example, the hand of a big clock sweeps over the dial. The area it has swept so far is a slice that grows at a steady rate when the hand turns at a steady rate.

The loop is like that hand. The sector of the loop inside the field grows at a steady rate.

Required equations

$A = \dfrac{1}{2}r^2\omega t$

$\varepsilon = \dfrac{d\Phi}{dt} = \dfrac{1}{2}B\omega r^2$

$I = \dfrac{\varepsilon}{R} = \dfrac{\frac{1}{2}B\omega r^2}{2}$

$I = \dfrac{1}{4}\omega r^2 B$

Applying it to this question

$\text{T} \times \text{rad s}^{-1} \times \text{m}^2 = \text{V}$, and $\dfrac{\text{V}}{\Omega} = \text{A}$, so the answer is a current.

$0.2$

$\dfrac{1}{2}r^2\omega = \dfrac{1}{2} \times 0.2^2 \times 10$

$0.1$

$\varepsilon = B \times 0.2 = 0.5 \times 0.2$

$0.05$

$I = \dfrac{\varepsilon}{R} = \dfrac{0.1}{2}$

Reasoning and calculation

The current is $I = \dfrac{1}{4}\omega r^2 B$.

That is option (1).

Checks

$w * r * r * b / 4$

Does it make sense?

The current gets bigger if $B$, $\omega$ or $r^2$ is bigger. A stronger field, a faster turn or a bigger loop all change the flux faster. That fits.

The current gets smaller if the resistance is bigger, as Ohm's law says.

This steady current flows while the loop is entering the field. When the loop leaves the field again, the flux shrinks at the same rate. The size of the current is the same and its direction is reversed.

Conclusion

$I = \dfrac{1}{4}\omega r^2 B$

The current in the loop is $I = \dfrac{1}{4}\omega r^2 B$.

The answer is (1).

Find how fast the area inside the field is changing. Multiply by $B$ to get the emf, then divide by the resistance to get the current.

Look at the figure to see where the field is. Here the field covers only one side of the straight edge.

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