2021 A/L Physics Past Paper · Question 44 (English Medium)

Answer: (3) $I = \dfrac{mg\tan\theta}{lB}$ and $Q$ to $P$

What the question asks

We want: how big the current $I$ must be, and which way it goes, to keep the wire still on the slope.

Think of a toy car on a ramp. It wants to roll down. To keep it still, you push it sideways into the ramp.

Here the magnetic field gives that push.

Relevant theory

The force acting on a unit current element of a current carrying conductor placed normally to a magnetic field is the magnetic flux density of the field at that point.

In the above energy conversion, there are three quantities normal to each other. They are the electric current, the magnetic field and the final result of the mechanical force or the motion.

Required equations

$F = BIl$

$mg\sin\theta$

$F\cos\theta$

$F\cos\theta = mg\sin\theta$

Reasoning and calculation

The symbols:

$m$ is the mass of the wire. $l$ is its length.

$B$ is the magnetic field. It points straight up.

$I$ is the current in the wire.

$\theta$ is the angle of the slope.

$F$ is the magnetic push on the wire.

Three things act on the wire: its weight $mg$ down, the push of the slope at right angles to the slope, and the magnetic push $F$.

The current, the field and the push are at right angles to each other. The field is up and the wire is level, so the push is level. It points into the hill.

Along the slope, gravity pulls the wire down with $mg\sin\theta$. The level push has a part along the slope, $F\cos\theta$, which pulls it up.

To keep still, the two are equal:

$$F\cos\theta = mg\sin\theta$$

So the magnetic push is:

$$F = mg\tan\theta$$

The wire is at right angles to the field, so $F = BIl$. Put it in:

$$BIl = mg\tan\theta$$

Divide both sides by $Bl$:

$$I = \dfrac{mg\tan\theta}{lB}$$

In words: a steeper slope, a heavier wire or a weaker field needs more current.

Now the direction. Use your left hand. Point the first finger up, along $B$. Point the thumb into the slope, the way the push must go.

The second finger then shows the way of the current. It points from $Q$ to $P$.

Reading the options

The size is $\dfrac{mg\tan\theta}{lB}$ and the direction is $Q$ to $P$. This is option (3).

Options (1) and (2) use $\sin\theta$ instead of $\tan\theta$. Option (4) has the right size but the wrong direction. Option (5) has no angle at all.

Checks

Unit check:

$$BIl = \text{T} \times \text{A} \times \text{m} = \text{N}$$

And $mg\tan\theta$ is also a force in newtons. The two sides match.

Two extreme slopes.

On a flat plane $\theta = 0$, so $\tan\theta = 0$. No current is needed, because nothing pulls the wire sideways.

On a very steep slope, $\tan\theta$ is large. A big current is needed.

$m * g * t / (l * b)$

Conclusion

$I = \dfrac{mg\tan\theta}{lB}, \ Q \to P$

The magnetic push must be level and into the slope, with $F = mg\tan\theta$.

With $F = BIl$ this gives $I = \dfrac{mg\tan\theta}{lB}$, and the current goes from $Q$ to $P$. The answer is (3).

On a slope with a vertical field, the magnetic push is level. Only $\cos\theta$ of it works along the slope.

Balance the parts along the slope. That gives $\tan\theta$, not $\sin\theta$.

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