2018 A/L Physics Past Paper · Question 49 (English Medium)

A toy car on a sinusoidal path — the e.m.f. induced in the axle

A toy car shown in figure (a) having an isolated metal axle $PQ$ , travels with a constant speed $v$ along a sinusoidal path whose vertical cross-section is in $zx$ plane as shown in figure (b). At time $t = 0$ , the axle $PQ$ coincides with the $y$ axis. If a uniform magnetic field of flux density $B$ exists throughout the region in the $+z$ direction and normal to the $xy$ plane, the variation of the induced e.m.f. $(e)$ at end $P$ of the axle with respect to end $Q$ , with time $(t)$ is best represented by, (Neglect the effect of the Earth magnetic field.)

(1) Figure (1)
(2) Figure (2)
(3) Figure (3)
(4) Figure (4)
(5) Figure (5)

Answer and explanation

Answer: (5) Figure (5)

What the question asks

We want: how the e.m.f. $e$ in the axle $PQ$ changes with time.

$e$ is at $P$ , measured from $Q$ .

Relevant theory

Electromagnetic induction can be briefly introduced as a phenomenon where an electromotive force is produced between the ends of a conducting coil or any other conductor when the magnetic flux intersecting is changing.

That is when the conductor is parallel to the magnetic field, no force acts on it even if a current passes through the conductor.

The axle is a conductor moving in a field. An e.m.f. appears between its ends.

$e = (\vec v \times \vec B) \cdot \vec L$

The part of the velocity along $\vec B$ is $v_z$ . It gives no e.m.f.

Required equations

$e = (\vec v \times \vec B)\cdot \vec L$

$e = B L v_x = B L v\cos\phi$

Reasoning and calculation

The graph has time $t$ across. It has $e$ at $P$ up the side.

At $t = 0$ the car is at $O$ , the bottom of the path. There it moves flat, along $+x$ .

So $\phi = 0$ and $v_x = v$ .

$e = BLv$

This is the largest $e$ .

At the tops and the bottoms, $\phi = 0$ . Here $e$ is largest.

At the steepest points, $\phi = \phi_{max}$ . Here $e$ is smallest.

From a top or a bottom onto a slope, $\cos\phi$ falls. So $e$ falls.

At the next top or bottom, $e$ rises again.

There is one peak of $e$ for every top and every bottom. So $e$ repeats twice as fast as the path.

The largest $e$ is $BLv$ .

The smallest is $BLv\cos\phi_{max}$ . It is more than $0$ . That is the broken line.

The force $q\vec v \times \vec B$ pushes positive charges towards $P$ . So $e$ is always positive.

So the graph starts at its highest point at $t = 0$ .

Then it swings between highs and lows. The lows never reach zero.

It looks like arches standing on the broken line.

Reading the options

Only figure (5) starts at its highest point. It stays above zero. Its lows never reach zero.

Why the other options fail

This sine wave changes sign.

It is below zero.

It starts from a low.

It touches zero.

Conclusion

$e = BLv\cos\phi$

It is largest at $t = 0$ and always positive. Its lows never reach zero.

The answer is option (5).

Explore

A toy car runs along a wavy road at a steady speed. The field $B$ points straight up. At $t = 0$ the car is at $O$ , the bottom of the road.

The metal axle $PQ$ moves with the car. When it cuts across $B$ , an e.m.f. $e$ appears between $P$ and $Q$ .

At the bottom and the top of the road the car moves flat. All its speed cuts across $B$ . So $e$ is largest.

On the slopes, part of the motion is up or down, along $B$ . That part gives nothing. So $e$ dips, but never to zero.

The car never goes backwards, so $e$ never changes sign. Largest at $t = 0$ , always above zero: option (5).

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