2019 A/L Physics Past Paper · Question 20 (English Medium)

Emf induced in a ring coil around a straight wire

A coil is made by winding $N$ number of turns around a plastic ring of mean radius $R$ and cross sectional area $A$ as shown in the figure. This coil is placed coaxially with a long straight wire carrying a current $i$ . If the rate of change of current through the straight wire is $i_0 \cos\omega t$ , which of the following expressions gives the electromotive force (emf) induced in the coil?

(1) $μ_0 ANi_0\cos ωt$
(2) $μ_0 AN^2 i_0\sin ωt$
(3) $\frac{μ_0 AN}{ω}\,i_0\sin ωt$
(4) $\frac{μ_0 AN}{2πR}\,i_0\cos ωt$
(5) $\frac{μ_0 AN}{4π^2R^2}\,i_0\cos ωt$

Answer and explanation

Answer: (4) $\frac{μ_0 AN}{2πR}\,i_0\cos ωt$

What the question asks

We want:

The emf induced in a ring coil round a straight wire.

The current in the wire changes.

We are given its rate of change, $\dfrac{di}{dt} = i_0\cos\omega t$ .

Relevant theory

The flux density of a long straight wire at the distance $R$ is:

$B = \dfrac{\mu_0 i}{2\pi R}$

The field lines are circles round the wire.

Faraday law: the emf is directly proportional to the rate of change of the flux linkage.

$\varepsilon = N\dfrac{d\Phi}{dt}$

We only want the size. So we leave out the minus sign.

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.

Required equations

$B = \frac{\mu_0 i}{2\pi R}$

$\Phi = BA$

$\varepsilon = N\frac{d\Phi}{dt}$

Reasoning and calculation

The symbols:

$R$ is the mean radius of the ring. It is the distance of the turns from the wire.

$A$ is the cross section of one turn.

$N$ is the number of turns.

$\dfrac{di}{dt} = i_0\cos\omega t$ is the rate of change of the current. It is not the current.

The ring has the same axis as the wire.

So the circular field lines go straight through every turn.

The area $A$ is small, so $B$ is the same over the whole turn.

$N\Phi = N A \frac{\mu_0 i}{2\pi R}$

$\varepsilon = \frac{\mu_0 A N}{2\pi R}\frac{di}{dt}$

$\varepsilon = \frac{\mu_0 A N}{2\pi R}\,i_0\cos\omega t$

Reading the options

Compare the forms.

Only option (4) divides by $2\pi R$ .

It has $N$ once.

It has $\cos\omega t$ .

Checks

Unit check:

$\mu_0\dfrac{di}{dt}$ has the unit $\text{T m A}^{-1} \cdot \text{A s}^{-1}$ .

$\dfrac{A}{R}$ has the unit m.

So the unit is $\text{T m}^2\text{ s}^{-1} = \text{Wb s}^{-1} = \text{V}$ .

Options (1) and (2) have no $R$ . They have an extra metre. Option (5) is short of one metre. They are not volts.

Limit check:

If $R$ is very large, the coil is far from the wire. The field is zero. So the emf is zero.

Option (4) gives this. Options (1), (2) and (3) do not depend on $R$ .

If $\dfrac{di}{dt} = 0$ the current is steady. The emf is zero.

Conclusion

$\varepsilon = \frac{\mu_0 A N}{2\pi R}\,i_0\cos\omega t$

The flux linkage is $N A \dfrac{\mu_0 i}{2\pi R}$ .

Its rate of change is:

$\varepsilon = \dfrac{\mu_0 AN}{2\pi R}\,i_0\cos\omega t$

So the answer is option (4).

Treating the given $i_0\cos\omega t$ as the current.

It is already the rate of change $\dfrac{di}{dt}$ .

The emf is directly proportional to it.

So we do not differentiate or integrate it again.

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