The Workbook

Eddy currents are closed loops of current that circulate inside a solid piece of conductor, like eddies in water. They are induced whenever the magnetic flux through the metal changes, either because the field changes or because the metal moves through a non-uniform field.

1. As the plate enters and leaves the gap, the flux through it changes and eddy currents flow. By Lenz's law these currents oppose the motion, so the plate feels a drag and its kinetic energy is turned into heat in the metal.

2. The slots cut across the paths the eddy currents would take, so only small loops can form and the currents are much weaker. The drag is small and the plate swings on.

Magnetic braking of trains and rotating machines. Induction furnaces, where eddy currents melt the metal. Damping of the coil in a dead-beat galvanometer. Also the speedometer of a vehicle and induction cooking.

A solid core would carry large eddy currents driven by the changing flux, wasting energy as heat. Laminations with insulating layers between them break the core into thin sheets, confining any eddy current to a small area. The currents and the losses are greatly reduced.

The alternating current makes a changing flux through the ring, inducing a current in it. By Lenz's law the induced current opposes the change, so the ring and coil act like two coils carrying opposite currents and repel. The upward force throws the ring off the core.

1. The flux through the coil is \(\Phi = NBA\cos\omega t\), since the angle between the normal and \(\vec B\) grows as \(\omega t\). Faraday: \(\varepsilon = -\dfrac{d\Phi}{dt} = NBA\omega\sin\omega t\), an alternating emf.

2. The slip rings turn with the coil and the brushes press on them. The rotating coil thus stays connected to the fixed external circuit without the leads twisting.

\(\omega = 2\pi \times 50 = 100\pi\) rad/s. Peak \(= NBA\omega = 200 \times 0.2 \times 0.05 \times 100\pi = 200\pi \approx 628\) V. Rms \(= \dfrac{628}{\sqrt2} \approx 444\) V.

\(\omega t = 100\pi \times \dfrac{1}{600} = \dfrac{\pi}{6}\). \(\varepsilon = 200\pi\sin\dfrac{\pi}{6} = 100\pi \approx 314\) V.

The emf is largest when the plane of the coil is parallel to the field: the flux is zero there but changing fastest. It is zero when the plane is perpendicular to the field: the flux is greatest but momentarily not changing.

Peak \(= NBA\omega\). Double the number of turns, the area, the field, or the speed of rotation. Doubling the frequency also halves the period, so the emf alternates twice as fast.

Self-inductance is the flux linkage per unit current, \(\Phi = LI\), or the emf induced per unit rate of change of current, \(\varepsilon = -L\dfrac{dI}{dt}\). Its unit is the henry: \(1\) H \(= 1\) Wb/A \(= 1\) V s/A.

\(L = \mu_0n^2Al = 4\pi \times 10^{-7} \times 10^{6} \times 10^{-3} \times 0.5 \approx 6.3 \times 10^{-4}\) H.

1. \(\dfrac{dI}{dt} = \dfrac{4}{0.1} = 40\) A/s. \(|\varepsilon| = L\dfrac{dI}{dt} = 0.2 \times 40 = 8\) V.

2. The current is steady: no emf.

3. \(\dfrac{dI}{dt} = \dfrac{4}{0.2} = 20\) A/s, so \(|\varepsilon| = 4\) V. The current is falling, so this emf tries to keep it going: opposite in sense to that of stage I.

\(U = \dfrac{1}{2}LI^2 = \dfrac{1}{2} \times 0.2 \times 16 = 1.6\) J.

\(L = \mu_r\mu_0n^2Al = 1000 \times 6.3 \times 10^{-4} \approx 0.63\) H.

\(u = \dfrac{B^2}{2\mu_0} = \dfrac{1}{2 \times 4\pi \times 10^{-7}} \approx 4 \times 10^{5}\) J/m\(^3\).

Opening the switch tries to stop the current almost instantly, so \(\dfrac{dI}{dt}\) is very large. The inductor's emf \(L\dfrac{dI}{dt}\) is then large enough to break down the air in the gap. The stored energy \(\dfrac{1}{2}LI^2\) is released in the spark.

Mutual inductance \(M\) is the flux linked with coil 2 per unit current in coil 1: \(\Phi_2 = MI_1\). When \(I_1\) changes, \(\varepsilon_2 = -M\dfrac{dI_1}{dt}\).

1. \(B_1 = \mu_0n_1I_1\) inside. Flux linkage with the coil per ampere: \(N_2B_1A = \mu_0n_1N_2A = 4\pi \times 10^{-7} \times 1000 \times 100 \times 10^{-3} \approx 1.26 \times 10^{-4}\) Wb.

2. \(M\) is this flux linkage per ampere: \(M = \mu_0n_1N_2A \approx 1.26 \times 10^{-4}\) H.

\(\varepsilon_2 = M\dfrac{dI_1}{dt} = 1.26 \times 10^{-4} \times 50 \approx 6.3 \times 10^{-3}\) V.

1. \(\varepsilon_2 = M\dfrac{\Delta I_1}{\Delta t} = 0.5 \times \dfrac{2}{0.01} = 100\) V.

2. \(\Delta\Phi_2 = M\,\Delta I_1 = 0.5 \times 2 = 1\) Wb.

Yes. Mutual inductance is a property of the pair of coils and their geometry, and \(M_{12} = M_{21}\) always. It is easier to compute using the solenoid's uniform field. A current in the small coil would link exactly the same flux per ampere with the solenoid.

\(M = k\sqrt{L_1L_2} = 0.5 \times \sqrt{0.36} = 0.3\) H. The largest possible value is for \(k = 1\), perfect coupling: \(\sqrt{L_1L_2} = 0.6\) H.

While the current grows, the source must push charge against the back emf \(L\dfrac{dI}{dt}\). Power supplied \(= \varepsilon I = LI\dfrac{dI}{dt}\), so the work done in a time \(dt\) is \(dW = LI\,dI\). Integrating from \(0\) to \(I\): \(W = \int_0^I LI\,dI = \dfrac{1}{2}LI^2\). This work is stored in the magnetic field and returned when the current falls.

Round the loop, \(E - IR - L\dfrac{dI}{dt} = 0\), so \(E = IR + L\dfrac{dI}{dt}\). With \(I = 0\) at \(t = 0\) the solution is \(I = \dfrac{E}{R}\left(1 - e^{-t/\tau}\right)\), where \(\tau = \dfrac{L}{R}\) is the time constant. It is the time in which the current reaches \(1 - e^{-1} \approx 63\%\) of its final value.

1. \(\tau = \dfrac{L}{R} = \dfrac{2}{4} = 0.5\) s. Final current \(= \dfrac{E}{R} = 3\) A.

2. \(I = 0\). Then \(E = L\dfrac{dI}{dt}\), so \(\dfrac{dI}{dt} = \dfrac{12}{2} = 6\) A/s.

3. \(t = \tau\): \(I = 3(1 - e^{-1}) = 3 \times 0.63 \approx 1.9\) A.

The inductor opposes any change of current. At the instant of closing, its back emf \(L\dfrac{dI}{dt}\) equals the full battery voltage, leaving nothing to drive current through \(R\). The current starts from zero and grows as the back emf falls.

Once the current is steady, \(\dfrac{dI}{dt} = 0\) and the inductor's emf is zero: it behaves as a plain piece of wire (or its own small resistance). The whole battery voltage then appears across \(R\).

\(\dfrac{1}{2} = 1 - e^{-t/\tau}\), so \(e^{-t/\tau} = \dfrac{1}{2}\) and \(t = \tau\ln 2 = 0.5 \times 0.69 \approx 0.35\) s.

1. \(1 - e^{-2} = 0.86\) and \(1 - e^{-3} = 0.95\) of the final current.

2. The slope is \(\dfrac{dI}{dt} = \dfrac{E - IR}{L}\). As \(I\) grows, the voltage left for the inductor, \(E - IR\), shrinks, so the current rises more and more slowly.

\(V_L = L\dfrac{dI}{dt} = Ee^{-t/\tau} = 12 \times 0.37 \approx 4.4\) V. \(V_R = IR = 1.9 \times 4 \approx 7.6\) V. Sum \(\approx 12\) V, as the loop equation requires.

Final: \(U = \dfrac{1}{2}LI^2 = \dfrac{1}{2} \times 2 \times 9 = 9\) J. At \(t = 0.5\) s the power into the inductor is \(V_LI = 4.4 \times 1.9 \approx 8.4\) W.

With no source, \(IR + L\dfrac{dI}{dt} = 0\), so \(\dfrac{dI}{I} = -\dfrac{R}{L}dt\). Integrating from \(I_0\) at \(t = 0\): \(\ln\dfrac{I}{I_0} = -\dfrac{t}{\tau}\), so \(I = I_0e^{-t/\tau}\) with \(\tau = \dfrac{L}{R}\). The inductor's emf now drives the current, keeping it going while it decays.

1. \(\tau = 0.5\) s, so \(t = \tau\): \(I = 3e^{-1} \approx 1.1\) A.

2. \(t = 2\tau\): \(I = 3e^{-2} \approx 0.42\) A.

It is dissipated as heat in the resistor as the current dies away. Total heat \(= \int I^2R\,dt = \dfrac{1}{2}LI_0^2 = \dfrac{1}{2} \times 2 \times 9 = 9\) J.

\(e^{-t/\tau} = 0.1\), so \(t = \tau\ln 10 = 0.5 \times 2.3 = 1.15\) s.

\(\tau = \dfrac{L}{R} = \dfrac{2}{8} = 0.25\) s: the current decays twice as fast. The same energy \(9\) J is dissipated, but in a shorter time.

\(I = I_0e^{-t/\tau}\), so \(\dfrac{dI}{dt} = -\dfrac{I_0}{\tau}e^{-t/\tau}\), which at \(t = 0\) is \(-\dfrac{I_0}{\tau}\). The tangent line is \(I = I_0 - \dfrac{I_0}{\tau}t\), which reaches zero at \(t = \tau\). This gives a quick way to read \(\tau\) off a graph.

1. Just after closing, the inductor carries no current and acts as an open branch. The battery sees \(R_1 + R_2 = 8\) Ω: \(I = \dfrac{12}{8} = 1.5\) A.

2. Finally the inductor is a short across \(R_2\), so the battery sees only \(R_1\): \(I = \dfrac{12}{4} = 3\) A.

(i) Inductor \(0\), \(R_2\) carries the full \(1.5\) A. (ii) The inductor short-circuits \(R_2\): inductor \(3\) A, \(R_2\) zero.

The inductor's current of \(3\) A cannot change instantly. With the battery branch open, it must flow round the loop of \(L\) and \(R_2\). So \(R_2\) carries \(3\) A, now in the opposite direction to before. Voltage across \(R_2\): \(3 \times 4 = 12\) V.

The decay loop contains only \(L\) and \(R_2\): \(\tau = \dfrac{L}{R_2} = \dfrac{2}{4} = 0.5\) s.

A sudden jump in current would mean an infinite \(\dfrac{dI}{dt}\) and an infinite back emf \(L\dfrac{dI}{dt}\), which no finite source can supply. The current therefore changes smoothly. The voltage across the inductor is \(L\dfrac{dI}{dt}\), which can change abruptly when the circuit is switched, as when the switch in Q17 is opened.

\(\tau = \dfrac{L}{R} = \dfrac{0.05}{10} = 5 \times 10^{-3}\) s \(= 5\) ms.

Final \(I = \dfrac{6}{2} = 3\) A. \(\tau = \dfrac{0.1}{2} = 0.05\) s. At \(t = \tau\): \(I = 3(1 - 0.37) \approx 1.9\) A.

\(L\) has units of V s/A (from \(\varepsilon = L\dfrac{dI}{dt}\)) and \(R\) has units of V/A. Their ratio is \(\dfrac{\text{V s/A}}{\text{V/A}} = \) s, a time.

At \(t = 0\) the whole \(10\) V is across the inductor: \(\dfrac{dI}{dt} = \dfrac{E}{L} = \dfrac{10}{0.5} = 20\) A/s.

\(IR = E\left(1 - e^{-Rt/L}\right)\). \(\dfrac{dI}{dt} = \dfrac{E}{R}\cdot\dfrac{R}{L}e^{-Rt/L} = \dfrac{E}{L}e^{-Rt/L}\), so \(L\dfrac{dI}{dt} = Ee^{-Rt/L}\). Adding: \(IR + L\dfrac{dI}{dt} = E - Ee^{-Rt/L} + Ee^{-Rt/L} = E\). It also gives \(I = 0\) at \(t = 0\), as required.