The Workbook

\(\Phi = \vec B \cdot \vec A = BA\cos\theta\), the number of field lines through the surface; unit weber (T m\(^2\)). \(\Phi = 0.5 \times 0.02 \times \cos 60^\circ = 5 \times 10^{-3}\) Wb.

1. Normal along \(\vec B\): \(\Phi = BA = 0.5 \times 0.02 = 10^{-2}\) Wb, the maximum.

2. Plane of the loop along \(\vec B\): no lines pass through, \(\Phi = 0\).

Flux linkage is the flux through the coil multiplied by the number of turns, \(N\Phi\), since each turn is threaded by the same flux. \(100 \times 5 \times 10^{-3} = 0.5\) Wb.

\(B = \mu_0nI = 4\pi \times 10^{-7} \times 1000 \times 2 = 2.5 \times 10^{-3}\) T. \(\Phi = BA = 2.5 \times 10^{-3} \times 10^{-3} = 2.5 \times 10^{-6}\) Wb.

Zero, always. Every field line that enters a closed surface also leaves it, because magnetic field lines form closed loops. There are no magnetic monopoles from which lines could start or end.

The emf induced in a circuit equals the rate of change of the flux linked with it: \(\varepsilon = -N\dfrac{d\Phi}{dt}\). The minus sign says that the emf drives a current whose own flux opposes the change, which is Lenz's law.

\(\varepsilon = N\dfrac{\Delta\Phi}{\Delta t} = 200 \times \dfrac{0.2 \times 0.01}{0.1} = 4\) V.

1. \(\dfrac{dB}{dt} = \dfrac{0.4}{2} = 0.2\) T/s. \(\varepsilon = NA\dfrac{dB}{dt} = 50 \times 0.1 \times 0.2 = 1\) V.

2. \(B\) is constant, so the flux does not change: \(\varepsilon = 0\).

3. \(\dfrac{dB}{dt} = \dfrac{0.4}{1} = 0.4\) T/s. \(\varepsilon = 50 \times 0.1 \times 0.4 = 2\) V, in the opposite sense to stage I.

The flux goes from \(+BA\) to \(-BA\), a change of \(2BA = 2 \times 0.1 \times 0.05 = 0.01\) Wb. \(\varepsilon = N\dfrac{\Delta\Phi}{\Delta t} = 100 \times \dfrac{0.01}{0.1} = 10\) V.

\(q = \dfrac{N\Delta\Phi}{R} = \dfrac{200 \times 0.002}{8} = 0.05\) C. It does not depend on the time: a slower collapse gives a smaller current for a longer time, and the same total charge.

\(\varepsilon = A\dfrac{dB}{dt} = 0.5 \times 0.4 = 0.2\) V. \(I = \dfrac{0.2}{2} = 0.1\) A. \(P = \varepsilon I = 0.02\) W, supplied by whatever agent is increasing the field.

\(\Delta\Phi = B\,\Delta A = 0.5 \times 0.1 = 0.05\) Wb. \(\varepsilon = \dfrac{0.05}{0.5} = 0.1\) V.

Lenz's law: the induced current flows in the direction that opposes the change of flux producing it. If it helped the change instead, the flux and the current would grow without limit, giving energy from nothing. Opposing the change means an external agent must do work, and that work is what appears as electrical energy.

1. The near face becomes a N pole, so that it repels the approaching N pole and opposes the increase of flux.

2. A face that is a N pole has anticlockwise current when viewed from outside, so seen from the magnet the current is anticlockwise.

3. Repulsion: the magnet must be pushed in against this force, and the work done becomes the electrical energy.

The flux is now decreasing, so the induced current reverses and the near face becomes a S pole, which attracts the departing N pole. The force on the magnet is now an attraction, again opposing its motion.

1. Less and less of the loop is in the field, so the flux (into the page) is decreasing.

2. The current must make its own flux into the page to replace what is lost: clockwise.

3. The force on the side still inside the field is to the left, opposing the pull. Work done against it becomes heat in the loop.

The flux into the page is growing, so the induced current makes a field out of the page inside the loop. That needs an anticlockwise current.

As the magnet falls, the flux through each ring of the tube changes, inducing circulating (eddy) currents in the copper. By Lenz's law these currents oppose the motion: the ring ahead repels the magnet and the ring behind attracts it. The magnet quickly reaches a small terminal speed.

Switching on increases the flux through the second coil, so its induced current flows opposite to the current in the first, to oppose the growth. Switching off decreases the flux. The induced current now flows in the same direction as the first coil's current was, trying to keep the flux.

1. The angle between the normal and \(\vec B\) is \(\omega t\): \(\Phi = NBA\cos\omega t\) per the whole coil, that is \(N\Phi_1\) with \(\Phi_1 = BA\cos\omega t\).

2. \(\varepsilon = -\dfrac{d}{dt}(NBA\cos\omega t) = NBA\omega\sin\omega t\). Peak \(= NBA\omega = 100 \times 0.1 \times 0.05 \times 100 = 50\) V.

\(\omega = 2\pi \times 50 = 100\pi\) rad/s. Peak \(= NBA\omega = 100 \times 0.1 \times 0.05 \times 100\pi = 50\pi \approx 157\) V.

\(\Delta\Phi = 2BA = 2 \times 5 \times 10^{-5} \times 0.1 = 10^{-5}\) Wb. \(\varepsilon = N\dfrac{\Delta\Phi}{\Delta t} = 500 \times \dfrac{10^{-5}}{0.2} = 2.5 \times 10^{-2}\) V.

No. The field is uniform and the coil's orientation is fixed, so the flux through it stays the same as it moves. Faraday's law needs a changing flux; motion alone is not enough.

\(I = \dfrac{\varepsilon}{R} = \dfrac{N}{R}\dfrac{d\Phi}{dt}\). The charge is \(q = \int I\,dt = \dfrac{N}{R}\int\dfrac{d\Phi}{dt}dt = \dfrac{N}{R}\int d\Phi = \dfrac{N\Delta\Phi}{R}\). The time cancels: only the total change of flux matters.

Each free electron moves with the rod at \(v\) and feels a magnetic force \(evB\) along the rod, pushing electrons to one end. Charge builds up until the electric field \(E\) inside the rod balances it: \(eE = evB\), so \(E = vB\). The potential difference across the length is \(\varepsilon = El = Blv\).

\(\varepsilon = Blv = 0.2 \times 0.5 \times 4 = 0.4\) V.

1. For a positive charge \(\vec v \times \vec B = \hat i \times (-\hat k) = \hat j\), towards \(P\). Electrons feel the opposite force and go to \(Q\), leaving \(P\) positive: \(P\) is at the higher potential.

2. Inside the rod, which acts as the source, current flows from \(Q\) to \(P\). In the external wire it flows from \(P\) to \(Q\).

Only the component of \(v\) perpendicular to \(B\) counts: \(\varepsilon = Blv\sin 30^\circ = 0.4 \times 0.5 = 0.2\) V.

\(\varepsilon = B_Vlv = 5 \times 10^{-5} \times 40 \times 250 = 0.5\) V.

The rod is an open circuit. The magnetic force moves electrons to one end only until the electric field of the separated charges balances it. After that there is no net force and no further flow. The emf is present as a potential difference, but a current needs a closed path.

An element \(dr\) at distance \(r\) from the pivot moves at \(v = \omega r\) and has emf \(d\varepsilon = B\omega r\,dr\). Adding along the rod: \(\varepsilon = \int_0^L B\omega r\,dr = \dfrac{1}{2}B\omega L^2\). Equivalently, the rod sweeps area at the rate \(\dfrac{1}{2}\omega L^2\).

\(\varepsilon = \dfrac{1}{2}B\omega L^2 = \dfrac{1}{2} \times 0.5 \times 10 \times 1 = 2.5\) V.

1. Every radius is a rotating rod: \(\varepsilon = \dfrac{1}{2}B\omega R^2 = \dfrac{1}{2} \times 0.4 \times 50 \times 0.04 = 0.4\) V.

2. No: all rim points are at the same potential, since every radius gives the same emf.

Each half is a rod of length \(\dfrac{L}{2}\) rotating about its end: emf \(\dfrac{1}{2}B\omega\left(\dfrac{L}{2}\right)^2 = \dfrac{1}{8}B\omega L^2\) between the centre and either end. Both ends are at the same potential relative to the centre, so the emf between the two ends is zero.

A positive charge at radius \(r\) moves tangentially at \(\omega r\). With \(\vec v\) tangential (anticlockwise) and \(\vec B\) out of the page, \(\vec v \times \vec B\) points radially outward. Positive charge is pushed to the rim, so the rim is at the higher potential.

1. \(\varepsilon = Blv = 0.5 \times 0.4 \times 5 = 1\) V. \(I = \dfrac{1}{2} = 0.5\) A.

2. The current in the field feels \(F = BIl = 0.5 \times 0.5 \times 0.4 = 0.1\) N, opposing the motion; the pull must equal it.

3. Mechanical: \(Fv = 0.1 \times 5 = 0.5\) W. Heat: \(I^2R = 0.25 \times 2 = 0.5\) W. They match.

The pull balances the magnetic force \(BIl\), so the power supplied is \(Fv = BIlv = I(Blv) = I\varepsilon\). With \(\varepsilon = IR\) this is \(I^2R\), the rate of heating in the resistor. All the mechanical work becomes heat.

Current \(I = \dfrac{Blv}{R}\) gives a retarding force \(BIl = \dfrac{B^2l^2}{R}v\): \(m\dfrac{dv}{dt} = -\dfrac{B^2l^2}{R}v\). The speed decays exponentially, \(v = v_0e^{-t/\tau}\) with \(\tau = \dfrac{mR}{B^2l^2} = \dfrac{0.2 \times 2}{0.25 \times 0.16} = 10\) s. Distance \(= v_0\tau = 5 \times 10 = 50\) m.

1. As the rod speeds up, the induced current and the magnetic drag \(\dfrac{B^2l^2}{R}v\) grow. When the drag equals \(F\) the net force is zero and the speed stays constant.

2. \(\dfrac{B^2l^2}{R}v_t = F\): \(v_t = \dfrac{FR}{B^2l^2} = \dfrac{0.4 \times 2}{0.25 \times 0.16} = 20\) m/s. Current \(= \dfrac{Blv_t}{R} = \dfrac{0.5 \times 0.4 \times 20}{2} = 2\) A; check \(BIl = 0.4\) N \(= F\).

At rest there is no back emf: \(I = \dfrac{2}{2} = 1\) A and \(F = BIl = 0.5 \times 1 \times 0.4 = 0.2\) N, which starts the rod moving. As it moves the motional emf opposes the battery; the current and force vanish when \(Blv = 2\) V, at \(v = \dfrac{2}{0.5 \times 0.4} = 10\) m/s. The rod then coasts at constant speed: a simple linear motor.

1. The capacitor charges to the emf \(\varepsilon = Blv = 1\) V: \(Q = C\varepsilon = 10 \times 10^{-6} \times 1 = 10\) μC.

2. Once charged, no current flows, so there is no magnetic force on the rod. With smooth rails no force is needed to keep it moving.

The rod's velocity along the incline has a horizontal component \(v\cos 37^\circ\) perpendicular to the vertical field; the vertical component is along \(\vec B\) and gives nothing. \(\varepsilon = Blv\cos 37^\circ = 0.5 \times 0.5 \times 2 \times 0.8 = 0.4\) V.

With no resistance, no energy is lost as heat. The moving rod drives a growing current through the inductor, storing energy in its magnetic field and slowing the rod. When the rod stops, the inductor's current is largest; it keeps flowing and pushes the rod back the other way. Energy passes back and forth between the rod's motion and the inductor, so the rod oscillates.

\(\varepsilon = Blv = 5 \times 10^{-5} \times 1.5 \times 20 = 1.5 \times 10^{-3}\) V.

No. The two sides perpendicular to the velocity each have a motional emf \(Blv\). The two are equal and drive current in opposite senses round the loop, so they cancel. Equivalently, the flux through the loop does not change.

Heat \(= I^2Rt = 0.25 \times 2 \times 10 = 5\) J. Work \(= Fvt = 0.1 \times 5 \times 10 = 5\) J. Equal, as Q13 requires.

The rod, its velocity and \(B_H\) are mutually perpendicular: \(\varepsilon = B_Hlv = 4 \times 10^{-5} \times 1 \times 10 = 4 \times 10^{-4}\) V.

\(m\dfrac{dv}{dt} = -\dfrac{B^2l^2}{R}v\). Write \(\dfrac{dv}{dt} = v\dfrac{dv}{dx}\): \(mv\dfrac{dv}{dx} = -\dfrac{B^2l^2}{R}v\), so \(m\,dv = -\dfrac{B^2l^2}{R}dx\). Integrating from \(v_0\) to \(0\): \(-mv_0 = -\dfrac{B^2l^2}{R}x\), so \(x = \dfrac{mRv_0}{B^2l^2}\), which is \(v_0\tau\).