IB Physics D.4: Induction worked solutions

The emf depends on how fast the flux linkage N\Phi changes, not on its size: a coil can hold a large steady flux linkage and have no emf at all. B leaves out the time; D leaves out the area, the angle and the number of turns.

\theta in \Phi = BA\cos\theta is measured from the normal, so \theta = 90^\circ - 30^\circ = 60^\circ. A = \pi(0.040)^2 = 5.03\times10^{-3}\,\mathrm{m^2} and \Phi = 0.25\times5.03\times10^{-3}\times\cos60^\circ = 6.3\times10^{-4}\,\mathrm{Wb}. B uses the angle to the plane; C ignores the angle; D takes the radius as 0.40\,\mathrm{m}.

\varepsilon = NA\,\Delta B/\Delta t = 50\times(0.030)^2\times0.30/0.15 = 0.090\,\mathrm{V}. A is the emf of one turn; B forgets to divide by the time; D uses the final field 0.40\,\mathrm{T} instead of the change.

Keep one normal for the whole turn. The flux goes from +BA = +1.0\times10^{-3}\,\mathrm{Wb} to -1.0\times10^{-3}\,\mathrm{Wb}, a change of 2.0\times10^{-3}\,\mathrm{Wb}, so \varepsilon = 2.0\times10^{-3}/0.10 = 0.020\,\mathrm{V}. A compares sizes only; C misses the factor of two.

An emf needs a change of flux linkage. In A the field, the area inside the field and the angle all stay the same, so the flux is constant. B changes the angle, C changes B, D changes the area inside the field.

The magnet's field at the ring points down and the downward flux falls as the magnet rises. The induced current replaces the lost downward flux, so it is clockwise seen from above. The current opposes the relative motion: the ring pulls the receding magnet back, a downward force.

The induced current always opposes the change that drives it, so the agent causing the change must do work. That work becomes the electrical energy of the induced current. If the current helped the change, energy would appear from nothing.

With the plane parallel to the field no field lines pass through the coil, so \Phi = 0, but the long sides are moving straight across the field and the flux is changing fastest. A pairs the largest flux with the largest emf, the classic slip: the emf follows the gradient of the flux, not its value.

Half the rotation rate halves the rate at which the flux changes, so the peak halves to 6.0\,\mathrm{V}; each turn takes twice as long, so the period doubles to 40\,\mathrm{ms}. A changes only the period; B halves both.

\varepsilon = BvL = 0.080\times4.0\times0.25 = 0.080\,\mathrm{V}. B leaves out the speed, D leaves out the length and A divides by the speed.

The magnetic force on a free electron, moving right with the rod in a field into the page, points down the rod, so electrons gather at the bottom and the top is left positive. They stop gathering when the electric force from the separated charge cancels the magnetic force. C uses the force on a positive charge for the electron.

The emf is proportional to N and to the rate of change of flux, which is proportional to the speed: 6.0\times3\times\tfrac12 = 9.0\,\mathrm{mV}. D multiplies by 2 for the speed instead of dividing; B and A turn the ratio of turns upside down.

(a) The steepest part of the flux graph is the fall: \Delta\Phi/\Delta t = 3.0\times10^{-3}/0.010 = 0.30\,\mathrm{Wb\,s^{-1}} (M1); \varepsilon = 200\times0.30 = 60\,\mathrm{V} (A1). The rise gives only 200\times3.0\times10^{-3}/0.020 = 30\,\mathrm{V}. (b) A constant -30\,\mathrm{V} from 0 to 20\,\mathrm{ms} and zero from 20 to 50\,\mathrm{ms}, where the flux is large but steady (A1); a constant +60\,\mathrm{V} from 50 to 60\,\mathrm{ms}, half as long as the first step (A1). (c) I = 60/12 = 5.0\,\mathrm{A} (A1).

(a) The field through the coil grows as the magnet comes closer, so the flux linkage changes and, by Faraday's law, an emf is induced (M1). By Lenz's law the current opposes the approach: the top of the coil becomes a north pole facing the falling north pole (A1). (b) Two pulses of opposite sign with zero between them (A1). The magnet has gained speed by the time it leaves, so the flux changes faster: the second peak is larger (A1) and the pulse lasts a shorter time (A1). The two pulses enclose equal areas, since the same flux change happens each time. (c) At the centre the flux linkage passes through its maximum, where its gradient is zero, so \varepsilon = 0 (A1).

(a) Each long side moves on a circle of radius 0.025\,\mathrm{m}: v = \omega r = 2\pi\times25\times0.025 = 3.93\,\mathrm{m\,s^{-1}} (A1). (b) One side gives at most BvL = 0.12\times3.93\times0.080 = 0.0377\,\mathrm{V} per turn; both long sides add and the short sides add nothing (M1); 150\times2\times0.0377 = 11.3\,\mathrm{V} (A1). (c) The emf is zero with the plane perpendicular to the field (A1): flux linkage is then at a maximum or minimum, so its gradient is zero (A1). (d) Doubling the rotation rate doubles the speed of each side, so the peak doubles and the period halves. At 25 rev/s a sine curve starting at zero, peak 11\,\mathrm{V}, one cycle in 40\,\mathrm{ms} (A1); at 50 rev/s peak 23\,\mathrm{V} and two cycles in 40\,\mathrm{ms} (A1).

(a) I_0 = 24/12 = 2.0\,\mathrm{A} (A1); peak power = V_0I_0 = 48\,\mathrm{W} (A1). (b) P = V^2/R = 48\sin^2(100\pi t)\,\mathrm{W}. This curve swings between 0 and 48\,\mathrm{W} and is symmetric about 24\,\mathrm{W}, repeating every 10\,\mathrm{ms}, so the mean is half the peak (M1): 24\,\mathrm{W} (A1). (c) V^2/12 = 24 (M1), so V = \sqrt{288} = 17\,\mathrm{V} (A1), which is 24/\sqrt2.

(a) Only the leading side cuts the field: \varepsilon = BvL = 0.60\times0.40\times0.10 = 0.024\,\mathrm{V} (M1); I = 0.024/0.50 = 0.048\,\mathrm{A} (A1). (b) F = BIL = 0.60\times0.048\times0.10 = 2.9\times10^{-3}\,\mathrm{N}, opposing the motion (A1). (c) Inside the region the flux through the loop stays at BA; the equal emfs of the two sides cancel round the loop (A1). (d) Current flows only while entering and leaving, 0.10/0.40 = 0.25\,\mathrm{s} each, 0.50\,\mathrm{s} in all (M1); E = I^2Rt = 0.048^2\times0.50\times0.50 = 5.8\times10^{-4}\,\mathrm{J} (A1). On the way out the current reverses, but the force still opposes the motion.

(a) I = 3.0/1.5 = 2.0\,\mathrm{A} (M1); F = BIL = 0.50\times2.0\times0.20 = 0.20\,\mathrm{N} (A1). (b) As the rod moves it cuts field lines and an emf BvL is induced; by Lenz's law it opposes the cell's emf (M1). The net emf 3.0 - BvL, the current and the force all fall as v rises; with smooth rails the speed becomes constant when the net emf, and so the force, reaches zero (A1). (c) BvL = 3.0, v = 3.0/(0.50\times0.20) = 30\,\mathrm{m\,s^{-1}} (A1).

(a) v_{\max} = \omega x_0 = (2\pi/0.50)\times0.040 = 0.503\,\mathrm{m\,s^{-1}} (M1); \varepsilon_{\max} = BvL = 0.25\times0.503\times0.30 = 0.038\,\mathrm{V} (A1). (b) The emf follows the velocity, not the displacement: zero at the extremes, where the rod is momentarily at rest (A1), and largest at the equilibrium position, where it moves fastest (A1). (c) The largest speed halves, so the peak emf halves to 0.019\,\mathrm{V} (A1); the period of SHM does not depend on amplitude, so the emf still repeats every 0.50\,\mathrm{s} (A1).

(a) As the current grows, the flux linkage of the coil, produced by that current, grows too (A1); this change induces an emf in the coil itself which, by Lenz's law, opposes the growth, so the current rises gradually (A1). (b) Once the current is steady there is no change of flux and no induced emf, so the current is set by the battery's emf and the circuit's resistance alone (A1).

(a) Run Faraday's law backwards: \Delta\Phi = -\varepsilon\Delta t/N = -0.60\times0.10/40 = -1.5\times10^{-3}\,\mathrm{Wb} (M1), so \Phi = 2.0 - 1.5 = 0.50\,\mathrm{mWb} (A1). A positive emf means the flux fell. (b) \Delta\Phi = -(-0.30)\times0.40/40 = +3.0\times10^{-3}\,\mathrm{Wb}, so \Phi = 3.5\,\mathrm{mWb} per turn (M1); flux linkage N\Phi = 40\times3.5\times10^{-3} = 0.14\,\mathrm{Wb} (A1). (c) Energy = (\varepsilon^2/R)\Delta t: 0.072\times0.10 = 7.2\times10^{-3}\,\mathrm{J} and 0.018\times0.40 = 7.2\times10^{-3}\,\mathrm{J} (M1); the two add whatever the sign of the emf, giving 0.014\,\mathrm{J} (A1).

(a) 0.40\times0.030 = 0.012\,\mathrm{Wb} and 0.60\times0.020 = 0.012\,\mathrm{Wb} (A1). (b) Both changes are steady, so at 1.0\,\mathrm{s}: B = 0.50\,\mathrm{T}, A = 0.025\,\mathrm{m^2}, \Phi = 0.0125\,\mathrm{Wb} (A1). (c) |\varepsilon| = N\Delta\Phi/\Delta t = 50\times5.0\times10^{-4}/1.0 (M1) = 0.025\,\mathrm{V} (A1). (d) The flux is not constant in between: \Phi = (0.40+0.10t)(0.030-0.0050t) rises to its largest value at t = 1.0\,\mathrm{s} and then falls, so an emf is induced (M1). The induced current flows one way for the first second and the other way for the second, by Lenz's law; equal flux at the two ends makes only the average emf zero (A1).

(a) The emf BvL, and so the current, increases with the speed (A1). The current in the rod feels a force BIL that opposes its fall, by Lenz's law (A1). The rod accelerates until BIL = mg; the resultant force is then zero and the speed stays constant (A1). (b) I = BLv/R, so B^2L^2v/R = mg (M1) and v = 0.020\times9.8\times0.50/(0.40^2\times0.25^2) = 9.8\,\mathrm{m\,s^{-1}} (A1). (c) mgv = 0.196\times9.8 = 1.9\,\mathrm{W}; \varepsilon = BLv = 0.98\,\mathrm{V} and \varepsilon^2/R = 0.98^2/0.50 = 1.9\,\mathrm{W} (A1). Gravity's work all becomes thermal energy: conservation of energy, of which Lenz's law is a consequence (A1).

(a) B_\perp = 5.0\times10^{-5}\sin65^\circ = 4.53\times10^{-5}\,\mathrm{T} (A1). (b) Only the vertical component passes through the horizontal coil: \Phi = 4.53\times10^{-5}\times\pi\times0.15^2 = 3.20\times10^{-6}\,\mathrm{Wb} per turn (A1). Turning over takes it from +3.20\times10^{-6} to -3.20\times10^{-6}\,\mathrm{Wb}, a change of 6.41\times10^{-6}\,\mathrm{Wb} (M1); \varepsilon = 25\times6.41\times10^{-6}/0.20 = 8.0\times10^{-4}\,\mathrm{V} (A1). (c) The wings move horizontally, so only the vertical component is cut: \varepsilon = BvL = 4.53\times10^{-5}\times250\times36 (M1) = 0.41\,\mathrm{V} (A1).

(a) Leading side out at 0.12/0.10 = 1.2\,\mathrm{s}; trailing side in at 0.20/0.10 = 2.0\,\mathrm{s} (A1). (b) Only a short side inside the field cuts it: |\varepsilon| = BvL = 0.80\times0.10\times0.050 = 4.0\times10^{-3}\,\mathrm{V} (M1). From 0 to 1.2\,\mathrm{s} the flux increases, so \varepsilon = -4.0\,\mathrm{mV} (A1); zero from 1.2 to 2.0\,\mathrm{s}, +4.0\,\mathrm{mV} from 2.0 to 3.2\,\mathrm{s} while the flux decreases, zero afterwards (A1). (c) Neither short side is in the field; the field region lies wholly inside the loop, so the flux stays at 0.80\times0.12\times0.050 = 4.8\times10^{-3}\,\mathrm{Wb} and its gradient is zero (A1). (d) I = 4.0\times10^{-3}/0.020 = 0.20\,\mathrm{A} for 2.4\,\mathrm{s} in all (M1); E = I^2Rt = 0.20^2\times0.020\times2.4 = 1.9\times10^{-3}\,\mathrm{J} (A1).

(a) C.1: T = 2\pi\sqrt{m/k} = 2\pi\sqrt{0.15/37} (M1) = 0.40\,\mathrm{s} (A1). (b) f = 1/T = 2.5\,\mathrm{Hz} (A1): the flux linkage of the coil rises as the magnet comes down and falls as it goes up, once in each oscillation (A1). (c) With the circuit closed a current flows, and by Lenz's law its field opposes the magnet's motion in both directions (A1). The magnet does work against this force, and the energy leaves as thermal energy in the resistor; with an open circuit there is an emf but no current and no force (A1). (d) C.1/A.3: E = \tfrac12kx_0^2 falls from \tfrac12\times37\times0.030^2 = 0.0167\,\mathrm{J} to \tfrac12\times37\times0.015^2 = 0.0042\,\mathrm{J} (M1); (0.0167-0.0042)/20 = 6.2\times10^{-4}\,\mathrm{W} (A1).

(a) D.3: F = evB = 1.6\times10^{-19}\times6.0\times0.20 = 1.9\times10^{-19}\,\mathrm{N} (A1). (b) The magnetic force acts along the rod and pushes free electrons to one end, leaving the other end positive (A1). The separated charge produces an electric field in the rod whose force on an electron opposes the magnetic force; the build-up stops when eE = evB (A1). (c) E = vB = 6.0\times0.20 = 1.2\,\mathrm{V\,m^{-1}} (A1); the field is uniform along the rod (D.2), so \varepsilon = EL = 1.2\times0.50 = 0.60\,\mathrm{V}, the same as BvL (A1). (d) I = 0.60/3.0 = 0.20\,\mathrm{A} (M1); F = BIL = 0.20\times0.20\times0.50 = 0.020\,\mathrm{N} (A1); Fv = 0.020\times6.0 = 0.12\,\mathrm{W} and I^2R = 0.20^2\times3.0 = 0.12\,\mathrm{W} (A1).

(a) \omega = 2\pi\times10 = 62.8\,\mathrm{rad\,s^{-1}} (M1); \varepsilon_0 = NBA\omega = 200\times0.15\times0.010\times62.8 = 18.8\,\mathrm{V} (A1). (b) B.5: the coil's resistance acts as an internal resistance, I_0 = 18.8/(6.0+2.0) = 2.36\,\mathrm{A} (A1). (c) Peak lamp power I_0^2R = 2.36^2\times6.0 = 33\,\mathrm{W} (M1); the mean is half the peak, 17\,\mathrm{W} (A1). (d) With the lamp connected a current flows in the coil, and the forces on it oppose the rotation (Lenz's law) (A1). A.3: the hand must supply at least the total electrical power, including that in the coil (M1): \tfrac12\times2.36^2\times8.0 = 22\,\mathrm{W} (A1).

(a) A = \pi\times0.040^2 = 5.03\times10^{-3}\,\mathrm{m^2} (M1); \varepsilon = A\,\Delta B/\Delta t = 5.03\times10^{-3}\times0.50 = 2.5\times10^{-3}\,\mathrm{V} (A1). (b) B.5: P = \varepsilon^2/R = (2.51\times10^{-3})^2/(2.0\times10^{-3}) = 3.2\times10^{-3}\,\mathrm{W} (A1). (c) B.1: Q = Pt = 0.095\,\mathrm{J} (M1); \Delta T = Q/mc = 0.095/(0.0020\times385) = 0.12\,\mathrm{K} (A1). (d) The induced current makes a downward field inside the ring to oppose the growing upward flux: clockwise seen from above (A1). (e) D.3: each short piece of the ring feels a force BIL at right angles to both the current and the vertical field, so in the plane of the ring along a radius (A1); pieces on opposite sides feel equal and opposite forces, so the resultant is zero (A1).