Practice Exam 8

Induction & Inductance

Questions10

Time90 min

Points100

CalculatorAllowed

Use \(\mu_0 = 4\pi\times10^{-7}\ \text{T}\cdot\text{m/A}\). For every direction question, say from which side you are looking.

1Lenz's law directions10 pts

Give the direction of the induced current in each case. (a) A bar magnet's N pole is pulled away from a loop (describe the current as seen from the magnet). (b) A loop lies in the page to the right of a long straight wire, also in the page, carrying an increasing current upward. (c) The same loop moves parallel to the wire (up the page) at constant distance, with steady current. (d) A square loop in the page is pulled out to the right of a region of uniform field pointing out of the page.

Solution

(a) External field at the loop points away from the N pole (away from the magnet); flux decreasing → induced \(\mathbf{B}\) in the same direction, away from the magnet → thumb pointing away from the viewer → clockwise as seen from the magnet. The loop attracts the retreating magnet.
(b) To the right of an upward current, the wire's field is into the page (grip rule). Increasing → induced \(\mathbf{B}\) out of the page → counterclockwise as seen from the front.
(c) The loop stays at the same distance in a steady field: the flux is constant, so no current.
(d) Flux out of the page is decreasing as the loop leaves → induced \(\mathbf{B}\) out of the page (same direction) → counterclockwise as seen from the front. The force on the loop pulls it back into the field.

2Flux and Faraday's law10 pts

A circular coil of \(50\) turns and radius \(3.0\) cm sits in a uniform field \(B(t) = (0.050\ \text{T/s}^2)\,t^2\) directed along the coil's axis. (a) Find the emf at \(t = 2.0\) s. (b) Repeat if the field instead makes \(30^\circ\) with the plane of the coil.

Solution

\[ A = \pi(0.030)^2 = 2.83\times10^{-3}\ \text{m}^2,\qquad \frac{dB}{dt} = 2(0.050)t = 0.20\ \text{T/s at } t = 2.0\ \text{s} \] \[ \text{(a)}\quad |\mathcal{E}| = NA\frac{dB}{dt} = (50)(2.83\times10^{-3})(0.20) = 2.83\times10^{-2}\ \text{V} = 28\ \text{mV} \]

(b) \(30^\circ\) with the plane means \(\phi = 60^\circ\) with the normal, so multiply by \(\cos60^\circ = 0.50\):

\[ |\mathcal{E}| = (2.83\times10^{-2})(0.50) = 1.41\times10^{-2}\ \text{V} = 14\ \text{mV} \]

Using \(\cos30^\circ\) here (24.5 mV) is the classic error: \(\phi\) is always measured from the normal.

3Rod on rails: terminal velocity12 pts

A rod of mass \(m = 20\) g and length \(L = 0.20\) m slides without friction on horizontal rails connected by \(R = 0.10\ \Omega\), in a vertical field \(B = 0.50\) T. Starting from rest, it is pulled by a constant force \(F = 0.10\) N. (a) Write Newton's second law and find the terminal speed. (b) Solve for \(v(t)\) and give the time constant. (c) Check the power balance at terminal speed.

Solution

(a) The magnetic force opposes the motion with size \(B^2L^2v/R\):

\[ m\frac{dv}{dt} = F - \frac{B^2L^2}{R}v,\qquad \frac{B^2L^2}{R} = \frac{(0.50)^2(0.20)^2}{0.10} = 0.10\ \text{N}\cdot\text{s/m} \] \[ v_t = \frac{FR}{B^2L^2} = \frac{0.10}{0.10} = 1.0\ \text{m/s} \]

(b) Same form as an RC or RL equation:

\[ v(t) = v_t\left(1 - e^{-t/\tau}\right),\qquad \tau = \frac{mR}{B^2L^2} = \frac{0.020}{0.10} = 0.20\ \text{s} \]

(c) At \(v_t\): \(\mathcal{E} = BLv = 0.10\) V, \(I = 1.0\) A, magnetic force \(ILB = 0.10\) N \(= F\) ✓. Power in: \(Fv = 0.10\) W. Power dissipated: \(I^2R = 0.10\) W ✓. At terminal speed all the work done by \(F\) becomes heat.

4Designing a generator · EE application10 pts

You need a peak emf of \(170\) V at \(60\) Hz from a coil of area \(0.010\ \text{m}^2\) rotating in a \(0.50\) T field. (a) How many turns? (b) At the instant the emf is maximum, what is the angle between the coil's normal and \(\mathbf{B}\), and what is the flux? (c) What happens to the peak emf if the speed is doubled?

Solution

\[ \text{(a)}\quad N = \frac{\mathcal{E}_{\max}}{BA\omega} = \frac{170}{(0.50)(0.010)(2\pi\cdot60)} = \frac{170}{1.885} = 90.2\;\Rightarrow\; N = 91\ \text{turns} \]

(Round up to guarantee at least 170 V.)

(b) \(\mathcal{E} = NBA\omega\sin\omega t\) is maximum when \(\omega t = 90^\circ\): the normal is perpendicular to \(\mathbf{B}\), the coil plane is parallel to the field, and the flux is zero — but changing fastest.

(c) \(\mathcal{E}_{\max}\propto\omega\), so it doubles (340 V) — and the frequency doubles to 120 Hz too. That is why grid generators are speed-regulated.

5Induced electric field8 pts

A long solenoid of radius \(R = 5.0\) cm has its field increasing at \(0.20\) T/s. Find the magnitude of the induced electric field at (a) \(r = 2.0\) cm and (b) \(r = 10\) cm from the axis. (c) Why can this field not be described by a potential?

Solution

By symmetry \(\mathbf{E}\) circles the axis; use a circle of radius \(r\) in \(\oint\mathbf{E}\cdot d\boldsymbol{\ell} = |d\Phi_B/dt|\).

\[ \text{(a)}\ r < R:\quad E(2\pi r) = \pi r^2\frac{dB}{dt}\;\Rightarrow\; E = \frac{r}{2}\frac{dB}{dt} = \frac{0.020}{2}(0.20) = 2.0\times10^{-3}\ \text{V/m} \] \[ \text{(b)}\ r > R:\quad E(2\pi r) = \pi R^2\frac{dB}{dt}\;\Rightarrow\; E = \frac{R^2}{2r}\frac{dB}{dt} = \frac{(0.050)^2}{2(0.10)}(0.20) = 2.5\times10^{-3}\ \text{V/m} \]

Note (b): the field outside the solenoid is (ideally) zero, yet there is an induced \(\mathbf{E}\) there, because the circle encloses changing flux.

(c) Its line integral around a closed loop is not zero. A potential would require \(\oint\mathbf{E}\cdot d\boldsymbol{\ell} = 0\) for every loop. Induced electric fields are non-conservative.

6Inductance of a toroid (derivation)10 pts

A toroid of \(N\) turns has a rectangular cross-section: inner radius \(a\), outer radius \(b\), height \(h\). (a) Derive \(L = \dfrac{\mu_0N^2h}{2\pi}\ln\dfrac{b}{a}\). (b) Evaluate it for \(N = 500\), \(a = 2.0\) cm, \(b = 3.0\) cm, \(h = 1.0\) cm.

Solution

(a) From Ampère's law (Unit 7), inside the windings \(B = \mu_0NI/(2\pi r)\). It varies across the cross-section, so integrate the flux through one turn using strips of height \(h\) and width \(dr\):

\[ \Phi_B = \int_a^b\frac{\mu_0NI}{2\pi r}\,h\,dr = \frac{\mu_0NIh}{2\pi}\ln\frac{b}{a},\qquad L = \frac{N\Phi_B}{I} = \frac{\mu_0N^2h}{2\pi}\ln\frac{b}{a} \]

\[ \text{(b)}\quad L = (2\times10^{-7})(500)^2(0.010)\ln1.5 = (5.0\times10^{-4})(0.405) = 2.03\times10^{-4}\ \text{H} \]

\(L\approx0.20\) mH. As always, \(L\propto N^2\) and depends only on geometry.

7RL circuit12 pts

A \(24\) V source, \(8.0\ \Omega\) resistor and \(0.40\) H inductor are in series; the switch closes at \(t = 0\). (a) Find \(\tau\) and the final current. (b) Find \(dI/dt\) at \(t = 0\). (c) Find \(I\) at \(t = 0.10\) s. (d) Find the energy stored once the current is steady. (e) Later the source is replaced by a wire (the current now circulates through \(R\) and \(L\)). How long until the current falls to \(0.30\) A?

Solution

\[ \text{(a)}\ \tau = \frac{L}{R} = \frac{0.40}{8.0} = 0.050\ \text{s},\quad I_\infty = \frac{24}{8.0} = 3.0\ \text{A}\qquad \text{(b)}\ \frac{dI}{dt}\Big|_0 = \frac{V}{L} = \frac{24}{0.40} = 60\ \text{A/s} \] \[ \text{(c)}\ I(0.10) = 3.0\left(1 - e^{-2}\right) = 3.0(0.865) = 2.59\ \text{A}\qquad \text{(d)}\ U = \tfrac12(0.40)(3.0)^2 = 1.8\ \text{J} \] \[ \text{(e)}\ 0.30 = 3.0\,e^{-t/\tau}\;\Rightarrow\; t = \tau\ln10 = (0.050)(2.303) = 0.115\ \text{s} \]

(b): at \(t = 0\) the current is zero, so there is no drop across \(R\) and the full 24 V is across the inductor: \(L\,dI/dt = 24\) V. (e): the current cannot jump when the source is removed — it starts at 3.0 A and decays.

8Magnetic energy in an MRI · EE8 pts

A 3.0 T MRI magnet has a uniform-field region of about \(0.50\ \text{m}^3\). (a) Find the energy density and the energy in that region. (b) What electric field would store the same energy density? Comment.

Solution

\[ \text{(a)}\quad u_B = \frac{B^2}{2\mu_0} = \frac{(3.0)^2}{2(4\pi\times10^{-7})} = 3.58\times10^{6}\ \text{J/m}^3,\qquad U\approx(3.58\times10^{6})(0.50) = 1.8\times10^{6}\ \text{J} \] \[ \text{(b)}\quad \tfrac12\varepsilon_0E^2 = \frac{B^2}{2\mu_0}\;\Rightarrow\; E = \frac{B}{\sqrt{\mu_0\varepsilon_0}} = cB = (3.00\times10^8)(3.0) = 9.0\times10^{8}\ \text{V/m} \]

About 1.8 MJ (roughly half a kilowatt-hour) in the bore alone; the field outside adds more. If the superconductor “quenches” (turns resistive) this energy is released as heat in seconds, boiling off liquid helium — MRI magnets have quench vents for that reason. The equivalent electric field, \(9\times10^8\) V/m, is hundreds of times the breakdown field of air: magnetic fields are the practical way to store field energy.

9Wireless charging & transformers · EE10 pts

(a) A charging pad and phone coil have mutual inductance \(M = 5.0\ \mu\text{H}\). The pad current is \(i_1 = (2.0\ \text{A})\sin(2\pi ft)\) with \(f = 150\) kHz. Find the peak emf induced in the phone coil. (b) An ideal transformer steps 240 V down to 12 V and has 1200 primary turns. Find the secondary turns, and the primary current when the secondary drives a \(4.0\ \Omega\) load. (c) Why would neither device work from a battery directly?

Solution

\[ \text{(a)}\quad \mathcal{E}_2 = -M\frac{di_1}{dt} = -M(2\pi f)(2.0)\cos(2\pi ft),\qquad \mathcal{E}_{\max} = (5.0\times10^{-6})(2\pi)(1.5\times10^{5})(2.0) = 9.4\ \text{V} \] \[ \text{(b)}\quad N_s = 1200\times\frac{12}{240} = 60,\qquad I_s = \frac{12}{4.0} = 3.0\ \text{A},\qquad I_p = I_s\frac{N_s}{N_p} = 3.0\times\frac{60}{1200} = 0.15\ \text{A} \]

Check (b): \(P = 12\times3.0 = 36\) W \(= 240\times0.15\) ✓.

(c) A battery gives a constant current, so the flux is constant and \(d\Phi_B/dt = 0\): nothing is induced. Both need AC — which is why chargers contain an inverter that chops DC into high-frequency AC. The high frequency in (a) is deliberate: the emf is proportional to \(f\), so a small \(M\) still transfers useful power.

10LC oscillations10 pts

A \(5.0\) nF capacitor charged to \(20\) V is connected at \(t = 0\) across a \(2.0\) mH inductor. (a) Find the angular frequency and frequency. (b) Find the maximum current. (c) At what time is the energy first entirely magnetic? (d) Verify energy conservation.

Solution

\[ \text{(a)}\quad \omega = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{(2.0\times10^{-3})(5.0\times10^{-9})}} = \frac{1}{\sqrt{1.0\times10^{-11}}} = 3.16\times10^{5}\ \text{rad/s},\qquad f = \frac{\omega}{2\pi} = 50.3\ \text{kHz} \] \[ \text{(b)}\quad Q = CV = 1.0\times10^{-7}\ \text{C},\qquad I_{\max} = \omega Q = (3.16\times10^{5})(1.0\times10^{-7}) = 3.16\times10^{-2}\ \text{A} = 31.6\ \text{mA} \] \[ \text{(c)}\quad t = \frac{T}{4} = \frac{1}{4f} = \frac{\pi}{2\omega} = 4.97\ \mu\text{s} \] \[ \text{(d)}\quad \tfrac12CV^2 = \tfrac12(5.0\times10^{-9})(20)^2 = 1.0\times10^{-6}\ \text{J},\qquad \tfrac12LI_{\max}^2 = \tfrac12(2.0\times10^{-3})(3.16\times10^{-2})^2 = 1.0\times10^{-6}\ \text{J}\ \checkmark \]

With \(q(t) = Q\cos\omega t\), the capacitor is empty (all energy in \(L\)) when \(\omega t = \pi/2\). A common slip is to use \(T/2\) — at half a period the capacitor is full again, with reversed polarity.

Scoring

Score	Where you are
90–100	Solid. On to AC circuits and electromagnetic waves.
75–89	Probably direction slips (Q1) or the angle in the flux (Q2b). Re-run the four-step recipe in Section 3 and the warning in Section 1.
60–74	Inductor behaviour is shaky (Q6, Q7, Q10). Re-read Section 6, Section 7 and Section 8.
< 60	Go back to Faraday's law and motional emf, then rework the worked examples.

Where the points usually go

Saying the induced field always opposes the external field, rather than its change (Q1a, Q1d); measuring \(\phi\) from the plane (Q2b); confusing maximum flux with maximum emf (Q4b); assuming \(E = 0\) outside the solenoid because \(B = 0\) there (Q5b); forgetting the inductor current is continuous when a switch moves (Q7e); and \(T/2\) instead of \(T/4\) in the LC timing (Q10c).