The Timpson Lyceum

1. (1 pt •) Two coils are wound on a common core. When the current in coil 1 changes at the rate $di_1/dt = 850\,\text{A/s}$ , an emf of magnitude $2.72\,\text{V}$ is induced in coil 2.

(a) Find the mutual inductance $M$ of the pair.

(b) When a steady current $i_1 = 6.0\,\text{A}$ flows in coil 1, what is the total flux linkage $N_2\Phi_{B2}$ through coil 2?

Solution

1.

(a) $|\mathcal{E}_2| = M\,|di_1/dt|$ , so $M = \dfrac{2.72}{850} = 3.2\times10^{-3}\,\text{H} = 3.2\,\text{mH}$ .

(b) The defining relation is $N_2\Phi_{B2} = M i_1 = (3.2\times10^{-3})(6.0) = 1.9\times10^{-2}\,\text{Wb}$ (weber-turns).

2. (2 pts ••) A long air-core solenoid has $N = 400$ turns wound over a length $l = 0.25\,\text{m}$ , with a circular cross-section of radius $r = 2.0\,\text{cm}$ .

(a) Compute its self-inductance $L$ .

(b) The current is $i = 5.0\,\text{A}$ and steady. Find the magnetic field $B$ inside the solenoid.

(c) Using the magnetic energy density $u = B^2/2\mu_0$ and the solenoid's interior volume, find the total stored energy. Check that it equals $\tfrac12 L i^2$ .

Solution

2.

(a) $L = \dfrac{\mu_0 N^2 A}{l}$ with $A = \pi r^2 = \pi(0.020)^2 = 1.26\times10^{-3}\,\text{m}^2$ .

L = \frac{(4\pi\times10^{-7})(400)^2(1.26\times10^{-3})}{0.25} \approx 1.0\times10^{-3}\,\text{H} = 1.0\,\text{mH}.

(b) Inside a long solenoid $B = \mu_0 n i$ with $n = N/l = 400/0.25 = 1600\,\text{m}^{-1}$ :

B = (4\pi\times10^{-7})(1600)(5.0) \approx 1.0\times10^{-2}\,\text{T}.

(c) Energy density $u = \dfrac{B^2}{2\mu_0} = \dfrac{(1.0\times10^{-2})^2}{2(4\pi\times10^{-7})} \approx 40\,\text{J/m}^3$ . Interior volume $V = A l = (1.26\times10^{-3})(0.25) = 3.14\times10^{-4}\,\text{m}^3$ , so

U = uV \approx (40)(3.14\times10^{-4}) \approx 1.3\times10^{-2}\,\text{J}.

Check: $\tfrac12 L i^2 = \tfrac12(1.01\times10^{-3})(5.0)^2 \approx 1.3\times10^{-2}\,\text{J}$ . ✓

3. (2 pts ••) A toroidal coil has $N = 500$ turns, a rectangular cross-section of height $h = 1.5\,\text{cm}$ , inner radius $a = 3.0\,\text{cm}$ , and outer radius $b = 5.0\,\text{cm}$ . Its self-inductance is

L = \frac{\mu_0 N^2 h}{2\pi}\ln\!\frac{b}{a}.

(a) Evaluate $L$ .

(b) An inductor is useful because it opposes changes in current. If the current in this toroid is ramped from $0$ to $3.0\,\text{A}$ in $4.0\,\text{ms}$ , what is the magnitude of the self-induced emf?

Solution

3.

(a) $L = \dfrac{\mu_0 N^2 h}{2\pi}\ln\dfrac{b}{a} = \dfrac{(4\pi\times10^{-7})(500)^2(0.015)}{2\pi}\ln\dfrac{0.050}{0.030}$ .

The factor $\dfrac{\mu_0}{2\pi} = 2.0\times10^{-7}$ , and $\ln(5/3) = 0.511$ , so

L = (2.0\times10^{-7})(500)^2(0.015)(0.511) \approx 3.8\times10^{-4}\,\text{H} = 0.38\,\text{mH}.

(b) $|\mathcal{E}| = L\left|\dfrac{di}{dt}\right| = (3.83\times10^{-4})\dfrac{3.0}{4.0\times10^{-3}} = (3.83\times10^{-4})(750) \approx 0.29\,\text{V}$ .

4. (2 pts ••) An inductor of $L = 0.15\,\text{H}$ carries a current that increases uniformly from $0$ to $4.0\,\text{A}$ in $20\,\text{ms}$ .

(a) Find the magnitude of the self-induced emf during the ramp, using $\mathcal{E} = -L\,di/dt$ .

(b) Find the energy stored in the inductor once the current reaches $4.0\,\text{A}$ .

Solution

4.

(a) $di/dt = \dfrac{4.0}{20\times10^{-3}} = 200\,\text{A/s}$ , so $|\mathcal{E}| = L\,|di/dt| = (0.15)(200) = 30\,\text{V}$ .

(b) $U = \tfrac12 L I^2 = \tfrac12(0.15)(4.0)^2 = 1.2\,\text{J}$ .

5. (2 pts ••) A series R–L circuit has $L = 0.30\,\text{H}$ , $R = 25\,\Omega$ , and a $12\,\text{V}$ battery. At $t = 0$ the switch is closed with zero initial current.

(a) Find the time constant $\tau$ and the final (steady-state) current.

(b) Find the current after one time constant has elapsed.

(c) How long after closing does the current reach $90\%$ of its final value?

Solution

5.

(a) $\tau = \dfrac{L}{R} = \dfrac{0.30}{25} = 0.012\,\text{s} = 12\,\text{ms}$ . Final current $I_\text{f} = \dfrac{\mathcal{E}}{R} = \dfrac{12}{25} = 0.48\,\text{A}$ .

(b) $i(\tau) = I_\text{f}(1 - e^{-1}) = 0.48(0.632) \approx 0.30\,\text{A}$ .

(c) Set $i = 0.9\,I_\text{f}$ : $1 - e^{-t/\tau} = 0.9 \Rightarrow e^{-t/\tau} = 0.1 \Rightarrow t = -\tau\ln(0.1) = (0.012)(2.303) \approx 0.028\,\text{s} = 28\,\text{ms}$ .

6. (2 pts ••) In the circuit of Problem 5, the current has reached its steady value $I_0 = 0.48\,\text{A}$ when, at a new $t = 0$ , the battery is removed and the inductor discharges through the same resistor.

(a) Write the current $i(t)$ during the decay.

(b) Find the time for the current to fall to half of $I_0$ .

(c) A student claims "the energy just disappears when the battery is removed." In one or two sentences, say where the stored magnetic energy actually goes.

Solution

6.

(a) With the source gone, the decay obeys $L\,di/dt + iR = 0$ , giving $i(t) = I_0\,e^{-t/\tau}$ with $\tau = L/R = 12\,\text{ms}$ , i.e. $i(t) = 0.48\,e^{-t/0.012}\,\text{A}$ .

(b) $i = I_0/2$ : $e^{-t/\tau} = \tfrac12 \Rightarrow t = \tau\ln 2 = (0.012)(0.693) \approx 8.3\times10^{-3}\,\text{s} = 8.3\,\text{ms}$ .

(c) The energy $\tfrac12 L I_0^2$ stored in the inductor's magnetic field is not lost — as the current decays it is dissipated as heat ( $i^2R$ ) in the resistor. (Integrating $\int_0^\infty i^2 R\,dt$ recovers exactly $\tfrac12 L I_0^2$ .)

7. (2 pts ••) An L–C circuit has $L = 0.40\,\text{H}$ and $C = 8.0\,\mu\text{F}$ . The capacitor is initially charged to $V_0 = 6.0\,\text{V}$ and released at $t = 0$ with no current.

(a) Find the angular frequency $\omega$ and period $T$ of the oscillation.

(b) Find the maximum charge $Q_0$ on the capacitor and the maximum current $I_\text{max}$ .

(c) At the instant the charge on the capacitor is $q = Q_0/2$ , what fraction of the total energy is stored in the capacitor? (In a real L–R–C series circuit the resistance would slowly drain this energy, damping the oscillation — but ignore resistance here.)

Solution

7.

(a) $\omega = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{(0.40)(8.0\times10^{-6})}} = \dfrac{1}{\sqrt{3.2\times10^{-6}}} \approx 5.6\times10^{2}\,\text{rad/s}$ .

$T = \dfrac{2\pi}{\omega} \approx \dfrac{2\pi}{559} \approx 1.1\times10^{-2}\,\text{s} = 11\,\text{ms}$ (frequency $f = \omega/2\pi \approx 89\,\text{Hz}$ ).

(b) $Q_0 = C V_0 = (8.0\times10^{-6})(6.0) = 4.8\times10^{-5}\,\text{C} = 48\,\mu\text{C}$ .

$I_\text{max} = \omega Q_0 = (559)(4.8\times10^{-5}) \approx 2.7\times10^{-2}\,\text{A} = 27\,\text{mA}$ .

(c) Capacitor energy is $U_C = \dfrac{q^2}{2C}$ and the total is $E = \dfrac{Q_0^2}{2C}$ , so the fraction is $\dfrac{U_C}{E} = \left(\dfrac{q}{Q_0}\right)^2$ . With $q = Q_0/2$ this is $\left(\tfrac12\right)^2 = \tfrac14$ , i.e. $25\%$ of the energy is on the capacitor (the remaining $75\%$ is in the inductor's magnetic field).

Extra Credit (3 pts •••)

Consider the series R–L circuit of Problem 5: a battery of emf $\mathcal{E}$ , resistor $R$ , and inductor $L$ in series, switch closed at $t = 0$ with $i(0) = 0$ .

(a) Apply the loop rule to write the first-order differential equation governing $i(t)$ , being careful with the sign of the inductor's voltage.

(b) Solve this differential equation from first principles (by separating variables and integrating) to obtain the closed-form current growth $i(t)$ . Show every step.

(c) From your $i(t)$ , derive the power delivered by the battery, the power dissipated in the resistor, and the rate at which energy is stored in the inductor, and show explicitly that these three balance at every instant (conservation of energy).

Solution

Extra Credit

(a) Traverse the loop in the direction of the current. Going through the battery is a rise $+\mathcal{E}$ ; through the resistor a drop $-iR$ ; the inductor opposes the increasing current, so it acts as a back-emf $-L\,di/dt$ . The loop rule gives

\mathcal{E} - iR - L\frac{di}{dt} = 0 \quad\Longrightarrow\quad L\frac{di}{dt} = \mathcal{E} - iR.

(b) Separate variables. Write $\dfrac{di}{dt} = \dfrac{\mathcal{E} - iR}{L}$ , so

\frac{di}{\mathcal{E} - iR} = \frac{dt}{L}.

Integrate from $t=0$ (where $i=0$ ) to time $t$ (current $i$ ). For the left side, let $u = \mathcal{E} - iR$ , $du = -R\,di$ :

\int_0^{i}\frac{di'}{\mathcal{E} - i'R} = -\frac{1}{R}\ln(\mathcal{E} - i'R)\Big|_0^{i} = -\frac{1}{R}\ln\!\frac{\mathcal{E} - iR}{\mathcal{E}}.

The right side is $t/L$ . Therefore

-\frac{1}{R}\ln\!\frac{\mathcal{E} - iR}{\mathcal{E}} = \frac{t}{L} \quad\Longrightarrow\quad \ln\!\frac{\mathcal{E} - iR}{\mathcal{E}} = -\frac{R}{L}\,t.

Exponentiate: $\dfrac{\mathcal{E} - iR}{\mathcal{E}} = e^{-Rt/L}$ . Solving for $i$ ,

\boxed{\,i(t) = \frac{\mathcal{E}}{R}\left(1 - e^{-t/\tau}\right), \qquad \tau = \frac{L}{R}.\,}

As a check: $i(0) = 0$ ✓ and $i(\infty) = \mathcal{E}/R$ ✓, matching Problem 5.

(c) With $i = \dfrac{\mathcal{E}}{R}(1 - e^{-t/\tau})$ , note $\dfrac{di}{dt} = \dfrac{\mathcal{E}}{R}\cdot\dfrac{1}{\tau}e^{-t/\tau} = \dfrac{\mathcal{E}}{L}e^{-t/\tau}$ (using $\tau = L/R$ ).

Power delivered by the battery:

P_\text{batt} = \mathcal{E}\,i = \frac{\mathcal{E}^2}{R}\left(1 - e^{-t/\tau}\right).

Power dissipated in the resistor:

P_R = i^2 R = \frac{\mathcal{E}^2}{R}\left(1 - e^{-t/\tau}\right)^2.

Rate of energy storage in the inductor, $\dfrac{d}{dt}\!\left(\tfrac12 L i^2\right) = L i\,\dfrac{di}{dt}$ :

P_L = L\cdot\frac{\mathcal{E}}{R}\left(1 - e^{-t/\tau}\right)\cdot\frac{\mathcal{E}}{L}e^{-t/\tau} = \frac{\mathcal{E}^2}{R}\left(1 - e^{-t/\tau}\right)e^{-t/\tau}.

Add the resistor and inductor terms:

P_R + P_L = \frac{\mathcal{E}^2}{R}\left(1 - e^{-t/\tau}\right)\Big[\left(1 - e^{-t/\tau}\right) + e^{-t/\tau}\Big] = \frac{\mathcal{E}^2}{R}\left(1 - e^{-t/\tau}\right) = P_\text{batt}.

So at every instant $P_\text{batt} = P_R + P_L$ : every joule the battery supplies is either dissipated in $R$ or stored in the inductor's magnetic field. This is precisely energy conservation for the circuit.

Chapter 30 — Inductance

Chapter 30 — Inductance

Extra Credit (3 pts •••)

Extra Credit