Homework Solutions

198

ℱ_{B} = B \cdot k π a^{2} = B_{z} π a^{2} = β t π a^{2}

is the ﬂux through the circuit upon which the charges are embedded.

ℰ = - \frac{d}{d t} ℱ_{B} = = - β π a^{2}

which creates an induced electric ﬁeld along the perimeter

\oint_{C} E_{e q u i v} \cdot d r = - β π a^{2} = - ∣ E_{i n d} ∣ 2 π a

that points counter-clockwise around the loop upon which the charges are embedded. This ﬁeld exerts a force and a torque on each charge;

N = r \times F = q r \times E_{i n d} = N q a (cos θ i + sin θ j) \times (\frac{β π a^{2}}{2 π a}) (sin θ i - cos θ j) = - \frac{q N β a^{2}}{2} k

This will provide an angular acceleration

N = I α (- k) = \frac{1}{2} m b^{2} α (- k) = \frac{1}{2} m b^{2} α, α = \frac{d ω}{d t} = \frac{q N β a^{2}}{m b^{2}}

and so the spin rate at time $t$ will be

ω = α t = \frac{q N β a^{2}}{m b^{2}} t

199
Let $i$ point right, $j$ point up and $k$ point out. At a distance $r$ from the long wire the ﬁeld of the wire is

B = \frac{μ_{0} I}{2 π r} (- k)

and the ﬂux through a horizontal strip of area $d A = a d r$ within the enclosed rectangle between $r$ and $r + d r$ is (taking its normal to be $n = - k$ )

d ℱ_{B} = \frac{μ_{0} I}{2 π r} (- k) \cdot (a d r (- k)), ℱ_{B} = \int_{x}^{x + b} \frac{μ_{0} I}{2 π r} k \cdot (a d r k) = \frac{μ_{0} I a}{2 π} ln \frac{x + b}{x}

The emf induced in the circuit is

ℰ = \oint_{C} E_{e q u i v} \cdot d r = - \frac{d}{d t} ℱ_{B} = - \frac{μ_{0} I a}{2 π} (\frac{v}{x + b} - \frac{v}{x}), \frac{d x}{d t} = v, x = r - \frac{b}{2}

and so we see that the induced electric ﬁeld within the wires opposes the directed arc lengths $d r$ , which with our choice of normal circulate counterclockwise. The induced electric ﬁeld and induced currents driven by it circulate around the rectangler clockwise, passing through the bar from right to left.
The force needed to move the bar at constant speed does work at a rate equal to the rate of heat creation in the resistant elements of the loop;

F \cdot v = F v = R I_{i n d u c e d}^{2} = \frac{ℰ^{2}}{R}

200
Let $i$ point right, $j$ point up and $k$ point out. At a distance $r$ from the long wire the ﬁeld of the wire is

B = \frac{μ_{0} I (t)}{2 π r} (- k)

and the ﬂux through a horizontal strip of area $d A = c d r$ within the enclosed rectangle between $r$ and $r + d r$ is (taking its normal to be $n = - k$ )

d ℱ_{B} = \frac{μ_{0} I (t)}{2 π r} (- k) \cdot (c d r (- k)), ℱ_{B} = \int_{a}^{a + b} \frac{μ_{0} N I (t)}{2 π r} k \cdot (c d r k) = \frac{μ_{0} N I (t) c}{2 π} ln \frac{a + b}{a}

The emf induced in the circuit is

ℰ = - \frac{d}{d t} ℱ_{B} = (- \frac{d I}{d t}) \frac{μ_{0} c}{2 π} ln \frac{a + b}{a} = - (100 A) (120 π \frac{1}{s}) cos (120 π t) \frac{μ_{0} N c}{2 π} ln \frac{a + b}{a} = - 101.32 V cos (\frac{120 π}{s} t)

204
Pick $n$ out, then $d r$ are CCW;

ℱ_{B} = B_{0} d^{2} cos ω t, E = - \frac{d}{d t} B_{0} d^{2} cos ω t = B_{0} d^{2} ω sin ω t, I_{i n d} = \frac{B_{0} d^{2} ω}{\sqrt{2} R} at t = 0.25 s

This current ﬂows CCW, to the right through the bar, so the magnetic force on the bar is (at $t = 0.25 s$ )

F = \frac{B_{0} d^{2} ω}{\sqrt{2} R} i \times B_{0} \frac{1}{\sqrt{2}} k d = \frac{B_{0}^{2} d^{3} ω}{2 R} (- j)

and you must apply the opposite force to hold the bar still. You were asked to use $R = 1.0 Ω$ in your calculation.
208
We know that the self inductance is proportional to $N^{2}$ ;

ℱ = N^{2} ℱ_{0}

so $ℱ_{0}$ is the ﬂux per turn, then

3 \times 1 0^{- 3} V = ∣ - L_{0} N^{2} (5.0 \frac{A}{s}) ∣, ℱ_{0} = 4 \times 1 0^{- 5} W b = \frac{ℱ}{N} = \frac{L I}{N} = L_{0} N (8.0 A)

therefore

5.0 \times 1 0^{- 6} \frac{W b}{A} = L_{0} N, 25 \times 1 0^{- 12} \frac{W b^{2}}{A^{2}} = L_{0}^{2} N^{2}

so divide

L_{0} = \frac{25 \times 1 0^{- 12} \frac{W b^{2}}{A^{2}}}{\frac{3 \times 1 0^{- 3} V s}{5 A}} = 4.17 \times 1 0^{- 8} H

210
Call the radius in question $c$ ; the ﬁeld at $r$ for $a \leq r \leq b$ is $∣ B ∣ = \frac{μ_{0} I}{2 π r}$ ;

\frac{1}{2} \int_{a}^{b} \frac{μ_{0} I^{2} ℓ}{4 π r^{2}} r d r = \int_{a}^{c} \frac{μ_{0} I^{2} ℓ}{4 π r^{2}} r d r, ln \sqrt{\frac{b}{a}} = ln \frac{c}{a}, c = \sqrt{a b}

Very long after closing the switch, all currents stop changing, so the loop equations

L_{1} \frac{d}{d t} (I_{0} - I_{2}) = 0 = I_{2} R_{2} - I_{5} R_{5}, I_{2} = \frac{R_{5}}{R_{2}} I_{5}

+ L_{2} \frac{d}{d t} (I_{0} - I_{2} - I_{5}) = 0 = I_{5} R_{5} + R_{4} I_{4}, I_{4} = - \frac{R_{5}}{R_{4}} I_{5}

(since we pass through the inductors $L_{i}$ in the senses opposing the currents through them for clockwise loop senses) and the junction rule $I_{2} + I_{5} = I_{4}$ overconstrain the three currents $I_{2}, I_{4}, I_{5}$ and so $I_{2} = I_{5} = I_{4} = 0$ , and so the big outer loop equation gives us

V_{0} = R_{2} I_{2} + R_{4} I_{4} + R_{0} I_{0} = R_{0} I_{0}, I_{0} = \frac{V_{0}}{R_{0}}

211
All currents are to the right or downwards, except we let $I_{0}$ be upwards. No cuurent can ﬂow through either inductor instantly after closing the switch, and so we discard any loop equation containing an inductor and the derivative of the current through it, and rely on junction equations

I_{0} = I_{2}, I_{5} = 0, I_{2} = I_{4}

and one loop equation

V_{0} = I_{2} R_{2} + I_{4} R_{4} + I_{0} R_{0} = (R_{0} + R_{2} + R_{4}) I_{0}, I_{0} = I_{2} = I_{4} = \frac{V_{0}}{R_{0} + R_{2} + R_{4}}

Very long after closing the switch, all currents stop changing, so the loop equations

L_{1} \frac{d}{d t} (I_{0} - I_{2}) = 0 = I_{2} R_{2} - I_{5} R_{5}, I_{2} = \frac{R_{5}}{R_{2}} I_{5}

+ L_{2} \frac{d}{d t} (I_{0} - I_{2} - I_{5}) = 0 = I_{5} R_{5} + R_{4} I_{4}, I_{4} = - \frac{R_{5}}{R_{4}} I_{5}

V_{0} = R_{2} I_{2} + R_{4} I_{4} + R_{0} I_{0} = R_{0} I_{0}, I_{0} = \frac{V_{0}}{R_{0}}

Very long after closing the switch, all currents stop changing, so the loop equations

L_{1} \frac{d}{d t} (I_{0} - I_{2}) = 0 = I_{2} R_{2} + I_{5} R_{5}, I_{2} = - \frac{R_{5}}{R_{2}} I_{5}

- L_{2} \frac{d}{d t} (I_{0} - I_{2} - I_{5}) = 0 = - I_{5} R_{5} - R_{4} I_{4}, I_{4} = - \frac{R_{5}}{R_{4}} I_{5}

V_{0} - L_{1} \frac{d}{d t} (I_{0} - I_{2}) = V_{0} + 0 = I_{4} R_{4} + I_{0} R_{0}

and the junction rule

I_{0} - I_{2} + I_{5} = I_{4}

results in

I_{5} = - \frac{I_{0}}{1 + \frac{R_{5}}{R_{4}} + \frac{R_{5}}{R_{2}}}, I_{2} = \frac{R_{5}}{R_{2}} \frac{I_{0}}{1 + \frac{R_{5}}{R_{4}} + \frac{R_{5}}{R_{2}}}, I_{4} = \frac{R_{5}}{R_{4}} \frac{I_{0}}{1 + \frac{R_{5}}{R_{4}} + \frac{R_{5}}{R_{2}}}

and

V_{0} = I_{0} (R_{0} + \frac{R_{5}}{1 + \frac{R_{5}}{R_{4}} + \frac{R_{5}}{R_{2}}})

213
With $S_{1}$ and $S_{2}$ closed for a long time, we have

V_{0} = R_{1} I_{1} + R_{2} I_{2}, - L \frac{d}{d t} (I_{2} - I_{1}) = 0 = - R_{1} I_{1}

and so $I_{1} = 0$ and $I_{2} = \frac{V_{0}}{R_{2}}$ . Now open $S_{1}$ . The current through $L$ cannot abruptly change, and so it remains at $I_{2} - I_{1} = \frac{V_{0}}{R_{2}}$ for an instant and steadily drops off. This current is forced by the emf of the inductor up through $R_{1}$ , creating a voltage

V_{e} - V_{b} = R_{1} \frac{V_{0}}{R_{2}} = 10.0 V \frac{1.0 \times 1 0^{6} Ω}{1 \times 1 0^{3} Ω} = 10, 000 V