Homework Solutions

193
The induced emf drives a current $I (t)$ through the galvanometer according to

ℰ = I R = R \frac{d q}{d t} = - \frac{d}{d t} ℱ_{B}

integrate;

\int_{0}^{t} \frac{d q}{d t} d t = q (t) - q (0) = q (t) = \frac{1}{R} \int_{0}^{t} - \frac{d}{d t} ℱ_{B} d t = \frac{1}{R} (ℱ_{B} (0) - ℱ_{B} (t))

If you have a galvanometer, you could in principle use it to measure a magnetic ﬁeld strength, or alternatively put a resistor on the loop instead of a galvanometer, connect an oscilloscope of very high impedance across the resistor, and ﬂip the coil over. You can integrate the oscilloscope trace (the voltage across $R$ ) and measure $∣ B ∣$ .

194
Let $i$ point right, $j$ point up and $k$ point out. The equivalent electric ﬁeld in the moving bar is then

E = (- v j) \times (B k) = - v B i

which creates emf

∣ ℰ ∣ = ∣ \int_{0}^{d} E_{e q u i v} \cdot d r ∣ = B v d

that forces induced current $I = \frac{ℰ}{R} = \frac{B v d}{R}$ to the left through the bar (in the direction of the electric ﬁeld by Ohm’s law). The force per unit length exerted by the magnetic ﬁeld on the bar is

\frac{d F}{d x} = I t \times B = (- \frac{B v d}{R} i) \times (B k) = \frac{B^{2} v d}{R} j

and so the magnetic force on the bar is

F = \int_{0}^{d} \frac{B^{2} v d}{R} j d x = \frac{B^{2} v d^{2}}{R} j

You must supply an equal and opposite force to move the bar. Note that the work which you do to move the bar at $v$ is dissipated as heat in the resistor

F \cdot v = \frac{B^{2} v^{2} d^{2}}{R} = R I^{2}

195
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

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

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}{d + x}), \frac{d x}{d t} = v

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}

196
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 vertical strip of area $d A = x d r$ within the enclosed rectangle between $r$ and $r + d r$ is

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

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 v}{2 π} ln \frac{d + a}{d}, \frac{d x}{d t} = v

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

197
With no ﬁeld, there is zero ﬂux through the loop. When the ﬁeld is turned on, the ﬂux jumps to

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

Suppose that this change in ﬂux occurs over the time $d t$ it takes to switch on the ﬁeld, then

ℰ = \frac{ℱ_{B} (b e f o r e) - ℱ_{B} (a f t e r)}{d t} = - \frac{B_{z} π a^{2}}{d t}

which creates an induced electric ﬁeld within the ring

\oint_{C} E_{e q u i v} \cdot d r = - \frac{B_{z} π a^{2}}{d t}

which must circulate contrary to $d r$ , and therefore ﬂows clockwise around the ring (with $n = k$ for the ring the $d r$ vectors circulate counter-clockwise). This drives an induced current

I_{i n d} = \frac{ℰ}{R} = \frac{B_{z} π a^{2}}{R d t}

clockwise around the ring. Let $θ$ be measured in a counter-clockwise sense starting at the $x$ axis, then the force on a length $a d θ$ of the ring between $θ$ and $θ + d θ$ is

d F = I_{i n d} (sin θ i - cos θ j) \times B a d θ = I_{i n d} (B_{r} k - B_{z} sin θ j - B_{z} cos θ i) a d θ

and the total force on the ring is

F = \int_{0}^{2 π} d F = 2 π a I_{i n d} B_{r} k

The ring jumps up. Suppose at $t = 0$ the ring is stationary and there is no ﬁeld, at time $d t$ the ﬁeld is turned on and the ring has velocity $v k$ . Then

m a = m (\frac{v k - 0}{d t}) = F = 2 π a (\frac{B_{z} π a^{2}}{R d t}) B_{r} k

The ring has been launched vertically at speed

v = \frac{2 π^{2} a^{3} B_{z} B_{r}}{R}

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}

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)

201
When the string is displaced by the wave $y = A sin (k x_{0} - ω t)$ at point $x = x_{0}$ where the pickup touches the string, the “N” end of the magnet is a distance $r = a + A sin (k x_{0} - ω t)$ from the coil, and so the ﬁeld of the bar magnet felt at the coil center is

∣ B ∣ = \frac{m}{{(a + A sin (k x_{0} - ω t))}^{3}} \approx \frac{m}{a^{3}} (1 - \frac{3 A}{a} sin (k x_{0} - ω t))

The ﬂux through the coil is then

ℱ_{B} \approx (\frac{m}{a^{3}}) (N π b^{2}) (1 - \frac{3 A}{a} sin (k x_{0} - ω t))

and the emf produced between the terminals of the coil is

ℰ = - \frac{d}{d t} ℱ_{B} = - (\frac{m}{a^{3}}) (N π b^{2}) (\frac{3 A ω}{a}) cos (k x_{0} - ω t))

which oscillates at the same frequency as the string, and has an amplitude proportional to the string’s amplitude.

202
I call this a dipole speaker. The current produces a dipole

m^{'} = I (t) N π b^{2} k

in the coil, which produces a magnetic ﬁeld. The potential energy of the bar-magnet magnetic moment $m = m (- k)$ in the ﬁeld of the coil is

U = - \frac{μ_{0}}{4 π} 2 \frac{∣ m ∣ ∣ m^{'} ∣}{x^{3}}

when they are a distance $x$ apart. The force exerted on the bar magnet is

F_{x} = - \frac{d}{d x} U = - \frac{μ_{0}}{4 π} 6 \frac{∣ m ∣ ∣ m^{'} ∣}{x^{4}} = - \frac{μ_{0}}{4 π} \frac{6 (I_{0} N π b^{2} sin ω t) (m)}{a^{4}}

which is transmitted into the diaphram that the magnet is attached to. This makes the speaker oscillate at the same frequency as the current through the coil.

203
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.

204

ℱ_{B} = (B_{0} j) \cdot (cos (ω t + \frac{3 π}{4}) i + sin (ω t + \frac{3 π}{4}) j) ℓ^{2} = B_{0} ℓ^{2} sin (ω t + \frac{3 π}{4})

ℰ = - \frac{d}{d t} B_{0} ℓ^{2} sin (ω t + \frac{3 π}{4}) = - B_{0} ω ℓ^{2} cos (ω t + \frac{3 π}{4}) = + \frac{B_{0} ω ℓ^{2}}{\sqrt{2}} at t = 0

so $I_{i n d} = \frac{B_{0} ω ℓ^{2}}{\sqrt{2} R}$ at $t = 0$ in the $d r$ direction, with $n$ as illustrated this means $C C W$ around the loop, passing through $R$ from big collar $a$ to small $b$ .
Without any further adieu we know that the torque would try to align the normal with the ﬁeld, so $F$ on the top bar points right;

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

Since we force the loop to rotate at $ω$ against this torque; the magnetic torque is opposite to that which we apply

℘ = N_{u s} \cdot ω = I_{i n d}^{2} R, N_{u s} = \frac{B_{0}^{2} ω ℓ^{4}}{2 R} k