Given: Square loop of side $l = 10 \text{ cm} = 0.10 \text{ m}$. Area $A = l^2 = (0.10 \text{ m})^2 = 0.01 \text{ m}^2$. Resistance $R = 0.5 \Omega$. The loop is vertical in the east-west plane. The magnetic field is uniform, initial magnitude $B_{initial} = 0.10 \text{ T}$, final magnitude $B_{final} = 0 \text{ T}$. Time interval $\Delta t = 0.70 \text{ s}$. The field is in the north-east direction, across the plane of the loop. This means the field is in the plane of the loop, but making an angle with the plane? No, "across the plane" usually means the field is perpendicular to the area vector of the loop. The loop is in the east-west plane. This means the loop is oriented vertically. Its area vector can be in the north or south direction (perpendicular to the plane). The magnetic field is in the north-east direction.

Let's assume "across the plane in the north-east direction" means the field vector $\vec{B}$ is in the North-East direction. The loop is in the East-West plane, vertically. The area vector of the loop can point North or South. The angle between the North direction and the North-East direction is 45°. So the angle $\theta$ between the area vector (pointing North, say) and the magnetic field (North-East) is 45°. The area vector can also point South, in which case the angle with North-East is $180^\circ - 45^\circ = 135^\circ$. The flux $\Phi_B = BA\cos\theta$ depends on $\cos\theta$. $\cos(45^\circ) = 1/\sqrt{2}$, $\cos(135^\circ) = -1/\sqrt{2}$. Let's take the area vector pointing North, $\theta=45^\circ$.

Initial magnetic flux through the loop: $\Phi_{B,initial} = B_{initial} A \cos\theta = (0.10 \text{ T}) \times (0.01 \text{ m}^2) \times \cos(45^\circ) = 0.001 \times (1/\sqrt{2}) \text{ Wb} = 0.001 / \sqrt{2} \text{ Wb}$.

Final magnetic flux: $\Phi_{B,final} = B_{final} A \cos\theta = 0 \times (0.01) \times \cos(45^\circ) = 0 \text{ Wb}$.

Change in magnetic flux: $\Delta\Phi_B = \Phi_{B,final} - \Phi_{B,initial} = 0 - 0.001/\sqrt{2} = -0.001/\sqrt{2} \text{ Wb}$.

The magnitude of the induced emf is $|\epsilon| = |\frac{\Delta\Phi_B}{\Delta t}|$ since the field is changing at a steady rate (constant rate of change). Time interval $\Delta t = 0.70 \text{ s}$.

$|\epsilon| = \left|\frac{-0.001/\sqrt{2} \text{ Wb}}{0.70 \text{ s}}\right| = \frac{0.001}{0.70 \sqrt{2}} \text{ V} = \frac{10^{-3}}{0.7 \times 1.414} \text{ V} \approx \frac{10^{-3}}{0.9898} \text{ V} \approx 1.01 \times 10^{-3} \text{ V} = 1.01 \text{ mV}$.

The text's calculation is $0.1 \times 10^{-2} / \sqrt{2} / 0.7 = 10^{-3} / (0.7 \sqrt{2}) \approx 1.0 \text{ mV}$. Let's assume the text's value is correct due to possible rounding in $\sqrt{2}$. $1.0 \text{ mV} = 1.0 \times 10^{-3} \text{ V}$.

The magnitude of the induced current is $I = |\epsilon| / R$.

$I = \frac{1.0 \times 10^{-3} \text{ V}}{0.5 \Omega} = 2.0 \times 10^{-3} \text{ A} = 2.0 \text{ mA}$.

The text's calculation $I = \frac{1.0 \times 10^{-3} \text{ V}}{0.5 \Omega} = 2 \text{ mA}$ matches.

Magnitudes of induced emf and current are approximately $1.0$ mV and $2.0$ mA.

Given: Circular coil radius $R = 10 \text{ cm} = 0.10 \text{ m}$. Area $A = \pi R^2 = \pi (0.10)^2 = 0.01\pi \text{ m}^2$. Number of turns $N = 500$. Resistance $R_{coil} = 2 \Omega$. Horizontal component of Earth's magnetic field $B_H = 3.0 \times 10^{-5} \text{ T}$.

Initially, the coil's plane is perpendicular to $B_H$. This means the area vector $\vec{A}$ is parallel or antiparallel to $B_H$. Let's assume $\vec{A}$ is parallel to $B_H$. Initial angle $\theta_{initial} = 0^\circ$. Initial flux through each turn $\Phi_{B,initial} = B_H A \cos(0^\circ) = B_H A$.

The coil is rotated about its vertical diameter through 180°. After rotation, the coil's plane is still perpendicular to the original direction of the horizontal field (as it rotated about a diameter in its plane). However, the area vector $\vec{A}$ is now pointing in the opposite direction relative to the horizontal field. Final angle $\theta_{final} = 180^\circ$. Final flux through each turn $\Phi_{B,final} = B_H A \cos(180^\circ) = -B_H A$.

Change in flux through each turn $\Delta\Phi_B = \Phi_{B,final} - \Phi_{B,initial} = -B_H A - B_H A = -2 B_H A$.

Time taken for rotation $\Delta t = 0.25 \text{ s}$. The induced emf is given by Faraday's law $\epsilon = -N \frac{\Delta\Phi_B}{\Delta t}$.

Magnitude of induced emf $|\epsilon| = \left|-N \frac{-2 B_H A}{\Delta t}\right| = \frac{2 N B_H A}{\Delta t}$.

$|\epsilon| = \frac{2 \times 500 \times (3.0 \times 10^{-5} \text{ T}) \times (0.01\pi \text{ m}^2)}{0.25 \text{ s}}$.

$|\epsilon| = \frac{1000 \times 3.0 \times 10^{-5} \times 0.01\pi}{0.25} \text{ V} = \frac{3.0 \times 10^{-2} \times 0.01\pi}{0.25} \text{ V} = \frac{3.0 \times 10^{-4} \pi}{0.25} \text{ V}$.

$|\epsilon| = 12 \pi \times 10^{-4} \text{ V} \approx 12 \times 3.14 \times 10^{-4} \text{ V} \approx 37.68 \times 10^{-4} \text{ V} = 3.768 \times 10^{-3} \text{ V}$.

The text gives $3.8 \times 10^{-3}$ V, which is close.

Magnitude of induced current $I = |\epsilon| / R_{coil}$.

$I = \frac{3.768 \times 10^{-3} \text{ V}}{2 \Omega} = 1.884 \times 10^{-3} \text{ A}$.

The text gives $1.9 \times 10^{-3}$ A, which is close.

Note on instantaneous values: The formula used calculates the average emf over the time interval $\Delta t$ assuming a steady rate of change of flux. If the coil rotates with constant angular speed $\omega$, the instantaneous emf is given by $\epsilon(t) = NBA\omega \sin(\omega t)$, and the rate of change of flux is not constant over the entire 180° rotation.

Magnitudes of induced emf and current are approximately $3.8$ mV and $1.9$ mA.

The magnetic field is normal to the plane of the loop and directed away from the reader (let's say this direction is "inward").

(i) Rectangular loop abcd moving into the field: As the loop enters the field, the magnetic flux through it increases (inward). By Lenz's law, the induced current must oppose this increase. It will create a magnetic field pointing outwards (opposite to the inward field). By the right-hand rule for current loops, a clockwise current creates an inward field, and a counter-clockwise current creates an outward field. Therefore, the induced current must be in the counter-clockwise direction.

Direction of induced current: **bcdab** (counter-clockwise).

(ii) Triangular loop abc moving out of the field: As the loop leaves the field, the magnetic flux through it decreases (inward). The induced current must oppose this decrease by creating a magnetic field pointing inwards (in the same direction as the external field). By the right-hand rule, a clockwise current creates an inward field.

Direction of induced current: **bacb** (clockwise).

(iii) Irregular shaped loop abcd moving out of the field: As the loop leaves the field, the magnetic flux through it decreases (inward). The induced current must oppose this decrease by creating an inward magnetic field. A clockwise current creates an inward field.

Direction of induced current: **cdabc** (clockwise).

Note: Induced current exists only when the magnetic flux through the loop is *changing*. Once the loop is completely inside or completely outside the uniform field region, the flux is constant, and induced emf/current is zero.

(a) Stationary loop in a strong magnetic field: Current is induced only when the magnetic flux through the loop **changes** with time ($\epsilon = -d\Phi_B/dt$). If the magnetic field is constant and the loop is stationary, the magnetic flux is constant, regardless of the field strength. Therefore, even with very strong fixed magnets, no current is generated in a stationary loop.

Answer: **No**, strength of magnets does not induce current; changing flux does.

(b) Loop moving in a constant electric field: Induced emf is caused by a changing magnetic flux ($\epsilon = -d\Phi_B/dt$). Magnetic flux is related to the magnetic field. The problem describes movement in an **electric field**, not a magnetic field. Electric fields do not induce magnetic fields in this context, and they do not directly induce electric current in a loop just by moving through them, unless there are specific relativistic effects or secondary magnetic fields involved (beyond the scope of this simple problem). Faraday's law applies to changes in *magnetic* flux.

(i) When the loop is wholly inside the uniform electric field: The electric flux through the loop is constant ($\Phi_E = \int \vec{E} \cdot d\vec{A}$). Also, there is no magnetic field or changing magnetic field mentioned. So, no induced emf or current due to changing magnetic flux.

(ii) When the loop is partially outside: The electric flux through the loop might change as the area within the field changes. But this does not induce *current* in the sense of electromagnetic induction (related to changing magnetic flux). While charge might redistribute on the conductor due to the electric field, leading to temporary currents, this is not the phenomenon of electromagnetic induction as described by Faraday's law (driven by $\Delta\Phi_B/\Delta t$).

Answer: **No** current is induced in either case by simply moving through a constant electric field in this context.

(c) Loops moving out of a uniform magnetic field: Both loops move with constant velocity $v$. The magnetic field is normal to the loops.

For the rectangular loop: As it moves out, the area of the loop *inside* the magnetic field changes linearly with distance. For example, if a side of length $l$ is perpendicular to the velocity, the rate of change of area inside the field is $l \times v = constant$. Since the field is uniform, the rate of change of flux ($\Phi_B = BA_{in}$) is $\frac{d\Phi_B}{dt} = B \frac{dA_{in}}{dt} = B (l \times v) = \text{constant}$. Therefore, the induced emf $|\epsilon| = |\frac{d\Phi_B}{dt}|$ is constant during the passage out of the field.

For the circular loop: As it moves out, the area of the loop *inside* the magnetic field does not change linearly with distance. The shape of the entering/leaving area is part of a circle, and its rate of change with linear displacement is not constant. Therefore, the rate of change of flux is not constant, and the induced emf will vary during the passage out.

Answer: The induced emf is expected to be constant during the passage out of the field region only in the case of the **rectangular loop**.

(d) Polarity of the capacitor: Figure 6.9 shows a metallic rod PQ moving to the left in a magnetic field pointing into the paper. The free charges (electrons) in the rod experience a Lorentz force $\vec{F}_B = q(\vec{v} \times \vec{B})$. The velocity $\vec{v}$ is to the left, and $\vec{B}$ is into the paper. For a positive charge ($q>0$), $\vec{v} \times \vec{B}$ is upwards (using right-hand rule). So positive charges are pushed towards P. For a negative charge (electron, $q<0$), the force $\vec{F}_B$ is in the direction opposite to $\vec{v} \times \vec{B}$, i.e., downwards, towards Q. Electrons accumulate at end Q, leaving end P deficient in electrons (positively charged).

If P and Q are connected to the plates of a capacitor, positive charges effectively move towards P, and negative charges towards Q. Plate A is connected to P, and Plate B is connected to Q.

Answer: The polarity of plate ‘A’ will be **positive** and the polarity of plate ‘B’ will be **negative**.

Given: Length of rod $l = 1$ m. This is also the radius of the circular ring $R = 1$ m. Frequency of rotation $f = 50$ rev/s. Angular frequency $\omega = 2\pi f = 2\pi (50 \text{ s}^{-1}) = 100\pi \text{ rad/s}$. Uniform magnetic field $B = 1$ T, parallel to the axis of rotation (and perpendicular to the plane of rotation). The emf is induced between the center (hinged end) and the rim (other end). This is a case of motional emf as the rod moves in the magnetic field.

The speed of a point on the rod depends on its distance $r$ from the center. The speed is $v = \omega r$. The velocity of each segment $dr$ is perpendicular to both the rod and the magnetic field (since $\vec{v}$ is tangential to the circle of radius $r$, $\vec{B}$ is parallel to the axis, and the rod is radial). The induced emf across an element $dr$ at distance $r$ is $d\epsilon = B v dr = B (\omega r) dr$.

To find the total emf between the center ($r=0$) and the rim ($r=R$), integrate $d\epsilon$ along the length of the rod from $r=0$ to $r=R$.

$\epsilon = \int_0^R d\epsilon = \int_0^R B \omega r dr = B \omega \int_0^R r dr = B \omega \left[\frac{r^2}{2}\right]_0^R = B \omega \left(\frac{R^2}{2} - 0\right) = \frac{1}{2} B \omega R^2$.

Substitute the values: $B=1$ T, $\omega = 100\pi$ rad/s, $R=1$ m.

$\epsilon = \frac{1}{2} \times (1 \text{ T}) \times (100\pi \text{ rad/s}) \times (1 \text{ m})^2 = 50\pi \text{ V}$.

Using $\pi \approx 3.14$: $\epsilon \approx 50 \times 3.14 \text{ V} = 157 \text{ V}$.

The text's answer $157$ V matches this calculation.

The emf between the center and the metallic ring is $157$ V.

Given: Number of spokes = 10. Length of each spoke (radius of wheel) $R = 0.5 \text{ m}$. Speed of rotation = 120 rev/min. Horizontal component of Earth's magnetic field $H_E = 0.4 \text{ G} = 0.4 \times 10^{-4} \text{ T} = 4 \times 10^{-5} \text{ T}$. The plane of rotation is normal to $H_E$, so the magnetic field component relevant for induction is $B = H_E$. The spokes are rotating in this field.

Convert speed from rev/min to rev/s: $120 \text{ rev/min} = \frac{120}{60} \text{ rev/s} = 2 \text{ rev/s}$. Frequency $f = 2 \text{ Hz}$. Angular speed $\omega = 2\pi f = 2\pi (2 \text{ s}^{-1}) = 4\pi \text{ rad/s}$.

This is the same situation as in Example 6.6, but with multiple spokes. Each spoke is a metallic rod rotating in a magnetic field. The emf induced between the center (axle) and the tip (rim) of one spoke is given by $\epsilon_{spoke} = \frac{1}{2} B \omega R^2$.

Substitute the values: $B = 4 \times 10^{-5}$ T, $\omega = 4\pi$ rad/s, $R = 0.5$ m.

$\epsilon_{spoke} = \frac{1}{2} \times (4 \times 10^{-5} \text{ T}) \times (4\pi \text{ rad/s}) \times (0.5 \text{ m})^2 = \frac{1}{2} \times 4 \times 10^{-5} \times 4\pi \times 0.25 \text{ V}$.

$\epsilon_{spoke} = 2 \times 10^{-5} \times \pi \text{ V} = 2\pi \times 10^{-5} \times 0.25 \text{ V} = 0.5\pi \times 10^{-5} \text{ V}$. Wait, $2 \times 10^{-5} \times 4\pi \times 0.25 = 8\pi \times 10^{-5} \times 0.25 = 2\pi \times 10^{-5}$. No. $ (1/2) \times 4 \times 10^{-5} \times 4\pi \times (0.5)^2 = 2 \times 10^{-5} \times 4\pi \times 0.25 = 8\pi \times 10^{-5} \times 0.25 = 2\pi \times 10^{-5}$ V.

Using $\pi \approx 3.14$: $\epsilon_{spoke} \approx 2 \times 3.14 \times 10^{-5} \text{ V} \approx 6.28 \times 10^{-5}$ V.

The spokes are all connected at the center (axle) and at the rim. Effectively, they are connected in parallel between the axle and the rim. When identical voltage sources are connected in parallel, the total voltage across the combination is the same as the voltage of a single source.

The number of spokes (10) is immaterial for the voltage between the axle and the rim. The induced emf between the axle and the rim of the wheel is the same as the emf induced in a single spoke.

The text's calculation is (1/2) $\times 4\pi \times 0.4 \times 10^{-4} \times (0.5)^2 = (1/2) \times 4\pi \times 0.4 \times 10^{-4} \times 0.25 = 2\pi \times 0.1 \times 10^{-4} = 0.2\pi \times 10^{-4} \approx 0.2 \times 3.14 \times 10^{-4} = 0.628 \times 10^{-4} = 6.28 \times 10^{-5}$ V.

Okay, the text used $\omega$ as $4\pi$, $B$ as $0.4 \times 10^{-4}$, $R$ as $0.5$. My calculation $2\pi \times 10^{-5}$ is wrong. $\omega = 4\pi$. $B = 4 \times 10^{-5}$. $R=0.5$. $\epsilon = (1/2) \times (4 \times 10^{-5}) \times (4\pi) \times (0.5)^2 = 2 \times 10^{-5} \times 4\pi \times 0.25 = 8\pi \times 10^{-5} \times 0.25 = 2\pi \times 10^{-5}$ V. No. $2 \times 4\pi \times 0.25 \times 10^{-5} = 2\pi \times 0.25 \times 10^{-5} = 0.5\pi \times 10^{-5}$. Wait, $(1/2) \times 4 \times 10^{-5} \times 4\pi \times (0.5)^2 = (1/2) \times 4 \times 4\pi \times 0.25 \times 10^{-5} = (1/2) \times 16\pi \times 0.25 \times 10^{-5} = 8\pi \times 0.25 \times 10^{-5} = 2\pi \times 10^{-5}$ V. Is the frequency 120 rev/min = 2 rev/s? Yes. $\omega = 4\pi$. $B_H = 0.4 \times 10^{-4}$. $R=0.5$. $\epsilon = (1/2) B_H \omega R^2 = (1/2) (0.4 \times 10^{-4}) (4\pi) (0.5)^2 = (1/2) (0.4 \times 10^{-4}) (4\pi) (0.25) = (0.2 \times 10^{-4}) (4\pi) (0.25) = 0.8\pi \times 0.25 \times 10^{-4} = 0.2\pi \times 10^{-4}$. $0.2\pi \times 10^{-4} \text{ V} \approx 0.2 \times 3.14 \times 10^{-4} \text{ V} = 0.628 \times 10^{-4} \text{ V} = 6.28 \times 10^{-5}$ V. Yes, the text's value is $6.28 \times 10^{-5}$ V. The initial calculation had a typo in the power of 10 for $\pi$.

The induced emf between the axle and the rim is $6.28 \times 10^{-5}$ V.

Let the length of the arm PQ be $l$ and its resistance be $r$. The magnetic field is uniform and perpendicular to the plane (into the paper), with magnitude $B$. The field exists for $0 \le x \le b$. The rod PQ moves along the x-axis with constant speed $v$. Assume the arm moves from $x=0$ to $x=2b$ and then back to $x=0$. Let the current direction be clockwise for positive emf.

Forward motion (0 to 2b):

• Magnetic Flux ($\Phi_B$): At position $x$, the area inside the field is $l \times x$ if $0 \le x \le b$. If $x > b$, the entire width $b$ is inside the field, so area is $l \times b$. $\Phi_B = Blx$ for $0 \le x \le b$. $\Phi_B = Blb$ for $b < x \le 2b$. Plot of $\Phi_B$ vs $x$ is linear increasing from 0 to $Blb$ in $0 \le x \le b$, then constant at $Blb$ in $b < x \le 2b$.
• Induced emf ($\epsilon$): $\epsilon = -d\Phi_B/dt = -B l (dx/dt) = -Blv$ for $0 \le x < b$ (since $dx/dt = v$). $\epsilon = -d(Blb)/dt = 0$ for $b < x \le 2b$ (flux is constant). Plot of $\epsilon$ vs $x$ is a constant negative value $-Blv$ for $0 \le x < b$, then zero for $b < x \le 2b$. Magnitude $|\epsilon| = Blv$ for $0 \le x < b$, 0 otherwise.
• Induced Current ($I$): $I = \epsilon/r = -Blv/r$ for $0 \le x < b$. $I = 0$ for $b < x \le 2b$. Plot of $I$ vs $x$ is a constant negative value $-Blv/r$ for $0 \le x < b$, then zero for $b < x \le 2b$.
• Force ($F$): Force on arm PQ is $F = IlB$ (magnitude). Direction opposes motion (to the left). $F = |I|lB = (|{-Blv/r}|)lB = (Blv/r)lB = B^2l^2v/r$ for $0 \le x < b$. $F = 0$ for $b < x \le 2b$ (since $I=0$). Plot of $F$ vs $x$ is a constant positive value $B^2l^2v/r$ for $0 \le x < b$, then zero for $b < x \le 2b$. This is the force external agent needs to apply to maintain constant velocity.
• Power ($P_{Joule}$): Joule heating power $P_J = I^2 r = (-Blv/r)^2 r = B^2l^2v^2/r$ for $0 \le x < b$. $P_J = 0$ for $b < x \le 2b$. Plot of $P_J$ vs $x$ is a constant positive value $B^2l^2v^2/r$ for $0 \le x < b$, then zero for $b < x \le 2b$. This power comes from the work done by the external agent $P_{mech} = F v = (B^2l^2v/r) v = B^2l^2v^2/r$, which matches $P_J$.

Inward motion (2b to 0): The arm moves from $x=2b$ back to $x=0$ with speed $v$. Here $dx/dt = -v$. The motion is in the negative x direction. Velocity $\vec{v}$ is to the right. Magnetic force $\vec{F}_B = q(\vec{v} \times \vec{B})$. $\vec{v}$ is right, $\vec{B}$ is into page. $\vec{v} \times \vec{B}$ is downwards. Force on positive charge is down, on electron is up (towards P). Electrons accumulate at P, leaving Q positive. Emf direction Q to P. Current direction clockwise. Or use flux. As it moves from $x>b$ to $x

Sketches (vs x):

• $\Phi_B$: Rises linearly from 0 to Blb (0 to b), then constant at Blb (b to 2b). Decreases linearly from Blb to 0 (b to 0).
• $\epsilon$: Constant negative -Blv (0 to b), then 0 (b to 2b). Constant positive Blv (b to 0).
• $I$: Constant negative -Blv/r (0 to b), then 0 (b to 2b). Constant positive Blv/r (b to 0).
• $F$: Constant positive $B^2l^2v/r$ (0 to b), then 0 (b to 2b). Constant positive $B^2l^2v/r$ (b to 0).
• $P_J$: Constant positive $B^2l^2v^2/r$ (0 to b), then 0 (b to 2b). Constant positive $B^2l^2v^2/r$ (b to 0).

The text's Fig. 6.12(b) shows sketches for forward motion (0 to 2b) only. The plots for induced emf, current, force, and power are non-zero only in the region where the magnetic field is present and changing (0 to b), then zero (b to 2b). The values are constant in the 0 to b region because the velocity and field are uniform. The plots show constant magnitude for emf, current, force, and power in the range 0 to b, and zero thereafter.

For inward motion, the magnitude plots would be the same, but emf and current would be positive in the region 0 to b.

Let's provide the sketches for both forward and inward motion (from $x=2b$ back to $x=0$). Let the length of PQ be $l$ and its resistance be $r$. Magnetic field $B$ is uniform from $x=0$ to $x=b$, and zero for $x>b$. Velocity is constant speed $v$. Assume outward velocity is $+v$ and inward velocity is $-v$ (i.e., $dx/dt = v$ for forward, $dx/dt = -v$ for inward). For consistent emf/current direction, use flux change. Outward motion increases inward flux (negative emf/current). Inward motion decreases inward flux (positive emf/current).

Forward motion ($x$: 0 $\to$ 2b):

• Flux $\Phi_B$: $Blx$ for $0 \le x \le b$, $Blb$ for $b < x \le 2b$. (Increases linearly then constant)
• Emf $\epsilon = -d\Phi_B/dt = -Bl(dx/dt) = -Blv$ for $0 \le x < b$. $\epsilon = 0$ for $b < x \le 2b$. (Constant negative then zero)
• Current $I = \epsilon/r = -Blv/r$ for $0 \le x < b$. $I = 0$ for $b < x \le 2b$. (Constant negative then zero)
• Force $F_{ext}$ (magnitude): $B^2l^2v/r$ for $0 \le x < b$. $F_{ext} = 0$ for $b < x \le 2b$. (Constant positive then zero)
• Power $P_J = P_{mech}$: $B^2l^2v^2/r$ for $0 \le x < b$. $P_J = 0$ for $b < x \le 2b$. (Constant positive then zero)

Inward motion ($x$: 2b $\to$ 0):

• Flux $\Phi_B$: $Blb$ for $b \le x \le 2b$, $Blx$ for $0 \le x < b$. (Constant then decreases linearly)
• Emf $\epsilon = -d\Phi_B/dt = -Bl(dx/dt) = -Bl(-v) = Blv$ for $0 \le x < b$. $\epsilon = 0$ for $b \le x \le 2b$. (Zero then constant positive)
• Current $I = \epsilon/r = Blv/r$ for $0 \le x < b$. $I = 0$ for $b \le x \le 2b$. (Zero then constant positive)
• Force $F_{ext}$: $B^2l^2v/r$ for $0 \le x < b$. $F_{ext} = 0$ for $b \le x \le 2b$. (Constant positive then zero)
• Power $P_J = P_{mech}$: $B^2l^2v^2/r$ for $0 \le x < b$. $P_J = 0$ for $b \le x \le 2b$. (Constant positive then zero)

Sketches (plots vs $x$ from 0 to 2b, then 2b to 0, representing forward and backward travel):

Flux ($\Phi_B$):

Induced Emf ($\epsilon$):

Induced Current ($I$):

Force ($F_{ext}$):

Power ($P_J$):

Given: Inner coil radius $r_1$, outer coil radius $r_2$, $r_1 \ll r_2$. Coils are concentric and co-axial.

Let's find the mutual inductance $M_{12}$ (flux linkage with coil 1 due to current in coil 2). Assume current $I_2$ flows in the outer coil (radius $r_2$). The magnetic field produced by the outer coil at its center is $B_2 = \frac{\mu_0 I_2}{2r_2}$. Since the inner coil (radius $r_1$) is much smaller and placed at the center, we can approximate the magnetic field due to the outer coil as uniform over the area of the inner coil, with magnitude $B_2$.

The flux through the inner coil (coil 1) is $\Phi_1 = B_2 \times (\text{Area of coil 1}) = \frac{\mu_0 I_2}{2r_2} \times (\pi r_1^2)$.

Assuming the inner coil has $N_1$ turns, the flux linkage with coil 1 is $N_1\Phi_1 = N_1 \frac{\mu_0 I_2 \pi r_1^2}{2r_2}$.

By definition, $N_1\Phi_1 = M_{12} I_2$.

$M_{12} I_2 = N_1 \frac{\mu_0 I_2 \pi r_1^2}{2r_2}$.

$M_{12} = \frac{\mu_0 N_1 \pi r_1^2}{2r_2}$.

By equality of mutual inductances, $M_{21} = M_{12}$. We can also calculate $M_{21}$ (flux linkage with coil 2 due to current in coil 1). Assume current $I_1$ flows in the inner coil (radius $r_1$, $N_1$ turns). The magnetic field produced by the inner coil varies with distance. Calculating the flux linkage with the outer coil (radius $r_2$) directly would be complex. However, knowing $M_{12} = M_{21}$, we have $M_{21} = \frac{\mu_0 N_1 \pi r_1^2}{2r_2}$. If the outer coil has $N_2$ turns, the definition $N_2\Phi_2 = M_{21} I_1$ means the flux linkage is $N_2 \times (\text{average flux through one turn of coil 2})$. The average flux through one turn of coil 2 due to $I_1$ in coil 1 is $\Phi_2 = M_{21} I_1 / N_2$. Using the formula with $N_1$ and $N_2$ turns, $M = \mu_0 n_1 n_2 \pi r_1^2 l$ where $n_1=N_1/l_1, n_2=N_2/l_2$. If we assume both are thin loops, $M = \frac{\mu_0 N_1 N_2 \pi r_1^2}{2 r_2}$. Yes, this formula holds if we consider the entire coils.

The mutual inductance is $M = \frac{\mu_0 N_1 \pi r_1^2}{2r_2}$.

(a) Magnetic energy in a solenoid: The magnetic energy stored in an inductor (like a solenoid) is $U_B = \frac{1}{2} LI^2$. We need to express this in terms of $B, A,$ and $l$. For a long solenoid, $B = \mu_0 n I$ (assuming air core, $\mu_r=1$). Also $L = \mu_0 n^2 Al$. The number of turns per unit length is $n$. Total length is $l$, area is $A$. So total turns $N = nl$. From $B = \mu_0 n I$, current $I = B/(\mu_0 n)$. Substitute into the energy formula: $U_B = \frac{1}{2} L I^2 = \frac{1}{2} (\mu_0 n^2 Al) \left(\frac{B}{\mu_0 n}\right)^2 = \frac{1}{2} \mu_0 n^2 Al \frac{B^2}{\mu_0^2 n^2} = \frac{1}{2} \frac{\mu_0 n^2 Al B^2}{\mu_0^2 n^2} = \frac{1}{2} \frac{Al B^2}{\mu_0}$.

The volume of the solenoid is $V = Al$. So the magnetic energy density (energy per unit volume) is $u_B = U_B / V = \frac{1}{2} \frac{B^2}{\mu_0}$.

Total magnetic energy stored in the solenoid is $\mathbf{U_B = \frac{1}{2} \frac{B^2}{\mu_0} Al}$.

(b) Comparison with electrostatic energy in a capacitor: The electrostatic energy stored in a capacitor is $U_E = \frac{1}{2} CV^2 = \frac{1}{2} \frac{Q^2}{C} = \frac{1}{2} QV$. For a parallel plate capacitor, the energy density of the electric field is $u_E = \frac{1}{2} \epsilon_0 E^2$. Total energy is $U_E = u_E \times V = \frac{1}{2} \epsilon_0 E^2 V$ (where V is the volume). Comparing the energy density formulas: Magnetic energy density: $u_B = \frac{1}{2} \frac{B^2}{\mu_0}$ Electrostatic energy density: $u_E = \frac{1}{2} \epsilon_0 E^2$

In both cases, the energy density is proportional to the square of the respective field strength. They involve the fundamental constants related to the medium's ability to store electric ($\epsilon_0$) or magnetic ($\mu_0$) energy.

Given: Number of turns $N = 100$. Area of coil $A = 0.10 \text{ m}^2$. Frequency of rotation $f = 0.5 \text{ rev/s}$. Uniform magnetic field $B = 0.01 \text{ T}$. The field is perpendicular to the axis of rotation, which is the standard setup for an AC generator.

The induced emf in the coil is given by $\epsilon(t) = NBA\omega \sin(\omega t)$, where $\omega = 2\pi f$. The maximum voltage generated is the amplitude of this alternating emf, $\epsilon_0 = NBA\omega$.

Calculate $\omega$: $\omega = 2\pi f = 2\pi (0.5 \text{ s}^{-1}) = \pi \text{ rad/s}$.

Calculate the maximum voltage $\epsilon_0$: $\epsilon_0 = N B A \omega$.

$\epsilon_0 = (100) \times (0.01 \text{ T}) \times (0.10 \text{ m}^2) \times (\pi \text{ rad/s})$.

$\epsilon_0 = 100 \times 0.01 \times 0.1 \times \pi \text{ V} = 1 \times 0.1 \times \pi \text{ V} = 0.1\pi \text{ V}$.

Using $\pi \approx 3.14$: $\epsilon_0 \approx 0.1 \times 3.14 \text{ V} = 0.314 \text{ V}$.

The text's answer $0.314$ V matches this calculation.

The maximum voltage generated in the coil is approximately 0.314 V.