Faraday's Law - Physics 2

1. What is the magnitude of the induced emf in the loop at $t = 0.20 \text{ s}$?
   Answer: $\mathcal{E} = B_0 \alpha A e^{-\alpha t} = (2.0)(5.0)(0.06) e^{-1.0} \approx 0.22 \text{ V}$, because Faraday's law gives the rate of change of flux through the loop area $A = ab$. Whenever you see a changing magnetic field through a conducting loop, your first instinct should be Faraday's Law: the induced emf equals the negative time derivative of the magnetic flux, $\mathcal{E} = -\frac{d\Phi_B}{dt}$. The magnitude is $\left|\frac{d\Phi_B}{dt}\right|$. Here, the flux through the loop is $\Phi_B = B(t) \cdot A = B_0 e^{-\alpha t} \cdot ab$. Taking the time derivative: $\frac{d\Phi_B}{dt} = -B_0 \alpha e^{-\alpha t} \cdot A$. So the magnitude of the emf is $\mathcal{E} = B_0 \alpha A e^{-\alpha t}$. Plugging in $B_0 = 2.0\ \text{T}$, $\alpha = 5.0\ \text{s}^{-1}$, $A = (0.20)(0.30) = 0.06\ \text{m}^2$, and $t = 0.20\ \text{s}$: $\mathcal{E} = (2.0)(5.0)(0.06)e^{-1.0} \approx 0.22\ \text{V}$. That confirms answer A is correct. Answer B confuses flux itself with emf — the emf is the rate of change of flux, not the instantaneous flux value. This omits the crucial factor of $\alpha$ from differentiation. Answer C drops the exponential entirely, incorrectly assuming $e^{-\alpha t} = 1$ at $t = 0.20\ \text{s}$. That's only valid at $t = 0$, and the problem explicitly asks about $t = 0.20\ \text{s}$. Answer D claims the $\alpha$ factors cancel during differentiation, which is mathematically wrong. Differentiating $e^{-\alpha t}$ brings down a factor of $\alpha$; nothing cancels it. Your key study tip: when differentiating an exponential $e^{-\alpha t}$, always remember it produces a multiplicative factor of $\alpha$ — that factor is physically important and cannot be dropped.
2. At the instant when the plane of the loop is parallel to $\vec{B}$ (i.e., the normal is perpendicular to $\vec{B}$), what is the magnitude of the induced emf?
   Answer: $\mathcal{E} = B_0 \pi R^2 \omega \sin(\omega t)$; at the specified instant $\sin(\omega t) = 1$, giving $\mathcal{E} = (1.5)(\pi)(0.0225)(20) \approx 2.12 \text{ V}$, because this instant corresponds to the maximum rate of change of flux through the loop. When a conducting loop rotates in a uniform magnetic field, the key quantity to track is the magnetic flux and, more importantly, its rate of change. The flux is $\Phi = B_0 \pi R^2 \cos(\omega t)$, and by Faraday's Law, the induced emf is: $\mathcal{E} = -\frac{d\Phi}{dt} = B_0 \pi R^2 \omega \sin(\omega t)$ When the plane of the loop is parallel to $\vec{B}$, the normal vector is perpendicular to $\vec{B}$, meaning $\theta = 90°$ and $\sin(\omega t) = 1$. This is the exact instant when flux is zero but changing most rapidly — like a pendulum at the bottom of its swing, where position is zero but speed is maximum. Plugging in: $\mathcal{E} = (1.5)(\pi)(0.15)^2(20) \approx 2.12 \text{ V}$. This confirms D as correct. A is wrong because it uses $\cos(\omega t)$ instead of $\sin(\omega t)$. The cosine form describes the flux itself, not the emf. The emf is the derivative of flux, which introduces the sine. B correctly identifies the sine formula but then contradicts itself with a fatal misconception: zero flux does not mean zero rate of change of flux. These are independent quantities, and confusing them is a classic trap. C is wrong because the emf in a rotating loop is sinusoidal, not constant. Only its maximum value equals $B_0 \pi R^2 \omega$, achieved at specific instants. Study tip: Always distinguish between the flux ($\cos$) and the emf ($\sin$, its negative derivative). Maximum emf occurs precisely when flux is zero — remembering this prevents the trap in B.
3. At what time $t$ is the induced current in the loop equal to zero?
   Answer: $t = 1/3 \text{ s}$, because the induced current is zero when $dB/dt = 0$, which occurs when the time derivative of the field polynomial vanishes: $6t - 2 = 0 \Rightarrow t = 1/3 \text{ s}$. When a conducting loop sits in a changing magnetic field, Faraday's Law tells you the induced EMF is $\mathcal{E} = -\frac{d\Phi_B}{dt}$. Since the field is perpendicular to the loop, the flux is simply $\Phi_B = B(t) \cdot A$, where $A = \pi r^2$ is constant. The induced current is then $I = \mathcal{E}/R$. The key insight: the current is zero when the EMF is zero, which happens when $d\Phi_B/dt = 0$, i.e., when $dB/dt = 0$ — not when $B$ itself is zero. Taking the derivative of $B(t) = 3t^2 - 2t + 1$ gives $dB/dt = 6t - 2$. Setting this equal to zero: $6t - 2 = 0 \Rightarrow t = \frac{1}{3} \text{ s}$. At this moment, the field is instantaneously not changing, so no EMF is induced and the current is zero. Answer A is correct. Answer B confuses the value of $B(t)$ with the rate of change of $B(t)$. A nonzero (even always-positive) field produces zero EMF if it momentarily stops changing — flux doesn't need to be zero for the current to be zero. Answer C makes the same error: it tries to find when $B(t) = 0$, which is irrelevant here, and even misidentifies where the quadratic minimum occurs ($t = 1/3$, not $2/3$). Answer D is essentially fabricated — there is no physical principle connecting $d^2B/dt^2$ to zero current, and the arithmetic it presents is meaningless. Your takeaway: zero induced current requires zero $dB/dt$, not zero $B$. Always differentiate the flux expression and set that derivative to zero.
4. What is the induced emf when the rod is at position $x = 1.0 \text{ m}$?
   Answer: $\mathcal{E} = B(x) \ell v = B_0(1+kx)\ell v = (0.50)(1 + 2.0)(0.80)(3.0) = 3.6 \text{ V}$, because for a rod moving in a spatially varying field, the motional emf uses the instantaneous local field at the rod's position. When a conducting rod moves through a spatially varying magnetic field, you need to think carefully about which field value drives the emf. The key principle: motional emf is generated by the magnetic force on charges in the rod itself, so it depends on the field at the rod's instantaneous location. The motional emf formula is $\mathcal{E} = B(x)\ell v$, where $B(x)$ is evaluated at the rod's current position. At $x = 1.0\text{ m}$, the field is $B(1.0) = 0.50(1 + 2.0 \cdot 1.0) = 1.5\text{ T}$. Plugging in: $\mathcal{E} = (1.5)(0.80)(3.0) = 3.6\text{ V}$, confirming B. Choice A makes the mistake of using only $B_0$, ignoring that the field at the rod's location is stronger than the background value. The spatial variation absolutely matters — it changes the force on charges in the rod right now. Choice C uses only the varying part $B_0 kx$, wrongly assuming the uniform component contributes nothing. Even a perfectly uniform field drives motional emf; what matters is the total field at the rod, not just its gradient. Choice D applies Faraday's law by differentiating the enclosed flux, but makes an algebra error — it incorrectly factors the derivative, generating a spurious factor of 2. Done correctly, $\frac{d\Phi}{dt} = B_0(1+kx)\ell v$, which reproduces answer B. Study tip: For a moving rod in a spatially varying field, always evaluate $B$ at the rod's current position and use $\mathcal{E} = B(x)\ell v$ directly — Faraday's law and the motional formula must agree when applied correctly.
5. What is the amplitude of the induced current in the inner loop?
   Answer: $I_{1,\text{max}} = \frac{\mu_0 I_0 \omega \pi r_1^2}{2 r_2 R_1} = \frac{(4\pi\times10^{-7})(10)(100\pi)(\pi)(0.02)^2}{2(0.50)(1.0)} \approx 5.0 \times 10^{-6} \text{ A}$, using the field at the center of the outer loop and differentiating the resulting flux through the inner loop. When two loops are coupled magnetically, the key chain of reasoning is: source current → magnetic field → flux through inner loop → EMF via Faraday's law → induced current. Don't shortcut any step. The outer loop produces a magnetic field at its center of $B = \frac{\mu_0 I(t)}{2r_2}$. Since $r_1 \ll r_2$, this field is approximately uniform over the inner loop's area, giving flux $\Phi = B \cdot \pi r_1^2 = \frac{\mu_0 I_0 \sin(\omega t)}{2r_2} \cdot \pi r_1^2$. Faraday's law requires you to differentiate this flux with respect to time: $\mathcal{E} = -\frac{d\Phi}{dt} = -\frac{\mu_0 I_0 \omega \pi r_1^2}{2r_2}\cos(\omega t)$. The amplitude of the induced EMF therefore carries a factor of $\omega$, and dividing by $R_1$ gives the current amplitude in answer A — approximately $5.0 \times 10^{-6}$ A. This is correct. Answer B makes the classic mistake of skipping the time derivative. It uses the peak flux itself (divided by resistance) rather than the peak rate of change of flux. Faraday's law demands differentiation — flux alone does not drive current. Answer C uses $r_2^2$ as the area instead of $r_1^2$. The flux through the inner loop depends on the inner loop's area, not the outer loop's. The outer loop's radius only appears in the denominator of the Biot–Savart field formula. Answer D correctly uses $r_1^2$ in the numerator but then places $r_1$ in the denominator of the field formula, replacing $r_2$. The field is produced by the outer loop, so its radius $r_2$ belongs in the denominator. Your takeaway: whenever a current is sinusoidal, the induced EMF amplitude always picks up a factor of $\omega$ from differentiation. If your final answer lacks $\omega$, you almost certainly forgot Faraday's law requires $d\Phi/dt$, not just $\Phi$.
6. What is the magnitude of the induced emf at the instant when the sliding wire has moved a distance $d = 0.10 \text{ m}$ past the center of the semicircle? (Treat the geometry carefully: the enclosed area is the semicircle plus a rectangle of width $2r$ and height $d$.)
   Answer: $\mathcal{E} = B \cdot (2r) \cdot v = (0.60)(0.50)(2.0) = 0.60 \text{ V}$, because the rate of change of enclosed area equals the length of the moving wire ($2r$) times its velocity, regardless of the fixed semicircular portion. When a conducting loop changes area in a uniform magnetic field, Faraday's Law tells you the induced emf is simply $\mathcal{E} = B \cdot \frac{dA}{dt}$. The key insight is that only the moving part of the boundary contributes to $\frac{dA}{dt}$ — fixed segments sweep no new area. Here, the sliding wire has length $2r = 0.50 \text{ m}$ and moves at $v = 2.0 \text{ m/s}$. The rate at which it sweeps area is just $\frac{dA}{dt} = (2r)(v)$, so: $\mathcal{E} = B \cdot (2r) \cdot v = (0.60)(0.50)(2.0) = 0.60 \text{ V}$ That's answer C — clean and direct. A is wrong because it differentiates the full area expression $\frac{1}{2}\pi r^2 + 2rd$ and keeps a $\pi r$ term. But $\frac{1}{2}\pi r^2$ is constant — the semicircle doesn't move! Its derivative is zero. Only the rectangular term $2rd$ changes with time. B makes the subtle error of using the arc length $\pi r$ of the semicircle as the "effective length." The semicircle is fixed; it contributes nothing to $\frac{dA}{dt}$. The curved boundary is irrelevant to motional emf. D incorrectly treats the wire's length as growing over time to $2(r+d)$. The sliding wire spans the full diameter $2r$ throughout its motion — it doesn't stretch or elongate as it moves outward. Study tip: When computing motional emf from Faraday's Law, always ask which parts of the boundary are actually moving? Only moving segments contribute to $\frac{dA}{dt}$; fixed geometry is a red herring.
7. What is the magnitude of the emf induced in the external rectangular loop?
   Answer: $\mathcal{E} = \mu_0 n (dI/dt) A_s = (4\pi \times 10^{-7})(2000)(3.0)(5.0 \times 10^{-4}) \approx 3.8 \times 10^{-6} \text{ V}$, because the magnetic flux through the external loop equals the field inside the solenoid multiplied by the solenoid's own cross-sectional area, not the loop area. Whenever you see an external loop wrapped around a solenoid, the key question is: where does the magnetic field actually exist? A solenoid confines its field almost entirely to its interior — outside the solenoid, $B \approx 0$. This is the central insight that unlocks the problem. Because the field is zero in the annular region between the solenoid and the external loop, the only flux threading the external loop comes from the field inside the solenoid. The flux through the external loop is therefore $\Phi = B \cdot A_s = \mu_0 n I \cdot A_s$, where $A_s$ is the solenoid's cross-sectional area, not the loop's. Applying Faraday's Law gives $\mathcal{E} = \mu_0 n (dI/dt) A_s = (4\pi \times 10^{-7})(2000)(3.0)(5.0 \times 10^{-4}) \approx 3.8 \times 10^{-6} \text{ V}$, confirming B is correct. A makes the critical error of using $A_L$, as if the magnetic field fills the entire loop area uniformly — it does not, because the field is zero outside the solenoid. C uses only the annular area $(A_L - A_s)$, which is precisely the region where the field is zero, so this area contributes nothing to the flux. D incorrectly applies the solenoid's self-inductance formula (which involves $n^2$) to a mutual-inductance problem; the external loop has only one turn, so no extra factor of $n$ appears. Study tip: Any time a loop surrounds a solenoid, always use the solenoid's area ($A_s$) to compute flux — the loop's size is irrelevant because the field is confined inside the solenoid.
8. Which of the following correctly describes the induced emf and the direction of the induced current in the loop as it enters the field region (before the trailing edge crosses $x = 0$)?
   Answer: $\mathcal{E} = Bav = (2.0)(0.30)(4.0) = 2.4 \text{ V}$; the induced current flows counterclockwise (when viewed from the front), because the flux into the page is increasing and by Lenz's law the induced current must oppose this increase by creating flux out of the page inside the loop. When a conducting loop moves into a uniform magnetic field, two fundamental tools help you analyze the situation: Faraday's law for the magnitude of the emf, and Lenz's law for the current direction. As the loop enters the field, only the leading edge sweeps through new flux. The rate of flux change is $\frac{d\Phi}{dt} = B \cdot a \cdot v$, since the area inside the field grows at rate $av$. Plugging in: $\mathcal{E} = Bav = (2.0)(0.30)(4.0) = 2.4 \text{ V}$. For direction, the flux into the page through the loop is increasing, so by Lenz's law the induced current must oppose that increase — it must generate flux out of the page inside the loop. By the right-hand rule, that means current flows counterclockwise when viewed from the front. This is exactly what C states, making it the correct answer. A is wrong on two counts: the emf formula $Ba^2/v$ is dimensionally incorrect and physically fabricated, and the current direction is wrong. B gets the emf magnitude right but misapplies Lenz's law. Lenz's law says the induced current opposes the change, meaning it creates flux opposing the increase — out of the page — not reinforcing it. Clockwise current (viewed from front) would create flux into the page, which would add to the increasing flux, violating Lenz's law. D also gets the emf right but reaches the wrong current direction. The force on positive charges in the leading edge is $\vec{F} = q\vec{v} \times \vec{B}$: with $\vec{v}$ in $+x$ and $\vec{B}$ in $-z$, the force points in $+y$ (upward), not $-y$ — so current in the leading edge flows upward, driving counterclockwise current, consistent with C. A reliable strategy: always determine current direction using both Lenz's law and the motional force cross product — if they agree, you're on solid ground.
9. Which expression correctly gives the induced emf as a function of time, and what is its maximum value?
   Answer: $\mathcal{E}(t) = NBA\omega \sin(\omega t)$; maximum value $= NBA\omega = (1)(0.80)(0.25)(60\pi) \approx 37.7 \text{ V}$, because when the normal starts parallel to $\vec{B}$, flux is maximum at $t = 0$ and the emf (its negative time derivative) is a sine function. When a conducting loop rotates in a magnetic field, the key is tracking how the flux changes over time and what that implies about the induced emf. Start by writing the flux: if the normal is parallel to $\vec{B}$ at $t = 0$, the angle between them is zero, so $\Phi(t) = NBA\cos(\omega t)$. This gives maximum flux at $t = 0$. Faraday's Law then gives the induced emf as $\mathcal{E} = -\frac{d\Phi}{dt} = NBA\omega\sin(\omega t)$. At $t = 0$, $\sin(0) = 0$, which makes physical sense — when flux is at its peak, it's momentarily not changing, so the emf is zero. The amplitude is $NBA\omega = (1)(0.80)(0.25)(60\pi) \approx 37.7 \text{ V}$. This is precisely what A states, making it the correct answer. B is wrong because it claims the emf is a cosine, meaning it would be maximum at $t = 0$. But that contradicts the setup — with flux already at its maximum at $t = 0$, its rate of change (and therefore the emf) must be zero there, not maximum. C correctly identifies the sine function but makes a critical error: it drops the $\omega$ from the amplitude. Since emf is the time derivative of flux, differentiating introduces a factor of $\omega$. The amplitude is $NBA\omega$, not $NBA$. D uses faulty phase logic, claiming the sine result "can equivalently be written as a cosine" — but a simple phase relabeling changes the physical meaning of the initial condition, which is fixed by the problem. Remember: Always write $\Phi(t)$ first based on the initial condition, then differentiate. The shape of the emf (sine vs. cosine) is determined by whether flux starts at its max or at zero.
10. What is the magnitude of the emf induced in the secondary coil at $t = 0.20 \text{ s}$?
    Answer: $\mathcal{E} = \frac{\mu_0 N N_s A}{C} \cdot \frac{I_0}{\tau} e^{-t/\tau} = \frac{(4\pi\times10^{-7})(500)(20)(2\times10^{-4})}{0.40} \cdot \frac{4.0}{0.10} e^{-2} \approx 3.4 \times 10^{-5} \text{ V}$, using the toroid field formula, the secondary flux linkage, and differentiation of the decaying exponential current. When a time-varying current flows through a primary coil (the toroid), it induces an emf in any secondary coil sharing the same magnetic flux — this is mutual inductance in action. The key formula chain is: toroid field → flux through secondary → differentiate with respect to time. The magnetic field inside a toroid is $B = \frac{\mu_0 N I}{C}$, where $N$ is the primary turn count and $C$ is the mean circumference. The flux through one secondary turn is $\Phi = BA = \frac{\mu_0 N I A}{C}$. Since the secondary has $N_s$ turns, total flux linkage is $\Lambda = N_s \Phi$. Faraday's law gives $\mathcal{E} = -d\Lambda/dt$, and differentiating $I(t) = I_0 e^{-t/\tau}$ yields $dI/dt = -(I_0/\tau)e^{-t/\tau}$. At $t = 0.20\text{ s}$, $e^{-0.20/0.10} = e^{-2}$. Plugging in: $\mathcal{E} = \frac{\mu_0 N N_s A}{C} \cdot \frac{I_0}{\tau} e^{-2} \approx 3.4 \times 10^{-5}\text{ V}$, confirming D. A is wrong because it drops $N = 500$ from the formula entirely — the primary turns are not absorbed elsewhere; they explicitly appear in the toroid field expression. B evaluates the exponential at $t = 0$ instead of $t = 0.20\text{ s}$, ignoring the problem's instruction to find the instantaneous value at a specific time. C incorrectly uses $N^2$ (the self-inductance formula) instead of $N \cdot N_s$ (the mutual inductance formula). Self-inductance and mutual inductance are distinct quantities. A reliable strategy: always build emf calculations from scratch — field → single-turn flux → total linkage → time derivative. Never guess which $N$ to use; trace every factor from its physical origin.

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