Lenz's Law - Physics 2

During the rotation, at the instant when the loop has turned $45°$ from its initial position, which of the following best describes the induced current in the loop?
Answer: The induced current is at an intermediate (non-zero, non-maximum) value at this instant, and its direction is such that the magnetic torque on the current-carrying loop opposes the rotation, consistent with Lenz's law acting as a braking torque on the rotating loop. Whenever you see a rotating loop in a magnetic field, anchor your thinking to Faraday's law: $\mathcal{E} = -\frac{d\Phi_B}{dt}$ . The EMF — and thus the induced current — depends on the rate of change of flux, not the flux itself. For a loop rotating at angular velocity $\omega$ in a uniform field, the flux is $\Phi = \Phi_0 \cos(\omega t)$ , so the induced EMF is $\mathcal{E} = \Phi_0 \omega \sin(\omega t)$ . At $45°$ , $\sin(45°) = \frac{\sqrt{2}}{2} \approx 0.707$ . This is neither zero nor one, meaning the current sits at an intermediate value — real, non-zero, but not at its peak. The maximum occurs at $90°$ (loop plane parallel to field), and the minimum (zero) occurs at $0°$ (loop plane perpendicular to field). As for direction: by Lenz's law, the induced current creates a magnetic torque that opposes the rotation — a braking effect. This makes D correct. A is wrong because the maximum EMF occurs at $90°$ , not $45°$ . The rate of change of flux is $\sin(\omega t)$ , which peaks at $90°$ , not halfway through. B is wrong and introduces a fabricated concept. There is no "cancellation" between field components — flux change is governed entirely by $\frac{d}{dt}[\Phi_0\cos(\omega t)]$ , with no subtraction between parallel and perpendicular components. C is wrong because it fundamentally misreads Faraday's law. EMF is proportional to the rate of change of flux, never to the total flux itself. Study tip: Always distinguish between the flux function $(\cos)$ and the EMF function $(\sin)$ — they are 90° out of phase, so their maxima occur at completely different moments in the rotation.
At time $t = \pi/(2\omega)$ , the current in the straight wire is at its maximum value $I_0$ and momentarily not changing. Which of the following correctly describes the induced current in the rectangular loop at this instant?
Answer: The induced current in the loop is zero at this instant, because the current in the wire is momentarily constant ( $dI/dt = 0$ ), so the flux through the loop is not changing, giving zero induced EMF by Faraday's law. Whenever you see a question involving electromagnetic induction, your first instinct should be to reach for Faraday's law: $\mathcal{E} = -\frac{d\Phi_B}{dt}$ . The induced EMF — and therefore the induced current — depends entirely on the rate of change of magnetic flux, not on the flux itself. Here, the wire carries $I(t) = I_0\sin(\omega t)$ , so the flux through the loop is proportional to $I(t)$ , which is also sinusoidal. At $t = \pi/(2\omega)$ , the current reaches its peak $I_0$ . This means flux is at its maximum — but a maximum is precisely where the derivative equals zero. Since $\frac{d\Phi_B}{dt} = 0$ at this instant, Faraday's law gives $\mathcal{E} = 0$ , and therefore the induced current is zero. Answer B is correct. Answer A commits the most common induction mistake: confusing large flux with large induced EMF. Faraday's law involves $d\Phi/dt$ , not $\Phi$ itself — flux being large tells you nothing about the induced current. Answer C incorrectly applies concepts from RL circuits with stored energy. The rectangular loop here has no self-sustaining inductance scenario being described; with zero EMF and a resistive loop, Ohm's law gives zero current immediately. Answer D invents a fictional rule. Faraday's law has only one term — $d\Phi_B/dt$ — and there is no "weighted combination" of flux and its derivative. Study tip: Treat induction questions like calculus: it's always the derivative of flux that matters, not the value. When flux is at a maximum or minimum, the induced EMF is zero — just like a function's slope at its peak.
As the north pole of the magnet approaches the ring from above, what is the direction of the induced current as seen by the observer above, and what happens to that direction immediately after the magnet passes through the plane of the ring and continues downward?
Answer: Counterclockwise as the north pole approaches; the current reverses to clockwise after the magnet passes through the plane of the ring, because the downward flux that was increasing now begins to decrease as the magnet recedes below the ring. Whenever you see a question combining Lenz's Law with a moving magnet, your job is to track how the magnetic flux through the loop is changing — not just what direction the field points. Here, the north pole points downward ( $-z$ ), so field lines exit the north pole and pass downward through the ring, meaning flux is negative by the right-hand rule (into the page for an observer above). As the magnet approaches, this downward flux is increasing in magnitude. By Lenz's Law, the induced current must oppose this increase — so it creates flux upward (out of the page) through the ring. Using the right-hand rule, an upward field through the ring corresponds to a counterclockwise current as seen from above. Once the magnet passes through the plane of the ring and continues downward, it begins moving away. The downward flux through the ring now decreases in magnitude. To oppose this decrease, the induced current must now create flux downward — meaning the current reverses to clockwise. This makes A correct. B is wrong because it reverses the initial direction — the north pole pointing down creates downward flux (not upward), so the opposing current is counterclockwise, not clockwise. C is wrong because the reversal is not gradual; the flux direction of change flips the instant the magnet crosses the plane, so the current reverses immediately. D is wrong on both counts — the initial direction is wrong, and the justification about the south pole "dominating" is a fabricated distraction. Study tip: Always anchor Lenz's Law to a two-step process — identify the flux direction, then identify whether it's increasing or decreasing. That tells you everything about the induced current.
As $I(t)$ increases, which of the following correctly describes the induced current directions in the two external loops, as viewed from the end of the solenoid from which the magnetic field points toward the observer?
Answer: Both the inner square loop and the outer circular loop carry induced currents in the clockwise direction; however, the EMF in the outer loop is determined by the flux through the solenoid's cross-section (area $A$ ), not through the full area $\pi\rho^2$ , because the field outside the solenoid is negligible. Whenever you see a question involving Faraday's law and loops inside or outside a solenoid, your first move should be to identify what flux actually threads each loop — that's what determines the induced EMF. For the inner square loop: the solenoid's field $B = \mu_0 n I(t)$ fills its interior, so the flux through the small square is $\Phi = B\ell^2$ . As $I$ increases, this flux increases (pointing toward the observer), so by Lenz's law the induced current must oppose it — generating a field pointing away from the observer inside the loop. Using the right-hand rule, that means a clockwise induced current as viewed from the observer's end. For the outer circular loop: the solenoid's magnetic field is essentially confined to its interior. Outside the solenoid, $B \approx 0$ . However, Faraday's law depends on the total flux through the loop's area, and because the field exists only inside the solenoid, the relevant flux is still $\Phi = B \cdot A$ (the solenoid's cross-section), not $B \cdot \pi\rho^2$ . This flux is increasing, so the outer loop also has an induced EMF and carries a clockwise current (same reasoning: oppose the increasing flux pointing toward you). This makes C correct. A is partially right about directions but wrong in its reasoning — it incorrectly implies both loops experience "increasing flux through their own areas," ignoring that the outer loop's flux is set by $A$ , not $\pi\rho^2$ . B is wrong because it claims no EMF exists in the outer loop — the flux through area $A$ still threads the outer loop's enclosed region and drives a real current. D incorrectly predicts counterclockwise for the outer loop; both loops oppose the same increasing flux, so both run clockwise. Remember: Faraday's law asks for flux through the loop, not flux times the loop's area. When a loop surrounds a solenoid, use the solenoid's area $A$ — the field is zero everywhere else.
In which direction does current flow in the outer loop, and what is the physical reason for this direction?
Answer: Counterclockwise in the outer loop, because the inner loop's decreasing current reduces the upward ( $+z$ ) flux threading the outer loop, so the outer loop's induced current must create upward flux — requiring a counterclockwise current by the right-hand rule. Whenever you see an electromagnetic induction question, your two anchors should be Faraday's law and Lenz's law — in that order. First, identify how the flux is changing; then, determine what direction of induced current would oppose that change. Here, the inner loop carries $I(t) = I_0 e^{-\alpha t}$ , a counterclockwise current that is exponentially decreasing. By the right-hand rule, a counterclockwise current (viewed from above) produces magnetic flux pointing in the $+z$ direction (upward). Because $I(t)$ is decreasing, the upward flux through the system is also decreasing over time. Lenz's law tells you the outer loop will induce a current that opposes this decrease — meaning it must generate its own upward ( $+z$ ) flux. Applying the right-hand rule again, a current flowing counterclockwise in the outer loop produces upward flux. Therefore, the induced current in the outer loop flows counterclockwise, confirming answer C. Choice A reaches the right physical diagnosis — decreasing upward flux — but then applies Lenz's law backwards. A clockwise current would create downward flux, which would worsen the flux loss, not oppose it. Choice B introduces the idea of "mirroring" through mutual inductance, which sounds technical but is physically meaningless here; Lenz's law, not current mirroring, determines direction. Choice D is tempting because its flux formula is roughly correct for the inner loop's self-flux, but it draws the wrong conclusion — the induced current opposes flux loss, so it must match (not blindly "compensate" via a vague formula) the original direction, which happens to be counterclockwise. Your go-to strategy: always ask "is flux increasing or decreasing?" before deciding direction. Draw a quick diagram, apply the right-hand rule twice — once for the source loop, once for the induced loop — and Lenz's law becomes mechanical.
Immediately after the switch is closed, which of the following correctly describes the behavior of the secondary coil according to Lenz's law, and what happens to the secondary current as the primary current approaches its steady-state value?
Answer: The secondary coil develops an induced EMF that drives current in a direction opposing the increase in flux from the primary; as the primary current approaches steady state, the rate of change of flux decreases toward zero, and the secondary induced EMF and current both approach zero — even though a large steady current flows in the primary. Whenever you see a transformer or mutual inductance question, ask yourself two things: what causes the induced EMF, and what happens to that cause over time? Faraday's law tells you that induced EMF depends on the rate of change of flux — $\mathcal{E} = -N\frac{d\Phi}{dt}$ — not the flux itself. Here's the core reasoning for why D is correct. When the switch closes, the primary current rises, so the flux through the shared core is increasing. By Lenz's law, the secondary coil induces an EMF that drives current in a direction opposing that flux increase — meaning the secondary's magnetic effect pushes back against the growing primary flux. As the primary current approaches its steady-state value $I_{ss} = V/r_P$ , the current stops changing, which means $d\Phi/dt \to 0$ . Because EMF depends on the rate of change, not the magnitude of flux, both the secondary EMF and secondary current decay to zero — even though a large, constant flux still threads the core. Answer A contains a seductive but false feedback claim: the step-up voltage ratio is real, but in this open-switch DC scenario, any secondary current cannot sustain itself once $d\Phi/dt = 0$ , and the described feedback loop misapplies Lenz's law at steady state. Answer B makes the classic error of confusing flux magnitude with flux change — EMF tracks $d\Phi/dt$ , so maximum flux corresponds to zero EMF, not maximum. Answer C reverses Lenz's law entirely; the secondary always opposes the change, never reinforces it. A reliable study tip: transformers only work with changing currents. The moment you have DC at steady state, mutual induction stops — that's why transformers run on AC.
Which of the following correctly describes the induced current in the loop as it passes through the region $z_1 < z < z_2$ where the horizontal magnetic field exists?
Answer: No induced current flows in the loop throughout the region $z_1 < z < z_2$ , because the magnetic field is horizontal and the loop's plane is also horizontal, meaning the field lies entirely within the plane of the loop and the flux through the loop is zero regardless of the loop's position or velocity. When tackling electromagnetic induction questions, always start with Faraday's law: $\mathcal{E} = -\frac{d\Phi_B}{dt}$ , where $\Phi_B = \int \vec{B} \cdot d\vec{A}$ . The key is evaluating the flux, not just whether a field exists. Here, the loop is horizontal, meaning its area vector $d\vec{A}$ points vertically. The magnetic field in the region $z_1 < z < z_2$ is horizontal, so $\vec{B} \perp d\vec{A}$ everywhere. This makes $\vec{B} \cdot d\vec{A} = BA\cos(90°) = 0$ . Since the flux is zero — and remains zero throughout the entire region regardless of the loop's speed or position — the rate of change of flux is also zero, meaning no EMF is induced and no current flows. That confirms B is correct. Choice A commits a critical geometric error: it claims the flux is maximized when the field is perpendicular to the loop's axis. In reality, flux is maximized when $\vec{B}$ is parallel to the area vector (i.e., perpendicular to the plane), not when it lies within the plane. Choice C is tempting but wrong — it imagines flux "jumping" at the boundaries, but since the field is always horizontal inside, the flux never becomes nonzero at any point, so there are no boundary spikes either. Choice D misapplies the motional EMF concept: while individual charges do experience $q\vec{v} \times \vec{B}$ forces, these forces push charges radially (inward or outward), not around the loop, so no net circulating EMF results. Your study tip: always draw the geometry. Dot product geometry — the angle between $\vec{B}$ and $\vec{A}$ — determines flux. A field parallel to a surface contributes zero flux, no matter how strong.
As the loop is pulled to the left, which of the following correctly describes the direction of the induced current in the loop and the direction of the magnetic force on the right side of the loop (the side still inside the field)?
Answer: The induced current flows counterclockwise (viewed from the front), and the magnetic force on the right side is directed to the right, opposing the loop's leftward motion in accordance with Lenz's law. Whenever you see a loop being pulled out of a magnetic field, your two-step toolkit is Lenz's law (to find current direction) followed by the **right-hand rule / $\vec{F} = I\vec{L} \times \vec{B}$ ** (to find the force on the current-carrying segment still inside the field). As the loop moves left, the magnetic flux through it — with $\vec{B}$ pointing out of the page — is decreasing. By Lenz's law, the induced current must oppose this decrease by trying to maintain flux out of the page inside the loop. Using the right-hand rule, a current flowing counterclockwise (viewed from the front) produces a magnetic field out of the page through the loop, which is exactly what's needed. That confirms A is correct. Now apply $\vec{F} = I\vec{L} \times \vec{B}$ to the right side of the loop: current flows upward in that segment, $\vec{B}$ points out of the page, so $\vec{F}$ points to the right — opposing the leftward pull, exactly as Lenz's law demands. B is wrong on the first claim: a clockwise current would create flux into the page, which worsens the decrease rather than opposing it. The force direction stated in B happens to be correct, but it's built on a false premise. C gets both parts wrong — clockwise current is incorrect, and a leftward force would aid the motion, violating Lenz's law entirely. D gets the current direction right but then claims the force is leftward. That contradicts Lenz's law; the force on the exiting segment must always resist the motion. A useful memory anchor: Lenz's law is always a braking law — the induced effects always fight the cause. If your force answer would accelerate the loop, something went wrong.
What is the direction of the induced current through the resistor, and which physical effect is primarily responsible for this direction?
Answer: Current flows from the bottom rail to the top rail through the resistor (in the $+y$ direction), because both the increasing circuit area and the increasing field at the rod's location contribute to growing upward flux; by Lenz's law the induced current must oppose this increase by generating downward flux, which determines the current direction. When a conducting rod moves through a non-uniform magnetic field, you need to apply Faraday's law carefully: the induced EMF depends on the total rate of change of magnetic flux through the circuit, not just one contributing factor. Here, the circuit area expands as the rod moves right, and the magnetic field $B(x) = B_0 x$ grows stronger at the rod's advancing position. Both effects increase the upward flux $\Phi = \int_0^x B_0 x' \cdot L \, dx' = \frac{1}{2}B_0 x^2 L$ . By Lenz's law, the induced current must oppose this growing upward flux by generating a downward ( $-z$ ) magnetic field inside the loop. Using the right-hand rule, this requires current to flow counterclockwise when viewed from above: through the rod in the $+y$ direction, along the top rail leftward, and through the resistor from top to bottom rail — wait, let's track carefully. Counterclockwise (top view) means current in the rod goes in $+y$ , returns along the top rail, down through the resistor in the $-y$ direction... Actually, counterclockwise viewed from $+z$ sends current through the resistor from bottom to top ( $+y$ ). This confirms answer A is correct: current flows in the $+y$ direction through the resistor, driven by both the expanding area and the strengthening field. Answer B is wrong because it attributes the effect to the increasing field alone, ignoring the area increase — both contributions are real and inseparable in Faraday's law. Answer C incorrectly dismisses the spatially varying field as irrelevant; the non-uniformity is precisely why the flux integral depends on $x^2$ , not $x$ . Answer D correctly identifies the motional force $q\vec{v}\times\vec{B}$ pushing positive charges in $+y$ , but then reverses the external current direction — if positive charges accumulate at the top of the rod, conventional current exits the top, travels through the external circuit (the resistor) from top to bottom, which is $-y$ . That makes D's reasoning internally contradictory with its conclusion about the resistor direction. Your strategy: always apply Faraday's law globally to the whole circuit rather than reasoning piecemeal about "field alone" or "area alone." When field and area both change, both matter.

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