Exercises — Lenz's law — opposing induced current

WHAT is changing? Nothing — the magnet is still, so ${\Phi }_B$ is constant, so $\frac{d{\Phi }_B}{dt}=0$. Therefore $\epsilon =-\frac{d{\Phi }_B}{dt}=0$ and the induced current is zero. The trap is thinking "there's a magnet nearby, so there must be current." Lenz responds to change, not to presence. Answer: no current.

Decreasing flux means the loop is losing what it had. To oppose the loss, it tries to keep the flux up, so the induced field points the same way as $\vec{B}$ (props it up). Answer: with $\vec{B}$.

The direction (sense) of the induced current — i.e. the physical meaning of the minus sign in $\epsilon =-\frac{d{\Phi }_B}{dt}$.

Step 1 (external field): into the page (the $\times$ symbols). Why $\times$? An "$\times$" is the tail of an arrow flying away from you — into the page. Step 2 (trend): $B$ increasing ⇒ into-page flux increasing. Step 3 (oppose): to oppose more into-page flux, the induced field inside the loop must point out of the page. Step 4 (Right-hand rule): point your right thumb out of the page; your curled fingers circle counterclockwise. So the induced current is CCW. Answer: counterclockwise.

Into-page flux now falling. To oppose the fall, the loop tries to maintain into-page flux ⇒ induced field points into the page. Right thumb into the page ⇒ fingers circle clockwise. Answer: clockwise — exactly the reverse of L2·Q1, because the trend reversed while the field direction did not. This is the heart of the parent note's first common mistake.

Motional EMF (see Motional EMF and sliding rod): as the rod moves, enclosed area grows, so ${\Phi }_B=BLx$ and $|\epsilon |=\left|\frac{d{\Phi }_B}{dt}\right|=BL\frac{dx}{dt}=BLv=(0.20)(0.50)(4.0)=0.40\text{V}.$ Current: $I=\frac{|\epsilon |}{R}=\frac{0.40}{2.0}=0.20\text{A}.$ Force on the current-carrying rod: $F=BIL=(0.20)(0.20)(0.50)=0.020\text{N}.$ By Lenz this force opposes the motion (drag). Answers: $0.40\text{V}$, $0.20\text{A}$, $0.020\text{N}$.

Mechanical power (you push against the drag $F$ at speed $v$): $P_{\text{mech}}=Fv=\frac{B^2{L}^2{v}^2}{R}.$ Plug in: $P_{\text{mech}}=\frac{(0.20{)}^2(0.50{)}^2(4.0{)}^2}{2.0}=\frac{0.04\times 0.25\times 16}{2.0}=\frac{0.16}{2.0}=0.080\text{W}.$ Electrical power (heat in the resistor): $P_{\text{elec}}=I^{2}R=(0.20{)}^2(2.0)=0.080\text{W}.$ They match. Every joule you push in reappears as resistor heat — this is Conservation of energy wearing the minus sign. Answer: both $0.080\text{W}$, equal.

Why is there drag at all? The falling magnet changes the flux through rings of the tube, inducing eddy currents. By Lenz these oppose the fall — a retarding force. Terminal condition: acceleration is zero, so drag = weight: $k{v}_t=mg\Rightarrow v_t=\frac{mg}{k}=\frac{(0.010)(9.8)}{0.20}=0.49\text{m/s}.$ Answer: $v_t=0.49\text{m/s}$. Faster than this ⇒ drag > weight ⇒ it slows; slower ⇒ weight > drag ⇒ it speeds up. It settles at $v_t$ — the drag is self-correcting but never reverses the fall (parent note's second mistake).

Flux at start ($\theta =0^{\circ }$, $\cos 0=1$): ${\Phi }_i=BA\cos 0^{\circ }=(0.50)(0.020)(1)=0.010\text{Wb}.$ Flux at end ($\theta =90^{\circ }$, $\cos 90^{\circ }=0$): ${\Phi }_f=0$. Average EMF: $|{\epsilon }_{\text{avg}}|=\left|\frac{{\Phi }_f-{\Phi }_i}{\Delta t}\right|=\frac{0.010}{0.10}=0.10\text{V}.$ By Lenz the current flows to oppose the loss of flux, i.e. it tries to keep the field threading the loop. Answer: $0.10\text{V}$.

(a) As the magnet leaves, distance grows, $B$ at the coil falls, so the flux (which pointed into the coil, since field lines leave the N-pole) decreases. (b) To oppose the loss, the coil tries to maintain the inward flux ⇒ its induced field points into the coil toward the magnet ⇒ the near face becomes a south pole (field lines enter S). (c) For the induced field to point into the coil (away from the observer on the magnet's side), the Right-hand rule gives a clockwise current as seen from the magnet. (d) A near-face S-pole facing the departing N-pole attracts it — the coil tries to pull the magnet back to keep the flux. Answers: decrease; south; clockwise; pulls (attracts).

Flux (face-on, $\cos \theta =1$): ${\Phi }_B=B(t)A=(B_0+\alpha t)a^2.$ Rate of change: $\frac{d{\Phi }_B}{dt}=\alpha a^2$ — the constant $B_0$ vanishes (its derivative is zero), only the changing part $\alpha t$ survives. $|\epsilon |=\alpha a^2=(2.0)(0.10{)}^2=0.020\text{V},I=\frac{|\epsilon |}{R}=\frac{0.020}{0.50}=0.040\text{A}.$ Because $\alpha$ is constant, $I$ is constant in time (independent of $t$). The offset $B_0$ and the elapsed time never enter the current. Answer: $I=0.040\text{A}$, time-independent.

Flux $=BA\cos 0^{\circ }=BA$. As the loop translates, $B$ is the same everywhere, $A$ is fixed, and the orientation is fixed, so ${\Phi }_B$ never changes: $\frac{d{\Phi }_B}{dt}=0$. Hence $\epsilon =0$ and no induced current. Motion alone is not enough — you need the flux to change. This is the classic degenerate case that separates real understanding from pattern-matching. Answer: zero. (Contrast: the sliding rod of L2·Q3 has current only because the enclosed area grows — there the loop's boundary is expanding, so flux changes.)

Entering the field: the area of the loop that is inside the field grows, so flux through the loop increases ⇒ the induced current opposes the increase (its field fights the incoming $\vec{B}$) ⇒ one definite sense, say CCW (for $\vec{B}$ out of the page). Exiting the field: now the inside-field area shrinks, flux decreases ⇒ the induced current reverses to oppose the loss (its field props up $\vec{B}$) ⇒ the opposite sense (CW). In both moments the induced force on the loop opposes the motion (drag on entry and on exit), so you must do work both times — Conservation of energy again. Answer: current direction is opposite at exit vs. entry; drag opposes motion in both.

The flux the coil makes through itself was steady; switching off tries to drop it to zero fast, so $\frac{d{\Phi }_B}{dt}$ is large and negative. By Lenz the coil opposes this loss of its own flux, so it drives a self-induced current in the same direction as the original current, trying to keep the flux alive — this is the back-EMF. Because the change is nearly instantaneous, $\left|\frac{d{\Phi }_B}{dt}\right|$ is huge, producing a large EMF that can push current across the tiny air gap of the opening switch — a spark. Answer: current continues in the original direction; the large back-EMF ionizes the gap and sparks.