Motional emf - Physics 2

1. What is the magnitude of the current through the resistor?
   Answer: $I = 0.375 \text{ A}$, because only the vertical component of $\vec{B}$ (i.e., $B\sin\theta$) contributes to the motional emf, giving $\mathcal{E} = BLv\sin\theta = 1.5 \text{ V}$, and $I = \mathcal{E}/R$. Whenever you see a motional emf problem with a tilted magnetic field, your first instinct should be to identify **which component of $\vec{B}$ is perpendicular to the plane of motion** — because only that component contributes to the flux change through the circuit loop. Here, the rod slides horizontally, so the circuit loop lies entirely in the horizontal plane. The motional emf is $\mathcal{E} = (\vec{v} \times \vec{B}) \cdot \vec{L}$, but a cleaner way to think about it: emf equals the rate of change of magnetic flux through the loop, $\mathcal{E} = \frac{d\Phi}{dt}$. Flux is $\Phi = B_\perp \cdot A$, where $B_\perp$ is the component of $\vec{B}$ perpendicular to the horizontal plane — that's the vertical component, $B\sin\theta$. So: $\mathcal{E} = B\sin\theta \cdot L \cdot v = (2.0)(0.5)(0.5)(3.0) = 1.5 \text{ V}$ $I = \frac{\mathcal{E}}{R} = \frac{1.5}{4.0} = 0.375 \text{ A}$ This confirms A is correct. Choice B incorrectly uses the horizontal component $B\cos\theta$. The horizontal part of $\vec{B}$ lies within the plane of the loop and doesn't contribute to flux change as the area sweeps out. Choice C uses the full magnitude of $\vec{B}$, ignoring that only the perpendicular component drives flux change — a very common trap. Choice D confuses "no work done by gravity" with "no emf," but emf depends on the magnetic force on charges, not gravitational work. Study tip: Always ask yourself, "What is the orientation of the circuit loop?" then use only the component of $\vec{B}$ normal to that loop in your emf calculation.
2. At the instant when the rod is at position $x = x_0$, what is the motional emf?
   Answer: $\mathcal{E} = B_0(1 + \alpha x_0)Lv$, because the motional emf depends on the local value of $B$ at the rod's instantaneous position, and since the field varies only with $x$ (not with $y$), the field is uniform along the rod's length at any given instant, so $\mathcal{E} = BLv$ applies with $B$ evaluated at $x_0$. Whenever you see a motional emf problem with a non-uniform magnetic field, your first move should be to write the fundamental definition: $\mathcal{E} = \int (\vec{v} \times \vec{B}) \cdot d\vec{\ell}$, then carefully evaluate what the field looks like along the rod's length at the instant in question. Here, the rod moves in $\hat{x}$ and is oriented along $\hat{y}$, so $\vec{v} \times \vec{B} = v\hat{x} \times B_0(1+\alpha x)\hat{z} = -B_0(1+\alpha x)v\,\hat{y}$. The critical observation is that $B$ depends only on $x$, not on $y$. At the instant the rod sits at position $x = x_0$, every point along the rod shares the same $x$-coordinate, so the field is perfectly uniform across the rod's length at that moment: $B = B_0(1+\alpha x_0)$. Integrating along $\hat{y}$ from 0 to $L$ gives $\mathcal{E} = B_0(1+\alpha x_0)Lv$, confirming D. A is wrong because it falsely claims the spatially varying part doesn't contribute. The field gradient is in $\hat{x}$, so it doesn't vary along the rod — but its full value at $x_0$, including the $\alpha x_0$ term, absolutely drives the emf. B is wrong because it integrates the $\alpha$-dependence in $\hat{y}$, but $B$ has no $y$-dependence. There is no variation to integrate along the rod. C is wrong because it mixes an instantaneous emf calculation with a time-accumulated correction. Motional emf is evaluated at a single instant — you don't add cumulative history terms. Study tip: Always ask, "Does the field vary along the rod's orientation?" If not, pull $B$ out of the integral evaluated at the rod's current position and use $\mathcal{E} = BLv$ directly.
3. What is the magnitude of the current flowing through the circuit formed by the two rods and the rails?
   Answer: $I = \dfrac{BLv}{2R}$, because the net emf in the closed loop equals the difference of the two motional emfs, $\mathcal{E}_{\text{net}} = BL(2v) - BLv = BLv$, and the two rods act as resistors in series giving total resistance $2R$. When two moving rods form a closed circuit on rails, think of each rod as a battery: a source of motional emf equal to $\mathcal{E} = BLv$. The key question is always how those "batteries" combine — do they add or oppose each other? Since both rods move in the same direction, they drive current around the loop in opposite senses — like two batteries connected positive-to-positive. This means their emfs oppose each other, and the net emf is the difference: $\mathcal{E}_{\text{net}} = BL(2v) - BLv = BLv$. Now for resistance: the current has only one path through the circuit — through Rod 2, along the rails, then through Rod 1, and back. That's a series connection, giving total resistance $2R$. Applying Ohm's Law: $I = \dfrac{BLv}{2R}$, confirming answer B. Answer A contains a subtle error: it correctly identifies the net emf but then misidentifies the rods as being in parallel. The rods are not parallel — there is only a single current loop, so the resistances add in series. Answer C adds the two emfs instead of subtracting them. This would only be correct if the rods moved in opposite directions, driving current the same way around the loop. Answer D wrongly ignores Rod 1's emf entirely and also mishandles the resistance. Both rods contribute emf and both contribute resistance — neither can be dropped from the analysis. Study tip: When multiple sources of emf share a single current loop, always determine whether they aid or oppose each other before computing net emf — this is the most common trap in motional-emf circuit problems.
4. Which evaluation of the student's two claims is correct?
   Answer: Both claims are correct: the motional emf $\mathcal{E} = BLv$ depends only on the rod's kinematics and field, not on circuit resistance; and since the braking force depends on current $I = \mathcal{E}/(R_{\text{rod}}+R_{\text{ext}})$, changing $R_{\text{ext}}$ changes the braking force and hence the terminal velocity under a constant applied force. Whenever you see a rail-and-rod (or any motional EMF) problem, anchor yourself to two separate questions: What generates the EMF? and What determines the current? Keeping these distinct prevents the most common mistakes here. The motional EMF arises purely from the magnetic force on charges in the moving rod: $\mathcal{E} = BLv$. This is a source property — like a battery's voltage — and it does not depend on what's connected to it. The rod's kinematics and the field alone determine it. So the student's first claim is correct, and answer D's first half is right. Now, if the rod moves at constant velocity, the net force on it is zero. That means an external applied force exactly cancels the magnetic braking force $F_{\text{brake}} = BIL$. The induced current is $I = \mathcal{E}/(R_{\text{rod}} + R_{\text{ext}}) = BLv/(R_{\text{rod}} + R_{\text{ext}})$. Because $R_{\text{ext}}$ appears in the denominator, changing it changes $I$, which changes $F_{\text{brake}}$, which changes the velocity at which braking force balances the applied force — i.e., the terminal velocity. The student's second claim is also correct, confirming D. A is wrong because it falsely claims EMF is reduced by back-reaction — that confuses EMF with terminal voltage. B is wrong because the braking current flows through the entire series circuit ($R_{\text{rod}} + R_{\text{ext}}$), not just $R_{\text{rod}}$. C is wrong because it mislabels the terminal voltage across $R_{\text{ext}}$ as the EMF itself — a classic source-vs-load confusion. Study tip: Always distinguish the source EMF ($BLv$, a kinematic quantity) from the terminal voltage across any element. On exam questions, distractors almost always blur this line.
5. What is the magnitude of the emf generated between the center and the rim of the disk?
   Answer: $\mathcal{E} = \dfrac{1}{2}B\omega R^2$, because a radial element $dr$ at distance $r$ from the center moves with speed $\omega r$, contributing $dV = B(\omega r)\,dr$, and integrating from $0$ to $R$ yields $\frac{1}{2}B\omega R^2$. When a conducting disk spins in a magnetic field, you're dealing with motional emf — not Faraday's flux rule. The key insight is that each tiny radial segment of the disk moves through the field and generates a small voltage, and you must integrate those contributions from center to rim. A radial element at distance $r$ from the center moves with speed $v = \omega r$. The motional emf across that element is $d\mathcal{E} = Bv\,dr = B\omega r\,dr$. Integrating from $0$ to $R$: $\mathcal{E} = \int_0^R B\omega r\,dr = B\omega \cdot \frac{R^2}{2} = \frac{1}{2}B\omega R^2$ This confirms A is correct. B is wrong because it treats the entire disk as if every part moves at the rim speed $\omega R$. Only the rim moves that fast — inner portions move slower, so you cannot shortcut the integration by using the maximum speed. C tempts students who reflexively apply Faraday's law ($\mathcal{E} = -d\Phi/dt$). While the flux through the disk is indeed constant, Faraday's law in integral form applies to a fixed circuit loop. Here, the conducting path itself is rotating, so motional emf arises from the magnetic force on charge carriers ($\vec{F} = q\vec{v} \times \vec{B}$), which Faraday's law in this form misses entirely. D incorrectly averages velocity using the centroid location. The centroid gives the geometric center of mass, not a valid substitute for integrating $r$ in an emf calculation. Study tip: Whenever a conductor moves through a magnetic field, reach for motional emf ($d\mathcal{E} = Bv\,dl$) and integrate — don't assume Faraday's flux rule applies directly to rotating or sliding conductors.
6. As the rod slides down with instantaneous speed $v$ (measured along the incline), what is the magnitude of the motional emf induced in the circuit?
   Answer: $\mathcal{E} = BLv\cos\phi$, because the component of the rod's velocity perpendicular to the magnetic field (which is vertical) is the horizontal component $v\cos\phi$, and the horizontal projection of the rod's length is still $L$, giving $\mathcal{E} = B(v\cos\phi)L$. Whenever you see a motional emf problem with a tilted surface and a non-perpendicular magnetic field, your go-to formula is $\mathcal{E} = \int (\vec{v} \times \vec{B}) \cdot d\vec{\ell}$. The key is identifying which component of velocity is perpendicular to both $B$ and the rod. Here, the rod lies horizontally across the incline (perpendicular to the page's slope direction), and $B$ points vertically upward. The rod slides down the incline at speed $v$, which has two components: a horizontal component $v\cos\phi$ and a vertical component $v\sin\phi$. The force on charges in the rod comes from $\vec{v} \times \vec{B}$. Since $B$ is vertical, only the horizontal component of velocity (perpendicular to $B$) produces a cross product with a component along the rod. The vertical component of velocity is parallel to $B$, so it contributes nothing to $\vec{v} \times \vec{B}$. The effective speed driving charges along the rod is therefore $v\cos\phi$. The rod's actual length along the rails is $L$ (the separation is $L$, and the rod lies horizontally with full length $L$), so the emf is $\mathcal{E} = B(v\cos\phi)L$, confirming choice C. Choice A incorrectly uses $v\sin\phi$ (the vertical component), confusing "perpendicular to the incline" with "perpendicular to $B$." Choice B ignores the geometry entirely — the angle always matters when $B$ is not perpendicular to the velocity. Choice D double-counts the cosine by foreshortening the rod's length to $L\cos\phi$, but the rod spans the full rail separation $L$ horizontally — its length is not projected. Your study tip: always decompose velocity relative to the field direction, not relative to the surface. Ask yourself, "which component of $v$ is perpendicular to $B$?" — that's your effective speed for emf calculations.
7. Which of the following correctly expresses the terminal velocity $v_T$?
   Answer: $v_T = \dfrac{F(R_{\text{ext}} + r)}{B^2 L^2}$, because at terminal velocity the net force is zero, so the applied force equals the braking force $F_{\text{brake}} = BIL$, where $I = \mathcal{E}/(R_{\text{ext}}+r) = BLv_T/(R_{\text{ext}}+r)$, yielding $F = B^2L^2v_T/(R_{\text{ext}}+r)$. When a conducting rod moves through a magnetic field, it generates an EMF and experiences a braking force that opposes its motion. Terminal velocity occurs when this magnetic braking force exactly balances the applied force — net force equals zero, so acceleration stops. This force-balance condition is the cleanest path to the answer. At terminal velocity, the rod moves at constant $v_T$, generating EMF $\mathcal{E} = BLv_T$. The circuit contains both the rod's internal resistance $r$ and the external load $R_{\text{ext}}$ in series, so the current is $I = \frac{BLv_T}{R_{\text{ext}} + r}$. The magnetic braking force on the rod is $F_{\text{brake}} = BIL = \frac{B^2L^2v_T}{R_{\text{ext}}+r}$. Setting $F = F_{\text{brake}}$ and solving gives $v_T = \frac{F(R_{\text{ext}}+r)}{B^2L^2}$, confirming B is correct. A is wrong because it ignores the rod's internal resistance $r$, treating the rod as ideal. This means the current formula $I = BLv_T/R_{\text{ext}}$ undercounts the total circuit resistance, giving an artificially low terminal velocity. C introduces a fabricated "impedance-matching factor" — no such correction exists in this DC resistive circuit. The condition $Fv_T = I^2 R_{\text{ext}}$ is also wrong as the terminal condition; total power input equals total dissipation, not just dissipation in $R_{\text{ext}}$. D incorrectly assumes all mechanical power goes into $r$ alone, ignoring $R_{\text{ext}}$ entirely, and uses a wrong current expression. As a study habit, whenever you see "terminal velocity" in an electromagnetic induction problem, immediately write $F_{\text{applied}} = BIL$ and use the full series resistance in the current expression — forgetting internal resistance is the most common trap on this topic.
8. Which expression correctly describes the magnitude of the motional emf induced in the loop at time $t$?
   Answer: $\mathcal{E} = \dfrac{\mu_0 I_0 w v}{2\pi} \left( \dfrac{1}{d + vt} - \dfrac{1}{d + h + vt} \right)$, because the emf arises from the difference in motional emf between the near and far sides of the loop, each at a different distance from the wire, with the loop's position increasing linearly in time. Whenever you see a moving loop near a current-carrying wire, your instinct should be to find the induced emf using either Faraday's law or the motional emf approach — both must agree. The magnetic field from the wire is non-uniform ($B = \frac{\mu_0 I_0}{2\pi r}$), so you must account for how each segment of the loop experiences a different field. The correct answer is A. As the loop moves away, only the two horizontal sides (parallel to the wire, of length $w$) contribute motional emf, because the force on charges in those sides is along the wire direction. The vertical sides move perpendicularly to themselves, so their contributions cancel. The near side sits at distance $d + vt$ from the wire, giving motional emf $\frac{\mu_0 I_0 w v}{2\pi(d+vt)}$. The far side sits at $d + h + vt$, giving a smaller emf in the opposite sense (it opposes the net emf). The net result is their difference, exactly as written in A — and this is consistent with $\mathcal{E} = -d\Phi/dt$ applied to the non-uniform flux through the loop. B is wrong because dropping the far side's contribution is an unjustified approximation — the problem gives no indication that $h \gg d + vt$, so you cannot ignore it. C is wrong in its formula's structure; while $d\Phi/dt$ is the right approach, the actual derivative of the flux through this geometry does not simplify to $wvh/(d+vt)^2$ — that form ignores the logarithmic nature of the flux integral. D is wrong because it mischaracterizes the cancellation: the far side does not cancel with the top/bottom sides; rather, it partially cancels the near side, and that residual difference is the emf. Your takeaway: always track every parallel segment in a moving loop near a non-uniform field source — dropping any segment requires explicit justification.
9. What is the motional emf between the wing tips, and which component of Earth's field is responsible?
   Answer: $\mathcal{E} \approx 0.65 \text{ V}$, produced by the vertical component of Earth's field $B_v = B\sin(60°)$, because only the vertical component of $\vec{B}$ contributes to the force on charges moving horizontally along the wingspan. When solving motional emf problems for a conductor moving through Earth's magnetic field, your first instinct should be to identify **which component of $\vec{B}$ actually exerts a force along the conductor**. The motional emf is $\mathcal{E} = BvL$, but only the field component perpendicular to both the velocity and the wingspan contributes. Here, the airplane moves forward (horizontally) and the wingspan runs left-to-right (also horizontal). The force on charges in the wing comes from $\vec{F} = q\vec{v} \times \vec{B}$. For a force component to push charges along the wingspan, it must point in the wingspan direction — meaning you need the cross product $\vec{v} \times \vec{B}$ to have a component along the wing axis. Only the vertical component of $\vec{B}$ produces a force in the horizontal wing direction when crossed with the forward horizontal velocity. The vertical component is $B_v = B\sin(60°) = (5.0 \times 10^{-5})\sin(60°) \approx 4.33 \times 10^{-5} \text{ T}$, giving $\mathcal{E} = B_v v L = (4.33 \times 10^{-5})(250)(60) \approx 0.65 \text{ V}$. This confirms C is correct. A is wrong because it uses $B\cos(60°)$, the horizontal component. The horizontal field is parallel to the velocity or the wing, so its cross product with $\vec{v}$ cannot drive charges along the span. B is wrong because not all field components contribute equally — only the geometrically correct component matters. D uses $\sin(60°)$ correctly numerically but misidentifies it as the horizontal component; $\sin(\text{dip angle})$ always gives the vertical component. Remember: the dip angle is measured from horizontal, so $\sin(\theta_{dip})$ → vertical, $\cos(\theta_{dip})$ → horizontal. Sketch the geometry before plugging in numbers.

Start this quiz