Physics 2 Quiz: Motional Emf

What this quiz covers

This quiz focuses on Motional Emf, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $I = \dfrac{BLv}{R}$, because the net emf in the loop equals the difference of the two motional emfs, $\mathcal{E}_{\text{net}} = BL(2v) - BLv = BLv$, and the two rods are in parallel (each provides an independent current path), so the effective resistance is $R/2$, giving $I = 2BLv/R$ total — but each rod carries $BLv/R$.
2. $I = \dfrac{BLv}{2R}$, 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 $2R$. (correct answer)
3. $I = \dfrac{3BLv}{2R}$, because the net emf is the sum of the two motional emfs, $\mathcal{E}_{\text{net}} = BL(2v) + BLv = 3BLv$, and the total resistance is $2R$.
4. $I = \dfrac{2BLv}{R}$, because only the faster rod contributes to the net emf since it dominates the circuit, giving $\mathcal{E}_{\text{net}} = BL(2v) = 2BLv$, and the effective resistance is $R$ (the slower rod's resistance is negligible in comparison).

Question 2

1. $\mathcal{E} = B_0 Lv$, because in a non-uniform field only the uniform baseline component $B_0$ contributes to the motional emf; the spatially varying part $\alpha x_0$ represents a field gradient that produces forces perpendicular to the rod rather than along it, and these do not contribute to the circuit emf.
2. $\mathcal{E} = B_0 L v + \frac{1}{2}\alpha B_0 L^2 v$, because the spatially varying field must be integrated along the rod's length $L$ in the $\hat{y}$ direction, giving an extra $\frac{1}{2}\alpha L$ correction term to the base emf $B_0 Lv$.
3. $\mathcal{E} = B_0(1 + \alpha x_0)Lv + \alpha B_0 v^2 t$, because the field at the rod's location changes as the rod moves, so a correction term $\alpha B_0 v^2 t$ must be added to account for the cumulative change in field experienced by the moving rod.
4. $\mathcal{E} = B_0(1 + \alpha x_0)Lv$, 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 $x_0$. (correct answer)

Question 3

1. $v_T = \dfrac{FR_{\text{ext}}}{B^2 L^2}$, because at terminal velocity the power delivered to the external load equals the mechanical input power, so $Fv_T = I^2 R_{\text{ext}}$, and solving with $I = BLv_T/R_{\text{ext}}$ (treating the rod as ideal) gives this result.
2. $v_T = \dfrac{F(R_{\text{ext}} + r)}{B^2 L^2}$, 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 $F = B^2L^2v_T/(R_{\text{ext}}+r)$. (correct answer)
3. $v_T = \dfrac{F}{B^2 L^2}(R_{\text{ext}} + r)^2 / R_{\text{ext}}$, because the terminal condition requires the electrical power dissipated in $R_{\text{ext}}$ to equal $Fv_T$, and the current through $R_{\text{ext}}$ includes an impedance-matching factor of $(R_{\text{ext}}+r)/R_{\text{ext}}$.
4. $v_T = \dfrac{Fr}{B^2 L^2}$, because at terminal velocity all the mechanical power is dissipated internally in the rod's own resistance $r$, and the condition $Fv_T = I^2 r$ with $I = BLv_T/r$ determines the terminal speed.

Question 4

1. $I = 0.375 \text{ A}$, 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 $I = \mathcal{E}/R$. (correct answer)
2. $I = 0.65 \text{ A}$, because only the horizontal component of $\vec{B}$ (i.e., $B\cos\theta$) contributes to the motional emf, giving $\mathcal{E} = BLv\cos\theta \approx 2.6 \text{ V}$, and $I = \mathcal{E}/R$.
3. $I = 0.75 \text{ A}$, because the full magnitude of $\vec{B}$ contributes to the motional emf regardless of field orientation, giving $\mathcal{E} = BLv = 3.0 \text{ V}$, and $I = \mathcal{E}/R$.
4. $I = 0$, because the rod moves horizontally and the vertical component of $\vec{B}$ is parallel to the gravitational force rather than to the rod's velocity, so no net work is done on the charges and no emf is induced.

Question 5

1. $\mathcal{E} = \dfrac{1}{2}B\omega R^2$, 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 $\frac{1}{2}B\omega R^2$. (correct answer)
2. $\mathcal{E} = B\omega R^2$, because the rim of the disk moves at speed $v = \omega R$ and the emf is simply $\mathcal{E} = BvR = B\omega R^2$, treating the disk as a single conductor of effective length $R$.
3. $\mathcal{E} = 0$, because the magnetic flux through the disk is constant (the disk's area does not change and $B$ is steady), so by Faraday's law no emf is induced.
4. $\mathcal{E} = \dfrac{2}{3}B\omega R^2$, because the average speed of a point on the disk is $\frac{2}{3}\omega R$ (the centroid of a disk is at $\frac{2}{3}R$), and the emf is $\mathcal{E} = B \cdot \frac{2}{3}\omega R \cdot R$.

Question 6

1. $\mathcal{E} \approx 0.375 \text{ V}$, produced by the horizontal component of Earth's field $B_h = B\cos(60°)$, because the horizontal field component is perpendicular to both the velocity and the wingspan, driving charges along the wing.
2. $\mathcal{E} \approx 0.75 \text{ V}$, produced by the full magnitude of Earth's field, because the airplane's horizontal motion means all components of $\vec{B}$ contribute equally to the Lorentz force on wing charges.
3. $\mathcal{E} \approx 0.65 \text{ V}$, 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. (correct answer)
4. $\mathcal{E} \approx 0.65 \text{ V}$, produced by the horizontal component of Earth's field $B_h = B\sin(60°)$, because the dip angle is measured from horizontal, so $\sin(60°)$ gives the horizontal component, which is perpendicular to the aircraft's forward velocity and drives current across the span.

Question 7

1. $\mathcal{E} = BLv\sin\phi$, because only the component of velocity perpendicular to the incline surface (i.e., the vertical component $v\sin\phi$) drives charges along the horizontal rod and contributes to the emf.
2. $\mathcal{E} = BLv$, because the motional emf depends on the rod's speed relative to the field, and since $B$ is uniform, the full incline speed $v$ contributes regardless of the tilt angle.
3. $\mathcal{E} = BLv\cos\phi$, 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 $\mathcal{E} = B(v\cos\phi)L$. (correct answer)
4. $\mathcal{E} = BLv\cos^2\phi$, because the rod has both a foreshortened effective length $L\cos\phi$ (horizontal projection) and a reduced effective velocity $v\cos\phi$, so these two cosine factors multiply to give $BLv\cos^2\phi$.

Question 8

1. Both claims are wrong: the motional emf $BLv$ is only valid in the open-circuit case; when the circuit is closed, back-reaction from the induced current reduces the effective emf, and changing $R_{\text{ext}}$ does not affect terminal velocity because the braking force is independent of the external load.
2. The first claim is correct but the second is wrong: the emf is indeed $BLv$, but terminal velocity is determined solely by $R_{\text{rod}}$ because the rod's internal resistance is the only resistance through which the braking current flows; $R_{\text{ext}}$ affects power delivery to the load but not the braking force.
3. The first claim is wrong and the second is correct: the motional emf depends on the external resistance through $\mathcal{E} = BLv \cdot R_{\text{ext}}/(R_{\text{rod}}+R_{\text{ext}})$ (since only the voltage across the external circuit drives the useful current), and changing $R_{\text{ext}}$ indeed changes the terminal velocity.
4. 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. (correct answer)

Question 9

1. $\mathcal{E} = \dfrac{\mu_0 I_0 w v}{2\pi} \left( \dfrac{1}{d + vt} - \dfrac{1}{d + h + vt} \right)$, 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. (correct answer)
2. $\mathcal{E} = \dfrac{\mu_0 I_0 v}{2\pi (d + vt)}$, because only the near side of the loop contributes a significant motional emf; the far side, being at greater distance, experiences a negligible field and its contribution can be dropped, and the top and bottom sides contribute nothing along the circuit.
3. $\mathcal{E} = \dfrac{\mu_0 I_0 w v h}{2\pi (d + vt)^2}$, because the emf is proportional to the rate of change of flux, which scales as the area $wh$ divided by the square of the distance from the wire.
4. $\mathcal{E} = \dfrac{\mu_0 I_0 w v}{2\pi (d + vt)}$, because the net emf equals the motional emf of the near side alone (evaluated at its current position), since contributions from the top and bottom sides cancel by symmetry and the far side's emf exactly cancels that of the top and bottom sides.