Physics 2 Quiz: Faradays Law
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Faradays LawQuestion 1 of 10

Two concentric circular loops lie in the same plane. The inner loop has radius r1=0.02 mr_1 = 0.02 \text{ m} and resistance R1=1.0 ΩR_1 = 1.0 \text{ }\Omega. The outer loop has radius r2=0.50 mr_2 = 0.50 \text{ m} and carries a current I(t)=I0sin⁡(ωt)I(t) = I_0 \sin(\omega t) where I0=10 AI_0 = 10 \text{ A} and ω=100π rad/s\omega = 100\pi \text{ rad/s}. Assume r1≪r2r_1 \ll r_2 so the field of the outer loop is approximately uniform over the inner loop's area.

What is the amplitude of the induced current in the inner loop?

I1,max=μ0I0ωπr122r2R1=(4π×10−7)(10)(100π)(π)(0.02)22(0.50)(1.0)≈5.0×10−6 AI_{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.
I1,max=μ0I0πr122r2R1=(4π×10−7)(10)(π)(0.02)22(0.50)(1.0)≈1.6×10−8 AI_{1,\text{max}} = \frac{\mu_0 I_0 \pi r_1^2}{2 r_2 R_1} = \frac{(4\pi\times10^{-7})(10)(\pi)(0.02)^2}{2(0.50)(1.0)} \approx 1.6 \times 10^{-8} \text{ A}, because the amplitude of the induced current equals the peak mutual flux divided by resistance, without including the ω\omega factor from differentiation.
I1,max=μ0I0ωπr222r2R1=(4π×10−7)(10)(100π)(π)(0.50)22(0.50)(1.0)≈9.9×10−3 AI_{1,\text{max}} = \frac{\mu_0 I_0 \omega \pi r_2^2}{2 r_2 R_1} = \frac{(4\pi\times10^{-7})(10)(100\pi)(\pi)(0.50)^2}{2(0.50)(1.0)} \approx 9.9 \times 10^{-3} \text{ A}, because the flux linking the inner loop should be computed using the outer loop's radius as the relevant area.
I1,max=μ0I0ωπr122r1R1=(4π×10−7)(10)(100π)(π)(0.02)22(0.02)(1.0)≈1.2×10−4 AI_{1,\text{max}} = \frac{\mu_0 I_0 \omega \pi r_1^2}{2 r_1 R_1} = \frac{(4\pi\times10^{-7})(10)(100\pi)(\pi)(0.02)^2}{2(0.02)(1.0)} \approx 1.2 \times 10^{-4} \text{ A}, because the denominator of the Biot–Savart center-field formula uses the radius of the loop producing the field, which here is taken to be r1r_1.
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Physics 2 Quiz

Physics 2 Quiz: Faradays Law

Practice Faradays Law in Physics 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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

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Question 1

Two concentric circular loops lie in the same plane. The inner loop has radius r1=0.02 mr_1 = 0.02 \text{ m} and resistance R1=1.0 ΩR_1 = 1.0 \text{ }\Omega. The outer loop has radius r2=0.50 mr_2 = 0.50 \text{ m} and carries a current I(t)=I0sin⁡(ωt)I(t) = I_0 \sin(\omega t) where I0=10 AI_0 = 10 \text{ A} and ω=100π rad/s\omega = 100\pi \text{ rad/s}. Assume r1≪r2r_1 \ll r_2 so the field of the outer loop is approximately uniform over the inner loop's area.

What is the amplitude of the induced current in the inner loop?

  1. I1,max=μ0I0ωπr122r2R1=(4π×10−7)(10)(100π)(π)(0.02)22(0.50)(1.0)≈5.0×10−6 AI_{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. (correct answer)
  2. I1,max=μ0I0πr122r2R1=(4π×10−7)(10)(π)(0.02)22(0.50)(1.0)≈1.6×10−8 AI_{1,\text{max}} = \frac{\mu_0 I_0 \pi r_1^2}{2 r_2 R_1} = \frac{(4\pi\times10^{-7})(10)(\pi)(0.02)^2}{2(0.50)(1.0)} \approx 1.6 \times 10^{-8} \text{ A}, because the amplitude of the induced current equals the peak mutual flux divided by resistance, without including the ω\omega factor from differentiation.
  3. I1,max=μ0I0ωπr222r2R1=(4π×10−7)(10)(100π)(π)(0.50)22(0.50)(1.0)≈9.9×10−3 AI_{1,\text{max}} = \frac{\mu_0 I_0 \omega \pi r_2^2}{2 r_2 R_1} = \frac{(4\pi\times10^{-7})(10)(100\pi)(\pi)(0.50)^2}{2(0.50)(1.0)} \approx 9.9 \times 10^{-3} \text{ A}, because the flux linking the inner loop should be computed using the outer loop's radius as the relevant area.
  4. I1,max=μ0I0ωπr122r1R1=(4π×10−7)(10)(100π)(π)(0.02)22(0.02)(1.0)≈1.2×10−4 AI_{1,\text{max}} = \frac{\mu_0 I_0 \omega \pi r_1^2}{2 r_1 R_1} = \frac{(4\pi\times10^{-7})(10)(100\pi)(\pi)(0.02)^2}{2(0.02)(1.0)} \approx 1.2 \times 10^{-4} \text{ A}, because the denominator of the Biot–Savart center-field formula uses the radius of the loop producing the field, which here is taken to be r1r_1.
Explanation: 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=μ0I(t)2r2B = \frac{\mu_0 I(t)}{2r_2}. Since r1≪r2r_1 \ll r_2, this field is approximately uniform over the inner loop's area, giving flux Φ=B⋅πr12=μ0I0sin⁡(ωt)2r2⋅πr12\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: E=−dΦdt=−μ0I0ωπr122r2cos⁡(ωt)\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 ω\omega, and dividing by R1R_1 gives the current amplitude in answer A — approximately 5.0×10−65.0 \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 r22r_2^2 as the area instead of r12r_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 r12r_1^2 in the numerator but then places r1r_1 in the denominator of the field formula, replacing r2r_2. The field is produced by the outer loop, so its radius r2r_2 belongs in the denominator. Your takeaway: whenever a current is sinusoidal, the induced EMF amplitude always picks up a factor of ω\omega from differentiation. If your final answer lacks ω\omega, you almost certainly forgot Faraday's law requires dΦ/dtd\Phi/dt, not just Φ\Phi.

Question 2

A square loop of side a=0.30 ma = 0.30 \text{ m} and resistance R=3.0 ΩR = 3.0 \text{ }\Omega is partially inside a region of uniform magnetic field B=2.0 TB = 2.0 \text{ T} directed into the page. The field exists only for x>0x > 0; the loop moves in the +x+x direction at constant velocity v=4.0 m/sv = 4.0 \text{ m/s}. At time t=0t = 0, the leading edge of the loop is at x=0x = 0 (just entering the field). The trailing edge is at x=−ax = -a.

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=0x = 0)?

  1. E=Ba2/v=(2.0)(0.09)/(4.0)=0.045 V\mathcal{E} = Ba^2 / v = (2.0)(0.09)/(4.0) = 0.045 \text{ V}; the induced current flows counterclockwise, because the emf is determined by the ratio of the total flux to the loop velocity rather than the rate at which the leading edge sweeps new area.
  2. E=Bav=2.4 V\mathcal{E} = Bav = 2.4 \text{ V}; the induced current flows clockwise (when viewed from the front), because as the loop enters the field the area inside the field region grows, and by Lenz's law the induced current must flow in the same sense as the external field (into the page) to sustain the increasing flux linkage.
  3. E=Bav=(2.0)(0.30)(4.0)=2.4 V\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. (correct answer)
  4. E=Bav=2.4 V\mathcal{E} = Bav = 2.4 \text{ V}; the induced current flows clockwise (when viewed from the front), because applying the right-hand rule to the loop's velocity in the +x+x direction and the field in the −z-z direction shows that the force on positive charges in the leading edge points in the −y-y direction, driving current downward in that edge and clockwise around the loop.
Explanation: 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 dΦdt=B⋅a⋅v\frac{d\Phi}{dt} = B \cdot a \cdot v, since the area inside the field grows at rate avav. Plugging in: E=Bav=(2.0)(0.30)(4.0)=2.4 V\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 Ba2/vBa^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 F⃗=qv⃗×B⃗\vec{F} = q\vec{v} \times \vec{B}: with v⃗\vec{v} in +x+x and B⃗\vec{B} in −z-z, the force points in +y+y (upward), not −y-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.

Question 3

A conducting rod of length ℓ=0.80 m\ell = 0.80 \text{ m} slides along two parallel conducting rails separated by distance ℓ\ell. The rails are connected at one end by a resistor R=5.0 ΩR = 5.0 \text{ }Ω. The entire apparatus lies in a horizontal plane. The magnetic field is directed vertically upward and varies in space as B(x)=B0(1+kx)B(x) = B_0 (1 + kx), where B0=0.50 TB_0 = 0.50 \text{ T}, k=2.0 m−1k = 2.0 \text{ m}^{-1}, and xx is measured from the resistor. The rod moves at constant velocity v=3.0 m/sv = 3.0 \text{ m/s} away from the resistor.

What is the induced emf when the rod is at position x=1.0 mx = 1.0 \text{ m}?

  1. E=B0ℓv=(0.50)(0.80)(3.0)=1.2 V\mathcal{E} = B_0 \ell v = (0.50)(0.80)(3.0) = 1.2 \text{ V}, because the motional emf formula uses the uniform background field B0B_0 and the spatial variation only affects the flux already enclosed, not the rate at which new flux is swept.
  2. E=B(x)ℓv=B0(1+kx)ℓv=(0.50)(1+2.0)(0.80)(3.0)=3.6 V\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. (correct answer)
  3. E=B0kvℓx=(0.50)(2.0)(3.0)(0.80)(1.0)=2.4 V\mathcal{E} = B_0 k v \ell x = (0.50)(2.0)(3.0)(0.80)(1.0) = 2.4 \text{ V}, because only the spatially varying part B0kxB_0 kx of the field contributes to the induced emf; the uniform component B0B_0 produces a constant background flux whose derivative is zero.
  4. E=ddt∫0xB0(1+kx′)ℓ dx′=B0ℓv+B0kℓvx=B0ℓv(1+2kx)=(0.50)(0.80)(3.0)(1+4.0)=6.0 V\mathcal{E} = \frac{d}{dt}\int_0^x B_0(1+kx')\ell\,dx' = B_0 \ell v + B_0 k \ell v x = B_0 \ell v (1 + 2kx) = (0.50)(0.80)(3.0)(1+4.0) = 6.0 \text{ V}, because differentiating the total enclosed flux with respect to time introduces an extra factor from the chain rule applied to the upper limit.
Explanation: 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 E=B(x)ℓv\mathcal{E} = B(x)\ell v, where B(x)B(x) is evaluated at the rod's current position. At x=1.0 mx = 1.0\text{ m}, the field is B(1.0)=0.50(1+2.0⋅1.0)=1.5 TB(1.0) = 0.50(1 + 2.0 \cdot 1.0) = 1.5\text{ T}. Plugging in: E=(1.5)(0.80)(3.0)=3.6 V\mathcal{E} = (1.5)(0.80)(3.0) = 3.6\text{ V}, confirming B. Choice A makes the mistake of using only B0B_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 B0kxB_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, dΦdt=B0(1+kx)ℓv\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 BB at the rod's current position and use E=B(x)ℓv\mathcal{E} = B(x)\ell v directly — Faraday's law and the motional formula must agree when applied correctly.

Question 4

A long solenoid has n=2000 turns/mn = 2000 \text{ turns/m}, cross-sectional area As=5.0×10−4 m2A_s = 5.0 \times 10^{-4} \text{ m}^2, and carries a current that increases at a constant rate dI/dt=3.0 A/sdI/dt = 3.0 \text{ A/s}. A single-turn rectangular loop of area AL=1.2×10−3 m2A_L = 1.2 \times 10^{-3} \text{ m}^2 is coaxially wound around the outside of the solenoid.

What is the magnitude of the emf induced in the external rectangular loop?

  1. E=μ0n(dI/dt)AL=(4π×10−7)(2000)(3.0)(1.2×10−3)≈9.0×10−6 V\mathcal{E} = \mu_0 n (dI/dt) A_L = (4\pi \times 10^{-7})(2000)(3.0)(1.2 \times 10^{-3}) \approx 9.0 \times 10^{-6} \text{ V}, because the field outside a solenoid is uniform and fills the loop area ALA_L.
  2. E=μ0n(dI/dt)As=(4π×10−7)(2000)(3.0)(5.0×10−4)≈3.8×10−6 V\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. (correct answer)
  3. E=μ0n(dI/dt)(AL−As)=(4π×10−7)(2000)(3.0)(7.0×10−4)≈5.3×10−6 V\mathcal{E} = \mu_0 n (dI/dt) (A_L - A_s) = (4\pi \times 10^{-7})(2000)(3.0)(7.0 \times 10^{-4}) \approx 5.3 \times 10^{-6} \text{ V}, because only the annular area between the solenoid and the loop contributes to the changing flux linkage.
  4. E=μ0n2(dI/dt)As=(4π×10−7)(2000)2(3.0)(5.0×10−4)≈7.5×10−3 V\mathcal{E} = \mu_0 n^2 (dI/dt) A_s = (4\pi \times 10^{-7})(2000)^2(3.0)(5.0 \times 10^{-4}) \approx 7.5 \times 10^{-3} \text{ V}, because mutual inductance between the solenoid and the loop involves n2n^2 from the standard solenoid inductance formula.
Explanation: 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≈0B \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 Φ=B⋅As=μ0nI⋅As\Phi = B \cdot A_s = \mu_0 n I \cdot A_s, where AsA_s is the solenoid's cross-sectional area, not the loop's. Applying Faraday's Law gives E=μ0n(dI/dt)As=(4π×10−7)(2000)(3.0)(5.0×10−4)≈3.8×10−6 V\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 ALA_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 (AL−As)(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 n2n^2) to a mutual-inductance problem; the external loop has only one turn, so no extra factor of nn appears. Study tip: Any time a loop surrounds a solenoid, always use the solenoid's area (AsA_s) to compute flux — the loop's size is irrelevant because the field is confined inside the solenoid.

Question 5

A circular conducting loop of radius r=0.10 mr = 0.10 \text{ m} and resistance R=4.0 ΩR = 4.0 \text{ }\Omega is placed in a uniform magnetic field B⃗\vec{B} directed perpendicular to the plane of the loop. The field magnitude varies as B(t)=3t2−2t+1B(t) = 3t^2 - 2t + 1 (in SI units).

At what time tt is the induced current in the loop equal to zero?

  1. t=1/3 st = 1/3 \text{ s}, because the induced current is zero when dB/dt=0dB/dt = 0, which occurs when the time derivative of the field polynomial vanishes: 6t−2=0⇒t=1/3 s6t - 2 = 0 \Rightarrow t = 1/3 \text{ s}. (correct answer)
  2. The induced current is never zero for t>0t > 0, because B(t)=3t2−2t+1B(t) = 3t^2 - 2t + 1 is always positive (its discriminant 4−12<04 - 12 < 0), meaning nonzero flux always threads the loop and therefore a nonzero emf is always present.
  3. t=2/3 st = 2/3 \text{ s}, because the induced current is zero when the total flux through the loop is zero, which requires solving 3t2−2t+1=03t^2 - 2t + 1 = 0; the minimum of this quadratic occurs at t=2/3 st = 2/3 \text{ s}.
  4. t=1/6 st = 1/6 \text{ s}, because the induced current is zero when the magnetic field reaches its instantaneous peak, found by setting d2B/dt2=6=0d^2B/dt^2 = 6 = 0, which has no solution, so the next best estimate uses dB/dt=6t−2dB/dt = 6t - 2 evaluated at half the coefficient: t=2/12=1/6 st = 2/12 = 1/6 \text{ s}.
Explanation: When a conducting loop sits in a changing magnetic field, Faraday's Law tells you the induced EMF is E=−dΦBdt\mathcal{E} = -\frac{d\Phi_B}{dt}. Since the field is perpendicular to the loop, the flux is simply ΦB=B(t)⋅A\Phi_B = B(t) \cdot A, where A=πr2A = \pi r^2 is constant. The induced current is then I=E/RI = \mathcal{E}/R. The key insight: the current is zero when the EMF is zero, which happens when dΦB/dt=0d\Phi_B/dt = 0, i.e., when dB/dt=0dB/dt = 0 — not when BB itself is zero. Taking the derivative of B(t)=3t2−2t+1B(t) = 3t^2 - 2t + 1 gives dB/dt=6t−2dB/dt = 6t - 2. Setting this equal to zero: 6t−2=0⇒t=13 s6t - 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)B(t) with the rate of change of B(t)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)=0B(t) = 0, which is irrelevant here, and even misidentifies where the quadratic minimum occurs (t=1/3t = 1/3, not 2/32/3). Answer D is essentially fabricated — there is no physical principle connecting d2B/dt2d^2B/dt^2 to zero current, and the arithmetic it presents is meaningless. Your takeaway: zero induced current requires zero dB/dtdB/dt, not zero BB. Always differentiate the flux expression and set that derivative to zero.

Question 6

A flat circular loop of radius R=0.15 mR = 0.15 \text{ m} is placed in a region where the magnetic field is uniform and given by B⃗=B0x^\vec{B} = B_0 \hat{x} with B0=1.5 TB_0 = 1.5 \text{ T} (constant in time). The loop is initially in the yzyz-plane. It then rotates about the yy-axis at a constant angular velocity ω=20 rad/s\omega = 20 \text{ rad/s}, so that its normal vector makes an angle θ(t)=ωt\theta(t) = \omega t with x^\hat{x}.

At the instant when the plane of the loop is parallel to B⃗\vec{B} (i.e., the normal is perpendicular to B⃗\vec{B}), what is the magnitude of the induced emf?

  1. E=B0πR2ωcos⁡(ωt)\mathcal{E} = B_0 \pi R^2 \omega \cos(\omega t); at the specified instant cos⁡(ωt)=0\cos(\omega t) = 0, giving E=0 V\mathcal{E} = 0 \text{ V}, because the emf is a cosine function that starts at its maximum when the normal is aligned with B⃗\vec{B} and passes through zero when the plane becomes parallel to B⃗\vec{B}.
  2. E=B0πR2ωsin⁡(ωt)\mathcal{E} = B_0 \pi R^2 \omega \sin(\omega t); at the specified instant sin⁡(ωt)=1\sin(\omega t) = 1, but since the flux through the loop is zero when the plane is parallel to B⃗\vec{B}, the emf is also zero — a zero flux cannot be changing, so E=0 V\mathcal{E} = 0 \text{ V}.
  3. E=B0πR2ω\mathcal{E} = B_0 \pi R^2 \omega; the magnitude equals the maximum possible value ≈2.12 V\approx 2.12 \text{ V} at every instant because the loop rotates at constant ω\omega and the field is uniform, so the emf is constant in time.
  4. E=B0πR2ωsin⁡(ωt)\mathcal{E} = B_0 \pi R^2 \omega \sin(\omega t); at the specified instant sin⁡(ωt)=1\sin(\omega t) = 1, giving E=(1.5)(π)(0.0225)(20)≈2.12 V\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. (correct answer)
Explanation: 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 Φ=B0πR2cos⁡(ωt)\Phi = B_0 \pi R^2 \cos(\omega t), and by Faraday's Law, the induced emf is: E=−dΦdt=B0πR2ωsin⁡(ωt)\mathcal{E} = -\frac{d\Phi}{dt} = B_0 \pi R^2 \omega \sin(\omega t) When the plane of the loop is parallel to B⃗\vec{B}, the normal vector is perpendicular to B⃗\vec{B}, meaning θ=90°\theta = 90° and sin⁡(ωt)=1\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: E=(1.5)(π)(0.15)2(20)≈2.12 V\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⁡(ωt)\cos(\omega t) instead of sin⁡(ωt)\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 B0πR2ωB_0 \pi R^2 \omega, achieved at specific instants. Study tip: Always distinguish between the flux (cos⁡\cos) and the emf (sin⁡\sin, its negative derivative). Maximum emf occurs precisely when flux is zero — remembering this prevents the trap in B.

Question 7

A rectangular conducting loop of dimensions a=0.20 ma = 0.20 \text{ m} by b=0.30 mb = 0.30 \text{ m} lies in the xyxy-plane. The magnetic field in the region is given by B⃗(t)=B0e−αtz^\vec{B}(t) = B_0 e^{-\alpha t} \hat{z}, where B0=2.0 TB_0 = 2.0 \text{ T} and α=5.0 s−1\alpha = 5.0 \text{ s}^{-1}.

What is the magnitude of the induced emf in the loop at t=0.20 st = 0.20 \text{ s}?

  1. E=B0αAe−αt=(2.0)(5.0)(0.06)e−1.0≈0.22 V\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=abA = ab. (correct answer)
  2. E=B0Ae−αt=(2.0)(0.06)e−1.0≈0.044 V\mathcal{E} = B_0 A e^{-\alpha t} = (2.0)(0.06) e^{-1.0} \approx 0.044 \text{ V}, because the induced emf equals the instantaneous flux, not its time derivative.
  3. E=B0αA=(2.0)(5.0)(0.06)=0.60 V\mathcal{E} = B_0 \alpha A = (2.0)(5.0)(0.06) = 0.60 \text{ V}, because the exponential factor equals unity when evaluated correctly at the initial decay rate.
  4. E=B0αAe−αt/α=B0Ae−αt=(2.0)(0.06)e−1.0≈0.044 V\mathcal{E} = B_0 \alpha A e^{-\alpha t} / \alpha = B_0 A e^{-\alpha t} = (2.0)(0.06) e^{-1.0} \approx 0.044 \text{ V}, because the α\alpha factors cancel when computing dΦ/dtd\Phi/dt for an exponential field.
Explanation: 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, E=−dΦBdt\mathcal{E} = -\frac{d\Phi_B}{dt}. The magnitude is ∣dΦBdt∣\left|\frac{d\Phi_B}{dt}\right|. Here, the flux through the loop is ΦB=B(t)⋅A=B0e−αt⋅ab\Phi_B = B(t) \cdot A = B_0 e^{-\alpha t} \cdot ab. Taking the time derivative: dΦBdt=−B0αe−αt⋅A\frac{d\Phi_B}{dt} = -B_0 \alpha e^{-\alpha t} \cdot A. So the magnitude of the emf is E=B0αAe−αt\mathcal{E} = B_0 \alpha A e^{-\alpha t}. Plugging in B0=2.0 TB_0 = 2.0\ \text{T}, α=5.0 s−1\alpha = 5.0\ \text{s}^{-1}, A=(0.20)(0.30)=0.06 m2A = (0.20)(0.30) = 0.06\ \text{m}^2, and t=0.20 st = 0.20\ \text{s}: E=(2.0)(5.0)(0.06)e−1.0≈0.22 V\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 α\alpha from differentiation. Answer C drops the exponential entirely, incorrectly assuming e−αt=1e^{-\alpha t} = 1 at t=0.20 st = 0.20\ \text{s}. That's only valid at t=0t = 0, and the problem explicitly asks about t=0.20 st = 0.20\ \text{s}. Answer D claims the α\alpha factors cancel during differentiation, which is mathematically wrong. Differentiating e−αte^{-\alpha t} brings down a factor of α\alpha; nothing cancels it. Your key study tip: when differentiating an exponential e−αte^{-\alpha t}, always remember it produces a multiplicative factor of α\alpha — that factor is physically important and cannot be dropped.

Question 8

A toroidal solenoid has N=500N = 500 turns, a mean circumference of C=0.40 mC = 0.40 \text{ m}, and a rectangular cross-section of area A=2.0×10−4 m2A = 2.0 \times 10^{-4} \text{ m}^2. The current through the toroid is I(t)=I0e−t/τI(t) = I_0 e^{-t/\tau} with I0=4.0 AI_0 = 4.0 \text{ A} and τ=0.10 s\tau = 0.10 \text{ s}. A secondary coil of Ns=20N_s = 20 turns is wound uniformly over the toroid.

What is the magnitude of the emf induced in the secondary coil at t=0.20 st = 0.20 \text{ s}?

  1. E=μ0NsAC⋅I0τe−t/τ=(4π×10−7)(20)(2×10−4)0.40⋅40 e−2≈6.8×10−8 V\mathcal{E} = \frac{\mu_0 N_s A}{C} \cdot \frac{I_0}{\tau} e^{-t/\tau} = \frac{(4\pi\times10^{-7})(20)(2\times10^{-4})}{0.40} \cdot 40 \, e^{-2} \approx 6.8 \times 10^{-8} \text{ V}, because only the secondary winding turn count NsN_s appears in the flux-linkage formula, as the primary turn count NN is already absorbed into the definition of the toroid's field.
  2. E=μ0NNsAC⋅I0τ=(4π×10−7)(500)(20)(2×10−4)0.40⋅40≈2.5×10−4 V\mathcal{E} = \frac{\mu_0 N N_s A}{C} \cdot \frac{I_0}{\tau} = \frac{(4\pi\times10^{-7})(500)(20)(2\times10^{-4})}{0.40} \cdot 40 \approx 2.5 \times 10^{-4} \text{ V}, because the exponential factor should be evaluated at t=0t = 0 (giving unity) since the question asks for the initial rate of decay, not the instantaneous value at t=0.20 st = 0.20 \text{ s}.
  3. E=μ0N2AC⋅I0τe−t/τ=(4π×10−7)(500)2(2×10−4)0.40⋅40 e−2≈8.5×10−4 V\mathcal{E} = \frac{\mu_0 N^2 A}{C} \cdot \frac{I_0}{\tau} e^{-t/\tau} = \frac{(4\pi\times10^{-7})(500)^2(2\times10^{-4})}{0.40} \cdot 40 \, e^{-2} \approx 8.5 \times 10^{-4} \text{ V}, because the mutual inductance between two coils on a toroid involves N2N^2 from the standard solenoid inductance formula rather than the product NNsN N_s.
  4. E=μ0NNsAC⋅I0τe−t/τ=(4π×10−7)(500)(20)(2×10−4)0.40⋅4.00.10e−2≈3.4×10−5 V\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. (correct answer)
Explanation: 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=μ0NICB = \frac{\mu_0 N I}{C}, where NN is the primary turn count and CC is the mean circumference. The flux through one secondary turn is Φ=BA=μ0NIAC\Phi = BA = \frac{\mu_0 N I A}{C}. Since the secondary has NsN_s turns, total flux linkage is Λ=NsΦ\Lambda = N_s \Phi. Faraday's law gives E=−dΛ/dt\mathcal{E} = -d\Lambda/dt, and differentiating I(t)=I0e−t/τI(t) = I_0 e^{-t/\tau} yields dI/dt=−(I0/τ)e−t/τdI/dt = -(I_0/\tau)e^{-t/\tau}. At t=0.20 st = 0.20\text{ s}, e−0.20/0.10=e−2e^{-0.20/0.10} = e^{-2}. Plugging in: E=μ0NNsAC⋅I0τe−2≈3.4×10−5 V\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=500N = 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=0t = 0 instead of t=0.20 st = 0.20\text{ s}, ignoring the problem's instruction to find the instantaneous value at a specific time. C incorrectly uses N2N^2 (the self-inductance formula) instead of N⋅NsN \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 NN to use; trace every factor from its physical origin.

Question 9

A conducting loop consists of a fixed semicircle of radius r=0.25 mr = 0.25 \text{ m} and a diameter wire that can slide. The diameter wire moves outward along the two straight sides of the semicircle (i.e., along the diameter direction) at speed v=2.0 m/sv = 2.0 \text{ m/s}, increasing the enclosed area. A uniform magnetic field B=0.60 TB = 0.60 \text{ T} is directed perpendicular to the plane of the loop.

What is the magnitude of the induced emf at the instant when the sliding wire has moved a distance d=0.10 md = 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 2r2r and height dd.)

  1. E=B⋅ddt(12πr2+2rd)=B⋅2r⋅v+B⋅πr⋅v=(0.60)(0.50+π×0.25)(2.0)≈1.54 V\mathcal{E} = B \cdot \frac{d}{dt}\left(\frac{1}{2}\pi r^2 + 2rd\right) = B \cdot 2r \cdot v + B \cdot \pi r \cdot v = (0.60)(0.50 + \pi\times0.25)(2.0) \approx 1.54 \text{ V}, because differentiating the total enclosed area requires contributions from both the rectangular and the semicircular portions of the boundary.
  2. E=B⋅πr⋅v=(0.60)(π×0.25)(2.0)≈0.94 V\mathcal{E} = B \cdot \pi r \cdot v = (0.60)(\pi \times 0.25)(2.0) \approx 0.94 \text{ V}, because the relevant length for the motional emf is the arc length of the semicircle (πr\pi r) rather than its diameter, since the semicircle forms the curved boundary of the loop.
  3. E=B⋅(2r)⋅v=(0.60)(0.50)(2.0)=0.60 V\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 (2r2r) times its velocity, regardless of the fixed semicircular portion. (correct answer)
  4. E=B⋅2(r+d)⋅v=(0.60)(2)(0.35)(2.0)=0.84 V\mathcal{E} = B \cdot 2(r + d) \cdot v = (0.60)(2)(0.35)(2.0) = 0.84 \text{ V}, because the effective length of the moving boundary increases as the wire slides outward, so the relevant length is 2(r+d)2(r + d) rather than the fixed diameter 2r2r.
Explanation: When a conducting loop changes area in a uniform magnetic field, Faraday's Law tells you the induced emf is simply E=B⋅dAdt\mathcal{E} = B \cdot \frac{dA}{dt}. The key insight is that only the moving part of the boundary contributes to dAdt\frac{dA}{dt} — fixed segments sweep no new area. Here, the sliding wire has length 2r=0.50 m2r = 0.50 \text{ m} and moves at v=2.0 m/sv = 2.0 \text{ m/s}. The rate at which it sweeps area is just dAdt=(2r)(v)\frac{dA}{dt} = (2r)(v), so: E=B⋅(2r)⋅v=(0.60)(0.50)(2.0)=0.60 V\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 12πr2+2rd\frac{1}{2}\pi r^2 + 2rd and keeps a πr\pi r term. But 12πr2\frac{1}{2}\pi r^2 is constant — the semicircle doesn't move! Its derivative is zero. Only the rectangular term 2rd2rd changes with time. B makes the subtle error of using the arc length πr\pi r of the semicircle as the "effective length." The semicircle is fixed; it contributes nothing to dAdt\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)2(r+d). The sliding wire spans the full diameter 2r2r 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 dAdt\frac{dA}{dt}; fixed geometry is a red herring.

Question 10

A square conducting loop of side L=0.50 mL = 0.50 \text{ m} and total resistance R=2.0 ΩR = 2.0 \text{ }\Omega rotates at angular frequency ω=60π rad/s\omega = 60\pi \text{ rad/s} in a uniform magnetic field B=0.80 TB = 0.80 \text{ T}. At t=0t = 0, the normal to the loop is parallel to the field.

Which expression correctly gives the induced emf as a function of time, and what is its maximum value?

  1. E(t)=NBAωsin⁡(ωt)\mathcal{E}(t) = NBA\omega \sin(\omega t); maximum value =NBAω=(1)(0.80)(0.25)(60π)≈37.7 V= NBA\omega = (1)(0.80)(0.25)(60\pi) \approx 37.7 \text{ V}, because when the normal starts parallel to B⃗\vec{B}, flux is maximum at t=0t = 0 and the emf (its negative time derivative) is a sine function. (correct answer)
  2. E(t)=NBAωcos⁡(ωt)\mathcal{E}(t) = NBA\omega \cos(\omega t); maximum value =NBAω≈37.7 V= NBA\omega \approx 37.7 \text{ V}, because when the normal starts parallel to B⃗\vec{B}, the flux is already at its maximum rate of change at t=0t = 0, making the emf a cosine that is maximum at t=0t = 0.
  3. E(t)=NBAωsin⁡(ωt)\mathcal{E}(t) = NBA\omega \sin(\omega t); maximum value =NBA=(1)(0.80)(0.25)=0.20 V= NBA = (1)(0.80)(0.25) = 0.20 \text{ V}, because the emf is sinusoidal but its amplitude equals the peak flux, not the peak flux multiplied by ω\omega.
  4. E(t)=NBAωcos⁡(ωt)\mathcal{E}(t) = NBA\omega \cos(\omega t); maximum value =NBAω≈37.7 V= NBA\omega \approx 37.7 \text{ V}, because differentiating Φ(t)=NBAcos⁡(ωt)\Phi(t) = NBA\cos(\omega t) gives −dΦ/dt=NBAωsin⁡(ωt)-d\Phi/dt = NBA\omega\sin(\omega t), and since sin⁡\sin and −cos⁡-\cos are phase-shifted by 90°90°, the emf can equivalently be written as a cosine with an appropriate phase offset.
Explanation: 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 B⃗\vec{B} at t=0t = 0, the angle between them is zero, so Φ(t)=NBAcos⁡(ωt)\Phi(t) = NBA\cos(\omega t). This gives maximum flux at t=0t = 0. Faraday's Law then gives the induced emf as E=−dΦdt=NBAωsin⁡(ωt)\mathcal{E} = -\frac{d\Phi}{dt} = NBA\omega\sin(\omega t). At t=0t = 0, sin⁡(0)=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 NBAω=(1)(0.80)(0.25)(60π)≈37.7 VNBA\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=0t = 0. But that contradicts the setup — with flux already at its maximum at t=0t = 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 ω\omega from the amplitude. Since emf is the time derivative of flux, differentiating introduces a factor of ω\omega. The amplitude is NBAωNBA\omega, not NBANBA. 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 Φ(t)\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.