Understand Induction - IB Physics

1. Two concentric circular loops have radii of 0.08 m and 0.12 m respectively. The magnetic field perpendicular to both loops varies as $B(t) = 0.20\cos(100t)$ T. What is the ratio of the induced EMF magnitudes in the larger loop to the smaller loop?
   Answer: 2.25 because the EMF depends on the area, which scales as radius squared When you encounter electromagnetic induction problems with changing magnetic fields, always think about Faraday's law: the induced EMF equals the negative rate of change of magnetic flux through the loop. The magnetic flux through each loop is $\Phi = B \cdot A = B(t) \cdot \pi r^2$. Since the same magnetic field passes through both concentric loops, the induced EMF becomes $|\varepsilon| = \left|\frac{d\Phi}{dt}\right| = \pi r^2 \left|\frac{dB}{dt}\right|$. For $B(t) = 0.20\cos(100t)$, we get $\frac{dB}{dt} = -20\sin(100t)$, so $|\varepsilon| = 20\pi r^2 |\sin(100t)|$. The key insight is that the EMF magnitude is proportional to the area of each loop, which scales as $r^2$. The ratio of EMF magnitudes is therefore: $\frac{|\varepsilon_{large}|}{|\varepsilon_{small}|} = \frac{r_{large}^2}{r_{small}^2} = \frac{(0.12)^2}{(0.08)^2} = \frac{0.0144}{0.0064} = 2.25$ Answer C correctly identifies that EMF depends on area, which scales as radius squared. Answer A incorrectly suggests linear proportionality to radius, missing the area relationship. Answer B mentions "different rates of flux change," but both loops experience the same $\frac{dB}{dt}$ since they're in the same field. Answer D references "field gradient" effects that don't exist in this uniform field scenario. Remember: in electromagnetic induction, when the magnetic field is uniform across different loop sizes, the induced EMF ratio equals the area ratio, not the radius ratio.
2. A square conducting loop of side 0.20 m and resistance 0.10 $\Omega$ is in a magnetic field directed perpendicular to its plane. The magnetic field strength decreases linearly from 0.50 T to 0 T in 4.0 s. What is the magnitude of the induced current in the loop?
   Answer: 0.050 A First, calculate the area $A = (0.20 \text{ m})^2 = 0.040 \text{ m}^2$. The rate of change of the magnetic field is $\frac{\Delta B}{\Delta t} = \frac{0.50 \text{ T}}{4.0 \text{ s}} = 0.125 \text{ T/s}$. The magnitude of the induced emf is $|\varepsilon| = |-\frac{\Delta\Phi}{\Delta t}| = |A \frac{\Delta B}{\Delta t}| = 0.040 \text{ m}^2 \times 0.125 \text{ T/s} = 0.0050 \text{ V}$. The induced current is $I = \frac{\varepsilon}{R} = \frac{0.0050 \text{ V}}{0.10 \Omega} = 0.050 \text{ A}$.
3. A long solenoid with 800 turns per meter carries a current that varies according to $I(t) = 2.0e^{-0.5t}$ A. A small circular coil with 50 turns and radius 0.03 m is placed inside the solenoid with its axis parallel to the solenoid axis. What is the magnitude of the induced EMF in the small coil at t = 2.0 s?
   Answer: 4.2 × 10⁻⁵ V accounting for the exponential decay rate properly When you encounter electromagnetic induction problems involving time-varying currents, you need to apply Faraday's law: the induced EMF equals the negative rate of change of magnetic flux. The key insight is recognizing that you must differentiate the current function with respect to time. The magnetic field inside a long solenoid is $B = \mu_0 n I$, where $n = 800$ turns/m is the turn density. Since the small coil is entirely within the uniform field, the flux through it is $\Phi = B \cdot A = \mu_0 n I \cdot N \cdot \pi r^2$, where $N = 50$ turns. By Faraday's law: $|\varepsilon| = \left|\frac{d\Phi}{dt}\right| = \mu_0 n N \pi r^2 \left|\frac{dI}{dt}\right|$ Given $I(t) = 2.0e^{-0.5t}$, we find $\frac{dI}{dt} = 2.0 \times (-0.5) \times e^{-0.5t} = -1.0e^{-0.5t}$ At $t = 2.0$ s: $\left|\frac{dI}{dt}\right| = 1.0e^{-1.0} = 0.368$ A/s Substituting: $|\varepsilon| = (4\pi \times 10^{-7})(800)(50)(\pi)(0.03)^2(0.368) = 4.2 \times 10^{-5}$ V Answer D correctly accounts for the exponential decay rate by properly differentiating the current function. Answer A incorrectly uses the instantaneous current value instead of its derivative. Answer B mentions mutual inductance, which is irrelevant here since we're dealing with flux linkage, not self-inductance. Answer C refers to a non-existent "magnetic field gradient" — the field inside a long solenoid is uniform. Remember: for induction problems, always differentiate the time-dependent quantity to find the rate of change that produces the EMF.
4. A bar magnet is dropped, north pole first, through a horizontal conducting ring. Let $a_1$ be the magnitude of the magnet's acceleration as it approaches the ring and $a_2$ be the magnitude of its acceleration as it moves away from the ring after passing through. Assume air resistance is negligible and let $g$ be the acceleration of free fall.
   Answer: $a_1 < g$ and $a_2 < g$ According to Lenz's law, the induced current creates a magnetic field that opposes the change in flux. As the north pole approaches, the ring induces a north pole on its top face to repel the magnet, creating an upward magnetic force. This makes the net downward force less than the gravitational force $mg$, so $a_1 < g$. As the magnet passes through and the north pole moves away, the flux is decreasing. The ring induces a south pole on its top face to attract the magnet, again creating an upward magnetic force. This makes the net downward force less than $mg$, so $a_2 < g$.
5. A conducting square loop is completely inside a uniform magnetic field directed out of the page. The loop is then pulled to the right with a constant velocity, remaining completely within the field region. What are the induced current and the net magnetic force on the loop?
   Answer: No induced current, no net magnetic force. Since the loop is moving entirely within a uniform magnetic field, the magnetic flux $\Phi = BA$ through the loop remains constant. According to Faraday's law of induction, an emf is induced only when the magnetic flux changes. Because $\Delta\Phi/\Delta t = 0$, the induced emf is zero, and therefore the induced current is zero. With no induced current, there is no magnetic force $F=BIL$ acting on the sides of the loop due to induction. Thus, the net magnetic force is zero.
6. A rectangular coil of 100 turns has dimensions 5.0 cm by 10.0 cm. It rotates at a frequency of 60 Hz in a uniform magnetic field of 0.25 T. What is the approximate peak emf generated?
   Answer: 47 V The peak emf is $\varepsilon_0 = NAB\omega$. First, convert units. Area $A = (0.050 \text{ m}) \times (0.10 \text{ m}) = 0.0050 \text{ m}^2$. Angular velocity $\omega = 2\pi f = 2\pi(60 \text{ Hz}) \approx 377 \text{ rad s}^{-1}$. Now, calculate $\varepsilon_0 = (100)(0.0050 \text{ m}^2)(0.25 \text{ T})(377 \text{ rad s}^{-1}) \approx 47.1 \text{ V}$.
7. A 500-turn coil experiences a magnetic flux of $2.0 \times 10^{-3}$ Wb. The direction of the magnetic field is then reversed over a time interval of 0.10 s. What is the magnitude of the average induced emf in the coil?
   Answer: 20 V The initial flux is $\Phi_i = 2.0 \times 10^{-3}$ Wb. When the field is reversed, the final flux is $\Phi_f = -2.0 \times 10^{-3}$ Wb. The change in flux is $\Delta\Phi = \Phi_f - \Phi_i = -4.0 \times 10^{-3}$ Wb. According to Faraday's law, the magnitude of the average induced emf is $|\varepsilon| = |-N \frac{\Delta\Phi}{\Delta t}| = | -500 \times \frac{-4.0 \times 10^{-3} \text{ Wb}}{0.10 \text{ s}} | = 500 \times \frac{4.0 \times 10^{-3}}{0.10} = 20$ V.
8. A conducting loop is moved with constant velocity out of a region of uniform magnetic field. An induced current flows, and an external force is required to maintain the constant velocity. What is the primary reason for this required external force?
   Answer: The magnetic field exerts a force on the induced current that opposes the loop's motion. As the loop exits the field, the changing flux induces a current (Faraday's Law). This current within the magnetic field experiences a magnetic force given by $F=BIL$. According to Lenz's law, this force opposes the motion that causes the induction. To maintain a constant velocity, an external force must be applied to counteract this magnetic braking force. The work done by this external force is converted into electrical energy in the loop.
9. A primary coil is connected to a DC power supply via a switch. A secondary coil is placed near the primary. The switch is initially open. What is observed in the secondary coil at the instant the switch is closed, and then a long time after it is closed?
   Answer: A brief pulse of current, then zero current. When the switch is closed, the current in the primary coil grows from zero, creating a changing magnetic field. This changing magnetic flux through the secondary coil induces a brief emf and a pulse of current. After a long time, the DC current in the primary becomes steady. The magnetic field is now constant, so the magnetic flux through the secondary is also constant. With no change in flux, there is no induced emf, and the current in the secondary coil is zero.
10. A rectangular conducting loop falls under gravity, entering a region of uniform magnetic field directed into the page. When the loop is partially inside the field, it falls at a constant terminal velocity. What is the correct explanation for this?
    Answer: The gravitational force is balanced by the upward magnetic force on the induced current. As the loop enters the field, the increasing downward flux induces a counter-clockwise current (Lenz's Law). The current in the top segment of the loop, which is inside the field, experiences an upward magnetic force ($F=BIL$). This force opposes gravity. As speed increases, the induced current and the upward magnetic force increase. Terminal velocity is reached when this upward magnetic force exactly balances the downward gravitational force, making the net force zero.
11. An ideal transformer has a primary coil with 200 turns and a secondary coil with 50 turns. The primary is connected to an AC source with an rms voltage of 120 V. The secondary is connected to a 10 $\Omega$ resistor. What is the peak power dissipated in the resistor?
    Answer: 180 W First, find the rms voltage in the secondary coil: $V_{s,rms} = V_{p,rms} \times (N_s/N_p) = 120 \text{ V} \times (50/200) = 30 \text{ V}$. Next, find the average power dissipated in the resistor: $P_{avg} = V_{s,rms}^2 / R = (30 \text{ V})^2 / 10 \Omega = 90 \text{ W}$. The power dissipated by a resistor in an AC circuit varies sinusoidally between 0 and a peak value. The average power is half the peak power. Therefore, the peak power is $P_{peak} = 2 \times P_{avg} = 2 \times 90 \text{ W} = 180 \text{ W}$.
12. A conducting rod is moving through a uniform magnetic field. In which orientation of the rod's velocity $\vec{v}$, the magnetic field $\vec{B}$, and the rod's length vector $\vec{L}$ will the induced motional emf be zero?
    Answer: $\vec{v}$ is parallel to $\vec{L}$, and $\vec{B}$ is perpendicular to both. Motional emf is generated by the magnetic force $\vec{F} = q(\vec{v} \times \vec{B})$ pushing charges along the rod's length. The emf is the work done per unit charge, which requires a component of this force to be parallel to $\vec{L}$. If $\vec{v}$ is parallel to $\vec{L}$, the magnetic force $\vec{F}$ is perpendicular to $\vec{v}$ and therefore also perpendicular to $\vec{L}$. The force pushes charges sideways across the rod, not along its length, so no potential difference is established along the rod, and the emf is zero.
13. A square loop of wire with side length $s$ is in a uniform magnetic field $B$. The plane of the loop is perpendicular to the field, and the magnetic flux is $\Phi$. A second square loop has side length $2s$ and is placed in a field of strength $B/2$. The plane of the second loop makes an angle of 60° with the field lines. What is the magnetic flux through the second loop?
    Answer: $2\Phi$ Initial flux: $\Phi = B s^2$ (since the plane is perpendicular to the field). Second loop: area $A' = (2s)^2 = 4s^2$, field strength $B' = B/2$. When the plane makes 60° with the field lines, the angle between the field and the normal to the plane is $\theta = 30°$. New flux: $\Phi' = B' A' \cos(\theta) = (B/2)(4s^2)\cos(30°) = 2Bs^2(\frac{\sqrt{3}}{2}) = \sqrt{3}Bs^2$. Since $\Phi = Bs^2$, we have $\Phi' = \sqrt{3}\Phi$. However, checking the calculation: $\Phi' = (B/2)(4s^2)(1/2) = Bs^2 = \Phi$. Actually, let me recalculate: $\cos(30°) = \sqrt{3}/2$, so $\Phi' = (B/2)(4s^2)(\sqrt{3}/2) = \sqrt{3}Bs^2 = \sqrt{3}\Phi$.
14. A solenoid with 500 turns and cross-sectional area 0.02 m² is placed in a time-varying magnetic field given by $B(t) = 0.3\sin(\omega t)$ T, where the field is parallel to the solenoid axis. If the peak induced EMF is 15 V, what is the angular frequency ω?
    Answer: 50 rad/s The flux through the solenoid is $\Phi = NBA\sin(\omega t) = 500 \times 0.3 \times 0.02 \times \sin(\omega t) = 3\sin(\omega t)$ Wb. The induced EMF is $\varepsilon = -\frac{d\Phi}{dt} = -3\omega\cos(\omega t)$. The peak EMF is $3\omega = 15$ V, so $\omega = 50$ rad/s. Choice A divides by 2 incorrectly. Choice C uses area incorrectly. Choice D confuses frequency with angular frequency.
15. The magnetic flux $\Phi$ through a single-turn coil varies with time $t$ according to the equation $\Phi = kt^2$, where $k$ is a positive constant. The magnitude of the induced emf in the coil is $\varepsilon$. Which statement is correct?
    Answer: $\varepsilon$ increases linearly with time. By Faraday's law of induction, the magnitude of the induced emf is given by $\varepsilon = |-\frac{d\Phi}{dt}|$. Given $\Phi = kt^2$, we differentiate with respect to time to find the emf: $\varepsilon = |-\frac{d}{dt}(kt^2)| = |-2kt| = 2kt$. This shows that the magnitude of the induced emf, $\varepsilon$, is directly proportional to time $t$ and therefore increases linearly with time.
16. A conducting rod of length 0.50 m rotates about one end at 120 rpm in a uniform magnetic field of 0.80 T perpendicular to the plane of rotation. What is the potential difference between the ends of the rod?
    Answer: 0.63 V with the free end positive relative to the pivot Converting to angular velocity: $\omega = 120 \times \frac{2\pi}{60} = 4\pi$ rad/s. For a rotating rod, the induced EMF is $\varepsilon = \frac{1}{2}B\omega L^2 = \frac{1}{2} \times 0.80 \times 4\pi \times (0.50)^2 = 0.63$ V. The free end moves faster and cuts more field lines, becoming positive. Choice B has wrong polarity. Choices C and D omit the factor of 1/2 in the rotation formula.
17. An AC generator consists of a coil with $N$ turns and area $A$ rotating with constant angular velocity $\omega$ in a uniform magnetic field $B$. The peak emf produced is $\varepsilon_0$. What is the new peak emf if the number of turns is doubled, the area is halved, and the angular velocity is doubled?
    Answer: $2\varepsilon_0$ The peak emf of an AC generator is given by the formula $\varepsilon_0 = NAB\omega$. The new parameters are $N' = 2N$, $A' = A/2$, and $\omega' = 2\omega$. The new peak emf $\varepsilon'_0$ is $\varepsilon'_0 = N'A'B\omega' = (2N)(A/2)B(2\omega) = 2(NAB\omega) = 2\varepsilon_0$.
18. A rectangular conducting loop with dimensions 0.20 m × 0.30 m is moving at constant velocity 2.0 m/s perpendicular to a uniform magnetic field of 0.15 T. The loop enters the field region completely in 0.10 s. What is the magnitude of the induced EMF during the time interval when the loop is partially inside the field?
    Answer: 0.090 V The induced EMF is given by $\varepsilon = Blv$, where B = 0.15 T, l is the width of the loop perpendicular to motion (0.30 m), and v = 2.0 m/s. Therefore, $\varepsilon = 0.15 \times 0.30 \times 2.0 = 0.090$ V. Choice A uses the wrong dimension (0.20 m). Choice C incorrectly uses the area flux change method with wrong time. Choice D incorrectly doubles the correct answer.
19. A circular loop of copper wire is in a uniform magnetic field. Which of the following actions will not induce an electromotive force (emf) in the loop?
    Answer: Moving the loop with constant velocity parallel to the magnetic field lines. An emf is induced when there is a change in magnetic flux, $\Phi = BA\cos\theta$. Option A involves changing the angle $\theta$. Option C involves changing the area $A$. Option D involves changing the magnetic field strength $B$. All of these will change the flux and induce an emf. Option B involves moving the loop parallel to the field lines, which keeps $B$, $A$, and $\theta$ constant. Therefore, there is no change in magnetic flux and no induced emf.
20. A flat coil rotates at a constant angular frequency $\omega$ in a uniform magnetic field. The induced emf is $\varepsilon$ and the magnetic flux linkage is $\Phi_N$. Which statement correctly describes the relationship between $\varepsilon$ and $\Phi_N$?
    Answer: $\varepsilon$ is maximum when $\Phi_N$ is zero. The induced emf is the negative rate of change of magnetic flux linkage: $\varepsilon = -d\Phi_N/dt$. The flux linkage is sinusoidal, $\Phi_N(t) = \Phi_{max}\cos(\omega t)$, and the emf is $\varepsilon(t) = \omega \Phi_{max}\sin(\omega t)$. The emf is maximum when $\sin(\omega t) = \pm 1$, which occurs when $\cos(\omega t) = 0$. Therefore, the emf is maximum when the flux linkage is zero. This corresponds to the coil being parallel to the field, cutting the flux lines at the fastest rate.

Start this quiz