IB Physics Quiz: Understand Induction

20 questions · exam conditions

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Understand InductionQuestion 1 of 20

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.

a_1 < g

and

a_2 < g

a_1 < g

and

a_2 > g

a_1 > g

and

a_2 < g

a_1 = g

and

a_2 = g

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IB Physics Quiz

IB Physics Quiz: Understand Induction

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

What this quiz covers

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

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.

All questions

Question 1

$a_1 < g$ and $a_2 < g$ (correct answer)
$a_1 < g$ and $a_2 > g$
$a_1 > g$ and $a_2 < g$
$a_1 = g$ and $a_2 = g$

Explanation: 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

Question 2

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?

$\varepsilon_0 / 2$
$\varepsilon_0$
$2\varepsilon_0$ (correct answer)
$4\varepsilon_0$

Explanation: 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

\varepsilon'_0 = N'A'B\omega' = (2N)(A/2)B(2\omega) = 2(NAB\omega) = 2\varepsilon_0

Question 3

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?

The induced current heats the loop, and the external force provides this thermal energy.
The magnetic field exerts a force on the induced current that opposes the loop's motion. (correct answer)
The change in magnetic flux creates an opposing mass, increasing the loop's effective inertia.
The external force is needed to overcome the viscous drag of the magnetic field itself.

Explanation: 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.

Question 4

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 $\Phi_N$ ?

$\varepsilon$ is maximum when $\Phi_N$ is maximum.
$\varepsilon$ is maximum when $\Phi_N$ is zero. (correct answer)
$\varepsilon$ is zero when $\Phi_N$ is zero.
$\varepsilon$ and $\Phi_N$ are always in phase with each other.

Explanation: 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.

Question 5

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?

4.7 V
47 V (correct answer)
94 V
296 V

Explanation: 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}

Question 6

A circuit contains an ideal inductor and a resistor in series with a DC power supply and a switch. When the switch is closed, the current increases from zero. What is the primary role of the inductor in this process?

To increase the total resistance of the circuit over time.
To generate an opposing emf that slows the rate of current increase. (correct answer)
To store charge and prevent it from reaching the resistor instantaneously.
To dissipate energy as heat, limiting the final current value.

Explanation: As the current increases, the magnetic flux created by the current through the inductor's coils also increases. According to Faraday's Law and Lenz's Law, this change in flux induces a 'back emf' across the inductor. This back emf opposes the change in current, meaning it acts against the voltage from the power supply. This opposition is why the current does not rise instantaneously but increases gradually.

Question 7

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?

$\varepsilon$ is constant and non-zero.
$\varepsilon$ is zero.
$\varepsilon$ is proportional to $t^2$ .
$\varepsilon$ increases linearly with time. (correct answer)

Explanation: 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.

Question 8

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?

45 W
90 W
180 W (correct answer)
360 W

Explanation: 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}

Question 9

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?

Clockwise current, net force to the left.
Counter-clockwise current, net force to the right.
No induced current, no net magnetic force. (correct answer)
Counter-clockwise current, no net magnetic force.

Explanation: 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.

Question 10

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?

$\vec{v}$ is perpendicular to $\vec{B}$ , and $\vec{L}$ is perpendicular to both $\vec{v}$ and $\vec{B}$ .
$\vec{v}$ is perpendicular to $\vec{L}$ , and $\vec{B}$ is at 45° to both $\vec{v}$ and $\vec{L}$ .
$\vec{v}$ is parallel to $\vec{L}$ , and $\vec{B}$ is perpendicular to both. (correct answer)
$\vec{v}$ is perpendicular to $\vec{B}$ , and $\vec{L}$ is parallel to $\vec{B}$ .

Explanation: 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.

Question 11

A conducting rod of length $L$ and resistance $R$ slides on frictionless parallel conducting rails that form a closed loop. A uniform magnetic field $B$ is perpendicular to the plane of the loop. The rod is pulled by a constant external force $F$ and accelerates from rest. What is the terminal velocity $v_t$ of the rod?

$F R / (B L)$
$B^2 L^2 / (F R)$
$F B L / R$
$F R / (B^2 L^2)$ (correct answer)

Explanation: As the rod moves at velocity

v

, a motional emf

\varepsilon = BvL

is induced. This drives a current

I = \varepsilon / R = BvL / R

. This current experiences a magnetic drag force

F_m = BIL = B(BvL/R)L = (B^2 L^2 v) / R

that opposes the motion. Terminal velocity

v_t

is reached when the pulling force

F

equals the drag force

F_m

. Setting

F = F_m

, we get

F = (B^2 L^2 v_t) / R

. Solving for

v_t

gives

v_t = F R / (B^2 L^2)

Question 12

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?

0.050 A (correct answer)
0.10 A
0.25 A
0.50 A

Explanation: 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}

Question 13

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?

The gravitational force is balanced by the upward magnetic force on the induced current. (correct answer)
The magnetic flux through the loop becomes constant, so the net force on the loop is zero.
The induced emf becomes equal to the potential difference from the battery powering the field.
The magnetic forces on the top and bottom segments of the loop become equal and opposite.

Explanation: 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.

Question 14

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?

A brief pulse of current, then a steady non-zero current.
A brief pulse of current, then zero current. (correct answer)
Zero current initially, then a steady non-zero current.
A steady non-zero current, which then decays to zero.

Explanation: 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.

Question 15

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?

1.0 V
2.0 V
10 V
20 V (correct answer)

Explanation: 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

Question 16

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?

$\Phi$
$\Phi/2$
$\sqrt{3}\Phi$
$2\Phi$ (correct answer)

Explanation: 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

Question 17

A magnet oscillates with simple harmonic motion of period $T$ along the axis of a stationary coil, inducing a current with maximum value $I_0$ . The oscillation is repeated with the same amplitude but at a period of $T/2$ . What is the new maximum induced current?

$I_0 / 2$
$I_0$
$2 I_0$ (correct answer)
$4 I_0$

Explanation: The induced emf

\varepsilon

is proportional to the rate of change of flux, which is proportional to the velocity of the magnet. In SHM, maximum velocity is

v_{max} = A\omega = A(2\pi/T)

. So,

\varepsilon_{max} \propto 1/T

. Since current

I = \varepsilon/R

, the maximum current

I_0

is also proportional to

1/T

. If the period is halved to

T/2

, the new maximum current

I'_0

will be proportional to

1/(T/2)

, which is twice the original proportionality. Therefore, the new maximum current is

2I_0

Question 18

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?

25 rad/s
50 rad/s (correct answer)
75 rad/s
100 rad/s

Explanation: 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.

Question 19

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?

0.63 V with the free end positive relative to the pivot (correct answer)
0.63 V with the free end negative relative to the pivot
1.26 V with the free end positive relative to the pivot
1.26 V with the free end negative relative to the pivot

Explanation: 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.

Question 20

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?

2.1 × 10⁻⁵ V using the instantaneous current value at t = 2.0 s
1.7 × 10⁻⁴ V including the mutual inductance between solenoid and coil
8.4 × 10⁻⁵ V considering the time-dependent magnetic field gradient
4.2 × 10⁻⁵ V accounting for the exponential decay rate properly (correct answer)

Explanation: 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}

t = 2.0

\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.