IB Physics Quiz: Apply Induction

20 questions · exam conditions

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Apply InductionQuestion 1 of 20

A straight conductor of length 0.50 m moves at a constant speed of 4.0 m s⁻¹ through a uniform magnetic field of strength 0.80 T. The velocity of the conductor is perpendicular to its length, and the magnetic field is directed at 60° to the velocity. What is the magnitude of the induced emf?

0.80 V

1.4 V

1.6 V

3.2 V

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IB Physics Quiz

IB Physics Quiz: Apply Induction

Practice Apply 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 Apply 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

0.80 V
1.4 V (correct answer)
1.6 V
3.2 V

Explanation: Motional emf is induced by the component of the magnetic field perpendicular to both the length of the conductor and its velocity. The general formula derived from the Lorentz force is ε = BvLsinθ, where θ is the angle between the velocity and the magnetic field. Here, ε = (0.80 T)(4.0 m s⁻¹)(0.50 m)sin(60°) = 1.6 × (√3 / 2) ≈ 1.39 V. The most common error is to use BvL (1.6 V) or BvLcos(60°) (0.80 V).

Question 2

A flexible conducting loop in a uniform magnetic field B (perpendicular to its plane) is stretched so that its radius increases linearly with time, r(t) = kt. How does the magnitude of the induced emf ε in the loop depend on time t?

ε is constant.
ε is proportional to 1/t.
ε is proportional to t².
ε is proportional to t. (correct answer)

Explanation: The area of the loop is A(t) = πr(t)² = π(kt)². The magnetic flux is Φ(t) = BA(t) = Bπk²t². According to Faraday's law, the magnitude of the induced emf is ε = |dΦ/dt|. Taking the derivative with respect to time: ε = d/dt (Bπk²t²) = Bπk²(2t). Therefore, ε = (2Bπk²)t. Since B, π, and k are constants, the induced emf ε is directly proportional to time t.

Question 3

A conducting loop is stationary in a non-uniform magnetic field that is directed into the page and becomes stronger with increasing height. If the loop is moved downwards at a constant velocity, what is the direction of the induced current?

Clockwise (correct answer)
Counter-clockwise
No current is induced.
The direction depends on the speed.

Explanation: The loop is moving downwards into a region where the magnetic field (directed into the page) is weaker. Therefore, the magnetic flux into the page is decreasing. According to Lenz's law, the induced current will flow in a direction that creates a magnetic field to oppose this change. To oppose a decrease, the induced field must be in the same direction as the original field, i.e., into the page. By the right-hand grip rule, a clockwise current produces a magnetic field into the page. Therefore, the induced current is clockwise.

Question 4

A conducting wire is formed into two concentric loops, a smaller loop of radius r and a larger loop of radius 2r. A current I in the outer loop is decreasing. What is the direction of the induced current in the inner loop?

In the same direction as the current in the outer loop. (correct answer)
In the opposite direction to the current in the outer loop.
There is no induced current in the inner loop.
Radially outwards from the center.

Explanation: The current I in the outer loop creates a magnetic field and thus a magnetic flux through the inner loop. As the current I decreases, this magnetic flux also decreases. According to Lenz's law, the induced current in the inner loop must flow in a direction that opposes this change. To oppose a decrease in flux, the induced current must create its own magnetic field in the same direction as the original field. A current in the same direction as the original current I will achieve this. Therefore, the induced current is in the same direction as the current in the outer loop.

Question 5

A flat coil of wire rotates at a constant frequency in a uniform magnetic field. The induced emf is sinusoidal. At the instant when the magnetic flux through the coil is at its positive maximum, the induced emf is:

zero. (correct answer)
at its positive maximum.
at its negative maximum.
half of its maximum value.

Explanation: The induced emf is proportional to the rate of change of magnetic flux (ε = -dΦ/dt). When the flux Φ is at a maximum or minimum (a peak or trough of the cosine function Φ(t) = BAcos(ωt)), its rate of change (the slope of the flux-time graph) is zero. Therefore, the induced emf is zero at that instant. Conversely, when the flux is zero, its rate of change is maximum, and the emf is maximum.

Question 6

A bar magnet with its north pole facing downwards is dropped through a horizontal conducting ring. Viewing the ring from above, what are the directions of the induced current as the magnet enters the ring and as it leaves the ring?

Clockwise, then counter-clockwise
Counter-clockwise, then clockwise (correct answer)
Clockwise, then clockwise
Counter-clockwise, then counter-clockwise

Explanation: As the magnet enters, the downward magnetic flux increases. By Lenz's law, the induced current will create an upward magnetic field to oppose this change. From the right-hand grip rule, a counter-clockwise current (viewed from above) produces an upward field. As the magnet leaves, the downward magnetic flux is decreasing. The induced current will create a downward magnetic field to oppose this decrease. A clockwise current (viewed from above) produces a downward field. Therefore, the direction is counter-clockwise, then clockwise.

Question 7

A metal aircraft with a wingspan of 40 m flies horizontally at a speed of 250 m s⁻¹ in a region where the vertical component of the Earth's magnetic field is 5.0 × 10⁻⁵ T. What is the potential difference induced between the wingtips?

0 V
50 V
2.0 V
0.50 V (correct answer)

Explanation: The aircraft's wings act as a single conductor of length L moving at velocity v through the vertical component of the Earth's magnetic field B. The motional emf (potential difference) induced is given by ε = BvL, where B, v, and L are mutually perpendicular. Here, L = 40 m, v = 250 m s⁻¹, and B is the vertical component, 5.0 × 10⁻⁵ T. So, ε = (5.0 × 10⁻⁵ T)(250 m s⁻¹)(40 m) = (5.0 × 10⁻⁵)(10000) = 0.50 V.

Question 8

A square conducting loop is pulled with a constant velocity out of a region of uniform magnetic field directed into the page. As the loop is exiting the field, a braking force acts on it. This braking force is a direct consequence of:

the interaction between the induced current and the external magnetic field. (correct answer)
the interaction of the loop's free electrons with the magnetic field, independent of any current.
the increase in the electrical resistance of the loop as it leaves the magnetic field.
the alignment of magnetic dipoles within the material of the conducting loop.

Explanation: As the loop exits the field, the magnetic flux through it decreases. By Faraday's law, this induces an emf and a current. According to Lenz's law, this current flows in a direction that creates a magnetic field opposing the change in flux. This induced current, flowing in the part of the loop still inside the external field, experiences a magnetic force (F = BIL) that opposes the motion of the loop. This is the braking force. The other options are incorrect. Option B describes the origin of the motional emf but not the macroscopic force. Options C and D are physically incorrect descriptions of the situation.

Question 9

A square coil of side length 0.40 m is in a uniform magnetic field of strength 1.5 T. The magnetic field lines make an angle of 30° with the plane of the coil. What is the magnetic flux through the coil?

0.12 Wb (correct answer)
0.21 Wb
0.24 Wb
0.60 Wb

Explanation: Magnetic flux is Φ = BAcosθ, where θ is the angle between the magnetic field vector and the normal to the area. The area of the coil is A = (0.40 m)² = 0.16 m². The question states the angle between the field and the plane of the coil is 30°. The angle θ required for the formula is between the field and the normal, which is 90° - 30° = 60°. Therefore, Φ = (1.5 T)(0.16 m²)cos(60°) = 0.24 × 0.5 = 0.12 Wb. Choosing the 30° angle directly would incorrectly yield 0.21 Wb.

Question 10

A conducting rod of resistance R slides without friction on two parallel conducting rails a distance L apart. A uniform magnetic field B is perpendicular to the plane of the rails. An external force pulls the rod, causing it to move at a constant velocity v. What is the rate at which thermal energy is dissipated in the rod?

(BvL)/R
B²L²v / R
(BvL)² / R (correct answer)
BvL

Explanation: The induced motional emf is ε = BvL. The rate of thermal energy dissipation is the power, P. Power can be calculated as P = ε²/R. Substituting the expression for emf, we get P = (BvL)² / R = B²L²v²/R. Option A is the current. Option B is the magnetic force. Option D is the emf.

Question 11

A flat coil with N turns and area A rotates at a constant angular velocity ω in a uniform magnetic field B. The peak induced emf is ε₀. What is the new peak induced emf if the number of turns is doubled, the area is halved, and the angular velocity is doubled?

ε₀ / 2
ε₀
2ε₀ (correct answer)
4ε₀

Explanation: The peak induced emf in a rotating coil (generator) is given by the formula ε₀ = NBAω. The new peak emf, ε_new, will be given by ε_new = (2N)B(A/2)(2ω) = 2(NBAω) = 2ε₀. The number of turns doubles (x2), the area halves (x0.5), and the angular velocity doubles (x2). The net effect is a factor of 2 × 0.5 × 2 = 2.

Question 12

The magnetic flux Φ through a coil of 50 turns is given by the expression Φ(t) = 0.10t - 0.02t², where Φ is in Webers and t is in seconds. What is the magnitude of the induced emf in the coil at time t = 2.0 s?

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

Explanation: According to Faraday's law of induction, the induced emf ε is given by ε = -N(dΦ/dt). First, we find the derivative of the flux with respect to time: dΦ/dt = d/dt (0.10t - 0.02t²) = 0.10 - 0.04t. At t = 2.0 s, dΦ/dt = 0.10 - 0.04(2.0) = 0.10 - 0.08 = 0.02 Wb s⁻¹. The magnitude of the induced emf is |ε| = N |dΦ/dt| = 50 × 0.02 = 1.0 V.

Question 13

A square loop with side length L and N turns is in a uniform magnetic field B. In a time Δt, it is rotated 180° from an initial position where its plane is perpendicular to the field. What is the magnitude of the average emf induced in the loop?

0
NBL² / Δt
2NBL² / Δt (correct answer)
NBL² / (2Δt)

Explanation: Initial position: plane perpendicular to B, so the normal is parallel to B (θ=0°). Initial flux Φᵢ = NBA = NBL². Final position: after 180° rotation, the normal is anti-parallel to B (θ=180°). Final flux Φ₟ = NBAcos(180°) = -NBL². The change in flux is ΔΦ = Φ₟ - Φᵢ = -NBL² - NBL² = -2NBL². The magnitude of the average induced emf is |ε| = |-ΔΦ/Δt| = |-(-2NBL²)/Δt| = 2NBL²/Δt.

Question 14

Two identical circular coils are placed coaxially with their centers separated by a distance equal to their radius $R$ . The current in the first coil is increasing at a rate $dI/dt = 2.0 \text{ A/s}$ . If each coil has $N = 100$ turns and the mutual inductance between the coils is $M = 5.0 \times 10^{-4} \text{ H}$ , what is the magnitude of the EMF induced in the second coil?

$1.0 \times 10^{-3} \text{ V}$ due to the changing flux from the first coil (correct answer)
$1.0 \times 10^{-3} \text{ V}$ due to self-inductance effects in the second coil
$1.0 \times 10^{-1} \text{ V}$ due to the changing flux from the first coil
$1.0 \times 10^{-1} \text{ V}$ due to self-inductance effects in the second coil

Explanation: The EMF induced in the second coil due to mutual inductance is

|\varepsilon| = M|dI/dt| = 5.0 \times 10^{-4} \times 2.0 = 1.0 \times 10^{-3} \text{ V}

. This is caused by the changing magnetic flux from the first coil linking with the second coil. Choice B gives the correct magnitude but incorrect reasoning (self-inductance would only occur if current in the second coil were changing). Choices C and D incorrectly multiply by the number of turns, but the mutual inductance already accounts for the geometry and turn count of both coils.

Question 15

An inductor with inductance $L = 0.25 \text{ H}$ and resistance $R = 8.0 \text{ Ω}$ is connected to a $12 \text{ V}$ battery through a switch. At $t = 0.050 \text{ s}$ after the switch is closed, the current is $1.0 \text{ A}$ . What is the magnitude of the self-induced EMF at this instant?

$4.0 \text{ V}$ opposing the applied voltage according to Lenz's law (correct answer)
$4.0 \text{ V}$ reinforcing the applied voltage to maintain current flow
$8.0 \text{ V}$ opposing the applied voltage according to Lenz's law
$8.0 \text{ V}$ reinforcing the applied voltage to maintain current flow

Explanation: From Kirchhoff's voltage law:

V_{battery} = IR + L(dI/dt)

, so

12 = 1.0 \times 8.0 + 0.25(dI/dt)

. Solving:

dI/dt = (12-8.0)/0.25 = 16 \text{ A/s}

. The self-induced EMF is

|\varepsilon_L| = L|dI/dt| = 0.25 \times 16 = 4.0 \text{ V}

. By Lenz's law, this EMF opposes the change in current (opposes the increase). Choice B has correct magnitude but wrong direction. Choices C and D use the voltage drop across the resistor instead of calculating the rate of change properly.

Question 16

A conducting disk of radius $R = 0.15 \text{ m}$ rotates about its center with angular velocity $\omega = 20 \text{ rad/s}$ in a uniform magnetic field $B = 0.40 \text{ T}$ parallel to the rotation axis. What is the potential difference between the center and rim of the disk?

$0.045 \text{ V}$ with the rim at higher potential due to charge separation (correct answer)
$0.045 \text{ V}$ with the center at higher potential due to charge separation
$0.090 \text{ V}$ with the rim at higher potential due to charge separation
$0.090 \text{ V}$ with the center at higher potential due to charge separation

Explanation: For a rotating disk in a magnetic field, the motional EMF between center and rim is

\varepsilon = \frac{1}{2}B\omega R^2 = \frac{1}{2} \times 0.40 \times 20 \times (0.15)^2 = \frac{1}{2} \times 0.40 \times 20 \times 0.0225 = 0.045 \text{ V}

. The rim moves faster than the center, so charges experience a greater magnetic force there, making the rim at higher potential. Choice B has correct magnitude but wrong polarity. Choices C and D use

\varepsilon = B\omega R^2

(missing the factor of 1/2) which would apply to the EMF across a diameter, not center to rim.

Question 17

A circular conducting loop of radius $0.15 \text{ m}$ is placed in a uniform magnetic field. The magnetic flux through the loop changes from $0.040 \text{ Wb}$ to $0.010 \text{ Wb}$ in $0.25 \text{ s}$ . If the loop has a resistance of $8.0 \text{ Ω}$ , what is the magnitude of the induced current and its direction relative to the original magnetic field?

$0.015 \text{ A}$ , opposing the change in flux according to Lenz's law (correct answer)
$0.015 \text{ A}$ , reinforcing the change in flux to maintain equilibrium
$0.0075 \text{ A}$ , opposing the change in flux according to Lenz's law
$0.0075 \text{ A}$ , reinforcing the change in flux to maintain equilibrium

Explanation: The induced EMF is

|\varepsilon| = |\Delta\Phi/\Delta t| = |0.010 - 0.040|/0.25 = 0.030/0.25 = 0.12 \text{ V}

. The induced current is

I = \varepsilon/R = 0.12/8.0 = 0.015 \text{ A}

. By Lenz's law, the induced current creates a magnetic field that opposes the change in flux. Since flux is decreasing, the induced current will create a field in the same direction as the original field to oppose this decrease. Choice B incorrectly states the current reinforces the change. Choices C and D use an incorrect current calculation (perhaps dividing by twice the resistance or making an error in the flux change calculation).

Question 18

Lenz's law, which determines the direction of an induced current, is a direct statement of the conservation of:

charge.
momentum.
angular momentum.
energy. (correct answer)

Explanation: Lenz's law states that the induced current flows in a direction that opposes the change in magnetic flux that produced it. This opposition ensures that work must be done to induce the current. For example, to push a magnet into a coil, one must push against the opposing magnetic force created by the induced current. The work done is converted into electrical energy in the coil, thus conserving energy. If the induced current assisted the change, it would lead to a runaway effect, creating energy from nothing, which violates the principle of conservation of energy.

Question 19

A conducting loop falls under gravity and passes through a horizontal region of uniform magnetic field. Air resistance is negligible. Which statement best describes the loop's acceleration 'a'?

a = g while entering, inside, and leaving the field.
a < g while entering and leaving, and a = g while fully inside the field. (correct answer)
a < g while entering and fully inside, and a = g while leaving the field.
a < g only while entering, and a = g at all other times.

Explanation: Induction occurs only when the magnetic flux through the loop changes. This happens as the loop enters the field (flux increases) and as it leaves the field (flux decreases). During these periods, an induced current creates an upward magnetic braking force, so the net downward force is less than mg, and a < g. While the loop is fully inside the uniform field, the flux is constant, so there is no induced current and no magnetic force. The only force is gravity, so a = g.

Question 20

A uniform magnetic field B is directed into the page. A conducting rod is oriented horizontally and moves vertically downwards with a constant speed v. What is the polarity of the induced emf?

The left end becomes positive, and the right end becomes negative.
The right end becomes positive, and the left end becomes negative. (correct answer)
The top surface of the rod becomes positive.
No emf is induced as the rod is not part of a closed circuit.

Explanation: We can use the right-hand rule for the Lorentz force on positive charge carriers in the rod: F = q(v × B). Point the fingers in the direction of velocity v (downwards). Curl the fingers into the direction of the magnetic field B (into the page). The thumb points to the right. This means positive charges are pushed to the right end of the rod, and negative charges (electrons) are pushed to the left end. This separation of charge creates an emf with the right end at a higher potential (positive) and the left end at a lower potential (negative). An emf is induced even without a closed circuit.