College Physics Quiz: Amperes Law
20 questions · exam conditions
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Amperes LawQuestion 1 of 20

Two parallel wires separated by 3.0 cm each carry currents of 5.0 A in the same direction. What is the magnitude of the magnetic field at a point exactly halfway between the wires?

Zero
3.3×10−53.3 \times 10^{-5} T
6.7×10−56.7 \times 10^{-5} T
1.3×10−41.3 \times 10^{-4} T
2.0×10−42.0 \times 10^{-4} T
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College Physics Quiz

College Physics Quiz: Amperes Law

Practice Amperes Law in College 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 Amperes Law, giving you a quick way to practice the rules, question types, and explanations that matter most for College 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

Two parallel wires separated by 3.0 cm each carry currents of 5.0 A in the same direction. What is the magnitude of the magnetic field at a point exactly halfway between the wires?

  1. Zero (correct answer)
  2. 3.3×10−53.3 \times 10^{-5} T
  3. 6.7×10−56.7 \times 10^{-5} T
  4. 1.3×10−41.3 \times 10^{-4} T
  5. 2.0×10−42.0 \times 10^{-4} T
Explanation: When you encounter parallel current-carrying wires, you're dealing with magnetic field superposition. Each wire creates a circular magnetic field around itself, and at any point in space, you must add these fields vectorially. For a long straight wire carrying current, the magnetic field at distance rr is given by B=μ0I2πrB = \frac{\mu_0 I}{2\pi r}, where the field direction follows the right-hand rule. At the midpoint between these wires (1.5 cm from each), each wire creates a field of magnitude B=(4π×10−7)(5.0)2π(0.015)=6.7×10−5B = \frac{(4\pi \times 10^{-7})(5.0)}{2\pi(0.015)} = 6.7 \times 10^{-5} T. Here's the crucial insight: since both currents flow in the same direction, the magnetic fields at the midpoint point in opposite directions. Using the right-hand rule, if both currents flow upward, the left wire's field points into the page at the midpoint, while the right wire's field points out of the page. These equal-magnitude fields cancel completely, giving zero net field. Answer A is correct because the fields cancel due to symmetry and same-direction currents. Answer B (3.3×10−53.3 \times 10^{-5} T) represents half the field from one wire—a mistake if you forgot about the second wire. Answer C (6.7×10−56.7 \times 10^{-5} T) is the field from just one wire, ignoring the other entirely. Answer D (1.3×10−41.3 \times 10^{-4} T) incorrectly adds the field magnitudes, forgetting that magnetic fields are vectors that can cancel. Remember: parallel same-direction currents create zero field at the midpoint due to cancellation. If the currents were opposite, they'd add constructively instead.

Question 2

A circular loop of radius 0.10 m carries a current of 2.0 A. Using Ampère's law concepts, what happens to the magnetic field at the center of the loop if both the radius is doubled and the current is halved?

  1. The magnetic field increases by a factor of 4
  2. The magnetic field decreases by a factor of 4 (correct answer)
  3. The magnetic field decreases by a factor of 2
  4. The magnetic field remains the same
  5. The magnetic field increases by a factor of 2
Explanation: When you encounter questions about magnetic fields and current loops, you're dealing with the relationship between current, geometry, and the resulting magnetic field strength. The key is understanding how each parameter affects the field. For a circular current loop, the magnetic field at the center is given by B=μ0I2RB = \frac{\mu_0 I}{2R}, where I is the current and R is the radius. This formula shows that the magnetic field is directly proportional to current and inversely proportional to radius. Let's analyze the changes: if the radius doubles (R becomes 2R) and current is halved (I becomes I/2), the new magnetic field becomes: Bnew=μ0(I/2)2(2R)=μ0I8RB_{new} = \frac{\mu_0 (I/2)}{2(2R)} = \frac{\mu_0 I}{8R} Comparing to the original field Boriginal=μ0I2RB_{original} = \frac{\mu_0 I}{2R}, we get: BnewBoriginal=1/8R1/2R=14\frac{B_{new}}{B_{original}} = \frac{1/8R}{1/2R} = \frac{1}{4} So the magnetic field decreases by a factor of 4, making B correct. A is wrong because it claims the field increases by 4 when both changes actually reduce the field strength. C incorrectly suggests only a factor of 2 decrease, which would occur if you only considered one of the two changes. D is incorrect because it ignores that the radius change (factor of 2 decrease) outweighs the current change (factor of 2 decrease), resulting in a net factor of 4 decrease. Remember: when multiple parameters change simultaneously, calculate their combined effect systematically rather than trying to reason through each change separately.

Question 3

Consider a solenoid with 500 turns per meter carrying a current of 3.0 A. What is the magnetic field inside the solenoid, far from the ends?

  1. 1.9×10−31.9 \times 10^{-3} T (correct answer)
  2. 9.4×10−49.4 \times 10^{-4} T
  3. 3.8×10−33.8 \times 10^{-3} T
  4. 6.3×10−46.3 \times 10^{-4} T
  5. 2.4×10−32.4 \times 10^{-3} T
Explanation: When you encounter solenoid problems, you're dealing with one of the most predictable magnetic field configurations in physics. The key insight is that inside a long solenoid (far from the ends), the magnetic field is uniform and depends only on the current and turn density. The magnetic field inside an ideal solenoid is given by B=μ0nIB = \mu_0 n I, where μ0=4π×10−7\mu_0 = 4\pi \times 10^{-7} T·m/A is the permeability of free space, nn is the number of turns per unit length, and II is the current. Substituting the given values: n=500n = 500 turns/m and I=3.0I = 3.0 A: B=(4π×10−7)(500)(3.0)=6π×10−4≈1.9×10−3B = (4\pi \times 10^{-7})(500)(3.0) = 6\pi \times 10^{-4} \approx 1.9 \times 10^{-3} T This confirms answer A is correct. Looking at the wrong answers: B (9.4×10−49.4 \times 10^{-4} T) appears to result from using 3π×10−43\pi \times 10^{-4} instead of 6π×10−46\pi \times 10^{-4}, possibly from forgetting to multiply by the current or making an arithmetic error. C (3.8×10−33.8 \times 10^{-3} T) is roughly double the correct answer, suggesting someone might have made an error with the π\pi factor or doubled something incorrectly. D (6.3×10−46.3 \times 10^{-4} T) is close to 2π×10−42\pi \times 10^{-4}, indicating a potential error in the coefficient calculation. Remember: solenoid field strength depends linearly on both current and turn density. Always double-check that you're using 4π×10−74\pi \times 10^{-7} for μ0\mu_0 and calculating n×In \times I correctly.

Question 4

A cylindrical conductor of radius 2.0 mm carries a uniformly distributed current of 8.0 A. Using Ampère's law, what is the magnetic field at a distance of 1.0 mm from the center (inside the conductor)?

  1. 1.0×10−31.0 \times 10^{-3} T
  2. 5.0×10−45.0 \times 10^{-4} T (correct answer)
  3. 2.0×10−32.0 \times 10^{-3} T
  4. 8.0×10−48.0 \times 10^{-4} T
  5. 4.0×10−44.0 \times 10^{-4} T
Explanation: When applying Ampère's law to find the magnetic field inside a current-carrying conductor, you need to consider that only the current enclosed by your Amperian loop contributes to the magnetic field at that point. For a cylindrical conductor with uniformly distributed current, the current density is J=ItotalπR2J = \frac{I_{total}}{\pi R^2}, where RR is the conductor's radius. At distance rr from the center (where r<Rr < R), the enclosed current is Ienc=J⋅πr2=Itotal⋅r2R2I_{enc} = J \cdot \pi r^2 = I_{total} \cdot \frac{r^2}{R^2}. Using Ampère's law with a circular path of radius r=1.0r = 1.0 mm: ∮B⃗⋅dl⃗=μ0Ienc\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc} Since the magnetic field is symmetric, B⋅2πr=μ0Itotalr2R2B \cdot 2\pi r = \mu_0 I_{total} \frac{r^2}{R^2} Solving for BB: B=μ0Itotalr2πR2B = \frac{\mu_0 I_{total} r}{2\pi R^2} Substituting values: B=(4π×10−7)(8.0)(1.0×10−3)2π(2.0×10−3)2=5.0×10−4B = \frac{(4\pi \times 10^{-7})(8.0)(1.0 \times 10^{-3})}{2\pi (2.0 \times 10^{-3})^2} = 5.0 \times 10^{-4} T This confirms answer B is correct. A (1.0×10−31.0 \times 10^{-3} T) likely results from using the full current instead of the enclosed current. C (2.0×10−32.0 \times 10^{-3} T) might come from incorrectly applying the formula for points outside the conductor. D (8.0×10−48.0 \times 10^{-4} T) could result from calculation errors in the geometry or current density. Remember: inside a conductor, the magnetic field increases linearly with distance from the center because you're only considering the current within your Amperian loop, not the total current.

Question 5

Two concentric circular Amperian loops of radii 3.0 cm and 6.0 cm surround a straight wire carrying current I. If ∮B⃗⋅dl⃗=2.4×10−6\oint \vec{B} \cdot d\vec{l} = 2.4 \times 10^{-6} T·m for the inner loop, what is ∮B⃗⋅dl⃗\oint \vec{B} \cdot d\vec{l} for the outer loop?

  1. 1.2×10−61.2 \times 10^{-6} T·m
  2. 4.8×10−64.8 \times 10^{-6} T·m
  3. 2.4×10−62.4 \times 10^{-6} T·m (correct answer)
  4. 6.0×10−66.0 \times 10^{-6} T·m
  5. 9.6×10−69.6 \times 10^{-6} T·m
Explanation: When you encounter Amperian loops and line integrals of magnetic fields, you're working with Ampère's Law, which states that ∮B⃗⋅dl⃗=μ0Ienc\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}, where IencI_{enc} is the current enclosed by the loop. The key insight here is that both circular loops enclose the same straight wire carrying current I. Since the enclosed current is identical for both loops, Ampère's Law tells us that the line integral ∮B⃗⋅dl⃗\oint \vec{B} \cdot d\vec{l} must be the same for both loops, regardless of their radii. This might seem counterintuitive since the magnetic field strength decreases with distance from the wire, but the path length increases proportionally, making the line integral constant. Therefore, if ∮B⃗⋅dl⃗=2.4×10−6\oint \vec{B} \cdot d\vec{l} = 2.4 \times 10^{-6} T·m for the inner loop, it must also equal 2.4×10−62.4 \times 10^{-6} T·m for the outer loop. This confirms answer C. Answer A (1.2×10−61.2 \times 10^{-6} T·m) incorrectly assumes the integral decreases with radius. Answer B (4.8×10−64.8 \times 10^{-6} T·m) wrongly suggests the integral doubles with doubled radius. Answer D (6.0×10−66.0 \times 10^{-6} T·m) incorrectly applies direct proportionality to the radius ratio. Remember: Ampère's Law depends only on enclosed current, not loop size. When you see concentric Amperian loops around the same current source, the line integral ∮B⃗⋅dl⃗\oint \vec{B} \cdot d\vec{l} remains constant regardless of loop radius.

Question 6

A wire bent into an equilateral triangle carries current I clockwise. A student wants to use Ampère's law to find the magnetic field at the center. Which statement about this approach is most accurate?

  1. Ampère's law applies directly since current flows through the triangle
  2. Ampère's law cannot be used because the current path lacks sufficient symmetry (correct answer)
  3. The calculation is straightforward because the field is uniform around the center
  4. Ampère's law works but requires integration over three separate segments
  5. The triangular geometry makes Ampère's law invalid for this configuration
Explanation: When you encounter Ampère's law problems, the crucial factor is whether the current configuration has sufficient symmetry to make the mathematical application feasible. Ampère's law states that ∮B⃗⋅dl⃗=μ0Ienclosed\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed}, but this is only practically useful when you can choose a path where the magnetic field has constant magnitude and is either parallel or perpendicular to your integration path. For a triangular current loop, the magnetic field at the center has no useful symmetry. The field contributions from each side of the triangle point in different directions and have different magnitudes depending on your position relative to each segment. You cannot choose any closed path around the center where B⃗\vec{B} remains constant in magnitude or direction, making the line integral mathematically intractable. Choice A is wrong because simply having current flow doesn't make Ampère's law applicable—you need the right symmetry. Choice C incorrectly assumes the field is uniform around the center, which it definitely isn't for a triangular geometry. Choice D suggests integrating over the three triangle segments, but this misunderstands how Ampère's law works—you integrate around a closed path in the field, not along the current-carrying conductor. For this triangle problem, you'd need to use the Biot-Savart law instead, calculating the field contribution from each straight segment and adding them vectorially. Study tip: Ampère's law only works with high-symmetry geometries like infinite straight wires, solenoids, and toroids. If the geometry lacks circular or linear symmetry, reach for Biot-Savart law instead.

Question 7

A hollow cylindrical conductor (inner radius 2.0 cm, outer radius 4.0 cm) carries 6.0 A uniformly distributed through its cross-section. What is the magnetic field at radius 3.0 cm from the center axis?

  1. 2.0×10−52.0 \times 10^{-5} T
  2. 1.0×10−51.0 \times 10^{-5} T (correct answer)
  3. 4.0×10−54.0 \times 10^{-5} T
  4. 3.0×10−53.0 \times 10^{-5} T
  5. Zero
Explanation: When you encounter a hollow cylindrical conductor problem, you need to apply Ampère's law while carefully considering which portion of the current contributes to the magnetic field at your point of interest. Since the radius of interest (3.0 cm) lies between the inner radius (2.0 cm) and outer radius (4.0 cm), you're inside the conducting material. Only the current enclosed by your Amperian loop contributes to the magnetic field. The current is uniformly distributed, so you need the current density and the area that actually contains current. The conducting area is π(0.042−0.022)=π(0.0016−0.0004)=0.0012π m2\pi(0.04^2 - 0.02^2) = \pi(0.0016 - 0.0004) = 0.0012\pi \text{ m}^2 The current density is J=6.0 A0.0012π m2=5000π A/m2J = \frac{6.0 \text{ A}}{0.0012\pi \text{ m}^2} = \frac{5000}{\pi} \text{ A/m}^2 The enclosed current at radius 3.0 cm includes only the annular region from 2.0 cm to 3.0 cm: Ienc=J×π(0.032−0.022)=5000π×π(0.0009−0.0004)=5000×0.0005=2.5 AI_{enc} = J \times \pi(0.03^2 - 0.02^2) = \frac{5000}{\pi} \times \pi(0.0009 - 0.0004) = 5000 \times 0.0005 = 2.5 \text{ A} Using Ampère's law: B(2πr)=μ0IencB(2\pi r) = \mu_0 I_{enc} B=μ0Ienc2πr=(4π×10−7)(2.5)2π(0.03)=1.0×10−5 TB = \frac{\mu_0 I_{enc}}{2\pi r} = \frac{(4\pi \times 10^{-7})(2.5)}{2\pi(0.03)} = 1.0 \times 10^{-5} \text{ T} Choice A (2.0×10−52.0 \times 10^{-5} T) uses the wrong enclosed current calculation. Choice C (4.0×10−54.0 \times 10^{-5} T) likely uses the full 6.0 A current incorrectly. Choice D (3.0×10−53.0 \times 10^{-5} T) represents another computational error. Remember: in hollow conductor problems, always calculate the current density first, then determine only the enclosed current within your radius of interest.

Question 8

Two identical solenoids are placed end-to-end with currents flowing in the same direction through their windings. At the junction between the solenoids, the magnetic field is:

  1. Zero due to cancellation of opposing fields
  2. Twice the field of a single solenoid due to superposition
  3. Equal to the field of a single solenoid by continuity (correct answer)
  4. Half the field of a single solenoid due to field sharing
  5. Undefined because Ampère's law doesn't apply at boundaries
Explanation: When analyzing magnetic fields at boundaries between current-carrying devices, you need to apply the principle of field continuity and superposition. The key insight is understanding how magnetic field lines behave at interfaces and how identical current configurations contribute to the total field. Inside a long solenoid, the magnetic field is uniform and given by B=μ0nIB = \mu_0 n I, where nn is the turn density and II is the current. At the junction between two identical solenoids with currents in the same direction, both solenoids contribute equally to the magnetic field at that point. Since the solenoids are identical and the currents flow in the same direction, each produces the same field magnitude in the same direction at the junction. However, the field doesn't simply double because each solenoid alone would produce its full field strength at its end. At the junction, you're not adding two complete solenoid fields—you're at the boundary where one solenoid's field transitions to the other's. The magnetic field must be continuous across this boundary, maintaining the same strength as within either individual solenoid. Answer A is wrong because the fields don't oppose each other—the currents flow in the same direction, creating fields in the same direction. Answer B incorrectly assumes simple addition of two complete solenoid fields, ignoring that we're at a transition point. Answer D has no physical basis—magnetic fields don't "share" or halve at junctions. Remember: magnetic field lines are continuous and cannot have sudden jumps in magnitude. At boundaries between similar current configurations, look for continuity rather than simple arithmetic addition.

Question 9

A student applies Ampère's law to a path that passes through a region where B⃗\vec{B} is not parallel to dl⃗d\vec{l}. In this region, the dot product B⃗⋅dl⃗\vec{B} \cdot d\vec{l} equals ∣B∣∣dl∣cos⁡θ|B||dl|\cos\theta where θ=60°\theta = 60°. If ∣B∣=2.0×10−4|B| = 2.0 \times 10^{-4} T and the path length through this region is 0.050 m, what is the contribution to ∮B⃗⋅dl⃗\oint \vec{B} \cdot d\vec{l}?

  1. 1.0×10−51.0 \times 10^{-5} T·m
  2. 5.0×10−65.0 \times 10^{-6} T·m (correct answer)
  3. 8.7×10−68.7 \times 10^{-6} T·m
  4. 1.7×10−51.7 \times 10^{-5} T·m
  5. Zero, because B⃗\vec{B} and dl⃗d\vec{l} are not parallel
Explanation: When applying Ampère's law, you're calculating the line integral ∮B⃗⋅dl⃗\oint \vec{B} \cdot d\vec{l} around a closed path. The key insight is that this integral sums up contributions from each segment of your path, and each contribution depends on both the magnetic field strength and how aligned the field is with your path direction. For this segment, you need to calculate B⃗⋅dl⃗=∣B∣∣dl∣cos⁡θ\vec{B} \cdot d\vec{l} = |B||dl|\cos\theta. With ∣B∣=2.0×10−4|B| = 2.0 \times 10^{-4} T, ∣dl∣=0.050|dl| = 0.050 m, and θ=60°\theta = 60°: B⃗⋅dl⃗=(2.0×10−4)(0.050)cos⁡(60°)=(1.0×10−5)(0.5)=5.0×10−6 T\cdotpm\vec{B} \cdot d\vec{l} = (2.0 \times 10^{-4})(0.050)\cos(60°) = (1.0 \times 10^{-5})(0.5) = 5.0 \times 10^{-6} \text{ T·m} This confirms answer B is correct. Let's see where the wrong answers come from: A (1.0×10−51.0 \times 10^{-5} T·m) is what you'd get if you forgot the cosine term entirely—a common mistake when the magnetic field isn't parallel to the path. C (8.7×10−68.7 \times 10^{-6} T·m) results from incorrectly using cos⁡(30°)=0.87\cos(30°) = 0.87 instead of cos⁡(60°)\cos(60°)—perhaps confusing the angle with its complement. D (1.7×10−51.7 \times 10^{-5} T·m) doesn't correspond to any obvious calculation error with these values. Remember: When B⃗\vec{B} isn't parallel to your path, the dot product automatically accounts for the projection. Always include the cosine factor—it reduces the contribution when the field and path aren't perfectly aligned, which is physically intuitive since only the component of B⃗\vec{B} along the path matters for Ampère's law.

Question 10

A finite solenoid has 400 turns over 0.20 m length and carries 1.5 A. A student calculates the field at the center using B=μ0nIB = \mu_0 nI and gets 3.8×10−33.8 \times 10^{-3} T. The actual field at the center of this finite solenoid will be:

  1. Exactly 3.8×10−33.8 \times 10^{-3} T because Ampère's law is exact
  2. Less than 3.8×10−33.8 \times 10^{-3} T because of finite-length end effects (correct answer)
  3. Greater than 3.8×10−33.8 \times 10^{-3} T due to field concentration
  4. Zero because the solenoid is not infinite
  5. Half of 3.8×10−33.8 \times 10^{-3} T due to the finite geometry
Explanation: When analyzing solenoid magnetic fields, you need to distinguish between the idealized infinite solenoid and real finite solenoids. The formula B=μ0nIB = \mu_0 nI applies to an infinitely long solenoid, where the field lines are perfectly uniform and contained within the solenoid. For a finite solenoid, the magnetic field at the center is weaker than this idealized calculation predicts. This happens because of "end effects" - the field lines near the ends of the solenoid curve outward and don't contribute as effectively to the field at the center. The shorter the solenoid relative to its diameter, the more pronounced these end effects become. Let's verify the student's calculation: n=400/0.20=2000 turns/mn = 400/0.20 = 2000 \text{ turns/m}, so B=(4π×10−7)(2000)(1.5)=3.77×10−3 TB = (4\pi \times 10^{-7})(2000)(1.5) = 3.77 \times 10^{-3} \text{ T}, which rounds to 3.8×10−3 T3.8 \times 10^{-3} \text{ T}. Answer A is wrong because while Ampère's law is exact, the formula B=μ0nIB = \mu_0 nI only applies to infinite solenoids. Answer C incorrectly suggests field concentration - finite length actually reduces the field strength at the center. Answer D is completely incorrect; finite solenoids do produce magnetic fields, just weaker ones than the infinite approximation predicts. The actual field will be less than 3.8×10−3 T3.8 \times 10^{-3} \text{ T}, making B correct. Study tip: Always check if a formula assumes ideal conditions (like infinite length). Real-world finite geometries typically produce weaker fields than idealized calculations predict.

Question 11

A toroidal solenoid has an inner radius of 8.0 cm, outer radius of 12.0 cm, 400 total turns, and carries 2.5 A. Using Ampère's law, what is the magnetic field at radius 10.0 cm from the center?

  1. 2.0×10−32.0 \times 10^{-3} T (correct answer)
  2. 1.0×10−31.0 \times 10^{-3} T
  3. 4.0×10−34.0 \times 10^{-3} T
  4. 5.0×10−45.0 \times 10^{-4} T
  5. Zero
Explanation: When you encounter a toroidal solenoid problem, you're dealing with a doughnut-shaped coil where Ampère's law provides an elegant solution. The key insight is that the magnetic field inside a toroidal solenoid has circular symmetry around the central axis. Applying Ampère's law with a circular path of radius r = 10.0 cm: ∮B⃗⋅dl⃗=μ0Ienc\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}. Since the magnetic field is tangent to the circular path and has constant magnitude, this becomes B⋅2πr=μ0NIB \cdot 2\pi r = \mu_0 N I, where N is the total number of turns and I is the current. Solving for B: B=μ0NI2πr=(4π×10−7)(400)(2.5)2π(0.10)=4×10−40.20=2.0×10−3B = \frac{\mu_0 N I}{2\pi r} = \frac{(4\pi \times 10^{-7})(400)(2.5)}{2\pi(0.10)} = \frac{4 \times 10^{-4}}{0.20} = 2.0 \times 10^{-3} T. Answer A (2.0×10−32.0 \times 10^{-3} T) is correct. Answer B (1.0×10−31.0 \times 10^{-3} T) results from incorrectly doubling the radius or halving another parameter. Answer C (4.0×10−34.0 \times 10^{-3} T) occurs if you forget to include the factor of 2π in the denominator. Answer D (5.0×10−45.0 \times 10^{-4} T) happens when you mistakenly use only a quarter of the turns or make an error with the current. Remember: For toroidal solenoids, the magnetic field depends only on the total number of turns, current, and radial distance from the center—the inner and outer radii are given to confirm you're calculating at a valid point within the coil.

Question 12

A long straight wire carries a current of 8.0 A. Using Ampère's law, what is the magnitude of the magnetic field at a distance of 2.0 cm from the wire? (μ0=4π×10−7\mu_0 = 4\pi \times 10^{-7} T·m/A)

  1. 8.0×10−58.0 \times 10^{-5} T (correct answer)
  2. 1.6×10−41.6 \times 10^{-4} T
  3. 3.2×10−53.2 \times 10^{-5} T
  4. 6.4×10−56.4 \times 10^{-5} T
  5. 2.0×10−42.0 \times 10^{-4} T
Explanation: When you encounter problems about magnetic fields around current-carrying wires, you're applying Ampère's law, which relates the magnetic field to the current that produces it. For a long straight wire, the magnetic field forms concentric circles around the wire, and the field strength depends on both the current and your distance from the wire. Using Ampère's law for a straight wire, the magnetic field magnitude is B=μ0I2πrB = \frac{\mu_0 I}{2\pi r}, where II is the current and rr is the distance from the wire. Substituting the given values: I=8.0I = 8.0 A, r=0.02r = 0.02 m, and μ0=4π×10−7\mu_0 = 4\pi \times 10^{-7} T·m/A. B=(4π×10−7)(8.0)2π(0.02)=32π×10−70.04π=32×10−70.04=8.0×10−5B = \frac{(4\pi \times 10^{-7})(8.0)}{2\pi(0.02)} = \frac{32\pi \times 10^{-7}}{0.04\pi} = \frac{32 \times 10^{-7}}{0.04} = 8.0 \times 10^{-5} T This confirms answer A is correct. Answer B (1.6×10−41.6 \times 10^{-4} T) is exactly double the correct value, suggesting you might have forgotten the factor of 2 in the denominator. Answer C (3.2×10−53.2 \times 10^{-5} T) could result from using the wrong distance or making an arithmetic error in the calculation. Answer D (6.4×10−56.4 \times 10^{-5} T) is close but likely comes from incorrectly manipulating the 2π2\pi factor. Remember that Ampère's law problems often involve careful attention to the geometric factor (like 2π2\pi for straight wires). Always double-check your formula setup before plugging in numbers, and convert distances to meters before calculating.

Question 13

A long straight wire carries current I upward. A student chooses an Amperian loop that is a square with one vertex touching the wire and the square extending to the right of the wire. Compared to using a circular loop centered on the wire, this choice:

  1. Makes the calculation easier because B⃗\vec{B} is constant along each side
  2. Makes the calculation impossible because B⃗\vec{B} varies along the loop
  3. Gives a different value for ∮B⃗⋅dl⃗\oint \vec{B} \cdot d\vec{l} because the geometry is different
  4. Makes the calculation harder but gives the same ∮B⃗⋅dl⃗\oint \vec{B} \cdot d\vec{l} value (correct answer)
  5. Is invalid because the loop must be symmetric about the current source
Explanation: When you encounter Ampère's law problems, remember that the law states ∮B⃗⋅dl⃗=μ0Ienc\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}, where the line integral depends only on the current enclosed by your loop, not the loop's shape. For a long straight wire, the magnetic field forms concentric circles around the wire. With a circular Amperian loop centered on the wire, B⃗\vec{B} has constant magnitude and is always tangent to the path, making B⃗⋅dl⃗=B dl\vec{B} \cdot d\vec{l} = B \, dl and the integral simply B×2πrB \times 2\pi r. The square loop makes the calculation much harder because the magnetic field varies in both magnitude and direction along each side. You'd need to account for how B⃗\vec{B} changes as you move along each segment, requiring complex integration. However, since the square still encloses the same current II, Ampère's law guarantees the integral equals μ0I\mu_0 I regardless of the loop shape. Choice A is wrong because B⃗\vec{B} is definitely not constant along each side of the square. Choice B incorrectly suggests the varying field makes calculation impossible—it's just much harder. Choice C represents a fundamental misunderstanding: Ampère's law ensures that ∮B⃗⋅dl⃗\oint \vec{B} \cdot d\vec{l} depends only on enclosed current, not loop geometry. The key insight is that Ampère's law is shape-independent. While we typically choose symmetric loops (like circles for straight wires) to make calculations manageable, any loop enclosing the same current gives the same result. Always pick the most symmetric loop possible to simplify your math.

Question 14

An Amperian loop is chosen as a circle of radius 5.0 cm centered on a long straight wire. If the magnetic field along this loop has magnitude 4.0×10−54.0 \times 10^{-5} T, what current flows through the wire?

  1. 10 A (correct answer)
  2. 5.0 A
  3. 2.5 A
  4. 20 A
  5. 15 A
Explanation: This question tests your understanding of Ampère's Law, which relates the magnetic field around a current-carrying wire to the current itself. When you see an Amperian loop around a straight wire, immediately think of the relationship ∮B⃗⋅dl⃗=μ0Ienc\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}. For a long straight wire, the magnetic field forms concentric circles around the wire. Since our circular Amperian loop is centered on the wire, the magnetic field is tangent to the loop everywhere and has constant magnitude. This simplifies Ampère's Law to B⋅2πr=μ0IB \cdot 2\pi r = \mu_0 I, where rr is the loop radius. Solving for current: I=B⋅2πrμ0I = \frac{B \cdot 2\pi r}{\mu_0} Substituting the given values: r=0.050 mr = 0.050 \text{ m}, B=4.0×10−5 TB = 4.0 \times 10^{-5} \text{ T}, and μ0=4π×10−7 T\cdotpm/A\mu_0 = 4\pi \times 10^{-7} \text{ T·m/A}: I=(4.0×10−5)(2π)(0.050)4π×10−7=4.0×10−64π×10−7=10 AI = \frac{(4.0 \times 10^{-5})(2\pi)(0.050)}{4\pi \times 10^{-7}} = \frac{4.0 \times 10^{-6}}{4\pi \times 10^{-7}} = 10 \text{ A} This confirms answer A is correct. Answer B (5.0 A) results from forgetting the factor of 2π2\pi in the circumference. Answer C (2.5 A) comes from both missing 2π2\pi and making an arithmetic error. Answer D (20 A) likely stems from using the radius in centimeters instead of converting to meters. Remember: Always convert units to SI base units first, and for circular Amperian loops around straight wires, the path length is always 2πr2\pi r.

Question 15

A current-carrying wire is bent into a single-turn circular loop of radius 8.0 cm carrying 3.0 A. If a student chooses an Amperian loop that is a circle of radius 12.0 cm concentric with the current loop, what is ∮B⃗⋅dl⃗\oint \vec{B} \cdot d\vec{l} around this Amperian path?

  1. μ0(3.0)=3.8×10−6\mu_0 (3.0) = 3.8 \times 10^{-6} T·m (correct answer)
  2. μ0(3.0)(8.0/12.0)=2.5×10−6\mu_0 (3.0)(8.0/12.0) = 2.5 \times 10^{-6} T·m
  3. μ0(3.0)(12.0/8.0)=5.7×10−6\mu_0 (3.0)(12.0/8.0) = 5.7 \times 10^{-6} T·m
  4. μ0(3.0)(12.0)2/(8.0)2=8.5×10−6\mu_0 (3.0)(12.0)^2/(8.0)^2 = 8.5 \times 10^{-6} T·m
  5. Zero, because the Amperian loop is larger than the current loop
Explanation: When you encounter Ampère's law problems, remember that ∮B⃗⋅dl⃗\oint \vec{B} \cdot d\vec{l} depends only on the current enclosed by your Amperian loop, not the loop's size or shape. Ampère's law states that ∮B⃗⋅dl⃗=μ0Ienclosed\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}. Here, your Amperian loop (radius 12.0 cm) is larger than the current-carrying wire loop (radius 8.0 cm), so the entire 3.0 A current is enclosed. Therefore: ∮B⃗⋅dl⃗=μ0(3.0)=3.8×10−6 T\cdotpm\oint \vec{B} \cdot d\vec{l} = \mu_0 (3.0) = 3.8 \times 10^{-6} \text{ T·m} This makes A correct. B incorrectly applies the ratio (8.0/12.0), suggesting the result depends on the relative sizes of the loops. This reflects a misunderstanding that Ampère's law somehow "scales" with loop geometry. C uses the inverse ratio (12.0/8.0), which might stem from incorrectly thinking that a larger Amperian loop somehow "captures more" of the magnetic field effect. D applies the ratio of areas (12.0)2/(8.0)2(12.0)^2/(8.0)^2, possibly confusing this with formulas where magnetic field strength depends on distance. This suggests mixing up Ampère's law with the Biot-Savart law or field calculations at specific points. Study tip: Ampère's law is beautifully simple—the line integral equals μ0\mu_0 times the enclosed current, period. The size and shape of your Amperian loop don't matter, only what current passes through it. Practice identifying what's "inside" versus "outside" your chosen loop.

Question 16

A student applies Ampère's law to a rectangular loop around a straight current-carrying wire. One side of the rectangle is parallel to the wire at distance 2.0 cm, and the opposite side is parallel at distance 6.0 cm. If the wire carries 12 A, what is ∮B⃗⋅dl⃗\oint \vec{B} \cdot d\vec{l} around this loop?

  1. 1.5×10−51.5 \times 10^{-5} T·m
  2. 3.0×10−53.0 \times 10^{-5} T·m
  3. 6.0×10−66.0 \times 10^{-6} T·m
  4. μ0(12 A)=1.5×10−5\mu_0 (12 \text{ A}) = 1.5 \times 10^{-5} T·m (correct answer)
  5. Zero, because the loop doesn't enclose current
Explanation: When you encounter Ampère's law problems, remember that the law states ∮B⃗⋅dl⃗=μ0Ienc\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}, where the line integral depends only on the current enclosed by your loop, not the loop's shape or size. For any closed loop around a current-carrying wire, the circulation of the magnetic field equals μ0\mu_0 times the enclosed current. Since your rectangular loop completely encloses the 12 A wire, the result is simply μ0(12 A)=1.5×10−5\mu_0 (12 \text{ A}) = 1.5 \times 10^{-5} T·m. The distances (2.0 cm and 6.0 cm) are irrelevant to this calculation—they're red herrings designed to make you overthink the problem. Answer D correctly applies Ampère's law and gives the right numerical value: μ0(12 A)=1.5×10−5\mu_0 (12 \text{ A}) = 1.5 \times 10^{-5} T·m. Answer A gives the correct numerical value but lacks the clear connection to Ampère's law that makes the physics transparent. Answer B (3.0×10−53.0 \times 10^{-5} T·m) suggests you might have doubled the current or the permeability constant—a common arithmetic error. Answer C (6.0×10−66.0 \times 10^{-6} T·m) is too small by a factor of 2.5, possibly from incorrectly using the given distances in your calculation or making an error with μ0\mu_0. Strategy tip: In Ampère's law problems, focus on what current the loop encloses, not the loop's geometry. Extra geometric information is often included to test whether you understand that only the enclosed current matters for the line integral.

Question 17

A long solenoid with cross-sectional area A=5.0×10−3A = 5.0 \times 10^{-3} m² and 300 turns/m carries 2.0 A. The magnetic flux through a single turn of the solenoid is:

  1. 3.8×10−63.8 \times 10^{-6} Wb (correct answer)
  2. 7.5×10−67.5 \times 10^{-6} Wb
  3. 1.9×10−61.9 \times 10^{-6} Wb
  4. 1.1×10−51.1 \times 10^{-5} Wb
  5. 2.4×10−52.4 \times 10^{-5} Wb
Explanation: When you encounter solenoid problems, remember that the magnetic field inside a long solenoid is uniform and depends only on the current and turn density, not the solenoid's dimensions. For a long solenoid, the magnetic field inside is B=μ0nIB = \mu_0 n I, where μ0=4π×10−7\mu_0 = 4\pi \times 10^{-7} T⋅m/A is the permeability of free space, nn is the turn density (turns per meter), and II is the current. First, calculate the magnetic field: B=(4π×10−7)(300)(2.0)=7.54×10−4B = (4\pi \times 10^{-7})(300)(2.0) = 7.54 \times 10^{-4} T. The magnetic flux through a single turn is Φ=BA\Phi = BA, since the field is perpendicular to the cross-sectional area: Φ=(7.54×10−4)(5.0×10−3)=3.77×10−6\Phi = (7.54 \times 10^{-4})(5.0 \times 10^{-3}) = 3.77 \times 10^{-6} Wb, which rounds to 3.8×10−63.8 \times 10^{-6} Wb. Answer A (3.8×10−63.8 \times 10^{-6} Wb) is correct using this straightforward approach. Answer B (7.5×10−67.5 \times 10^{-6} Wb) likely comes from doubling the correct answer or making an error in the magnetic field calculation. Answer C (1.9×10−61.9 \times 10^{-6} Wb) represents roughly half the correct value, possibly from using an incorrect formula or missing a factor of 2 somewhere. Answer D (1.1×10−51.1 \times 10^{-5} Wb) is too large by about a factor of 3, suggesting a significant computational error or wrong approach. Remember: for solenoid flux problems, always start with B=μ0nIB = \mu_0 n I for the internal field, then multiply by the cross-sectional area. The turn density and current are the key parameters—the solenoid's length doesn't directly affect the field inside.

Question 18

A solenoid has 200 turns uniformly distributed over a length of 0.50 m and carries a current of 4.0 A. If the solenoid is cut in half (keeping the same turn density), what happens to the magnetic field inside each half?

  1. It doubles because the length is halved
  2. It remains the same because turn density is unchanged (correct answer)
  3. It is halved because there are fewer total turns
  4. It becomes zero because the solenoid is no longer infinite
  5. It increases by a factor of 2\sqrt{2} due to geometry changes
Explanation: When analyzing magnetic fields in solenoids, you need to focus on the key factors that determine field strength. The magnetic field inside a long solenoid depends on the number of turns per unit length (turn density) and the current, following the formula B=μ0nIB = \mu_0 n I, where nn is turns per unit length and II is current. The original solenoid has 200 turns over 0.50 m, giving a turn density of 400 turns/m. When cut in half while maintaining the same turn density, each half-solenoid still has 400 turns/m - just over a shorter 0.25 m length. Since the turn density nn and current II remain unchanged, the magnetic field strength inside each half remains the same. Choice A incorrectly assumes that halving the length doubles the field, confusing length with turn density. The absolute length doesn't directly affect field strength - only the turns per unit length matters. Choice C falls into the trap of thinking fewer total turns means weaker field, but it's the concentration of turns (density) that counts, not the absolute number. Choice D wrongly suggests the field becomes zero, misunderstanding that while "infinite" solenoids are used in idealized calculations, real finite solenoids still produce substantial internal fields as long as they're reasonably long compared to their diameter. Remember: for solenoid problems, always think "turns per unit length," not total turns or absolute length. The magnetic field depends on how tightly packed the coils are, not how many total coils exist.

Question 19

An infinite current sheet carries surface current density K=3.0K = 3.0 A/m. A student attempts to apply Ampère's law using a rectangular Amperian loop with sides parallel and perpendicular to the sheet. If the loop has dimensions 0.20 m × 0.10 m, what should be the result of ∮B⃗⋅dl⃗\oint \vec{B} \cdot d\vec{l}?

  1. μ0(3.0)(0.20)=7.5×10−7\mu_0 (3.0)(0.20) = 7.5 \times 10^{-7} T·m (correct answer)
  2. μ0(3.0)(0.10)=3.8×10−7\mu_0 (3.0)(0.10) = 3.8 \times 10^{-7} T·m
  3. μ0(3.0)(0.02)=7.5×10−8\mu_0 (3.0)(0.02) = 7.5 \times 10^{-8} T·m
  4. Zero, no current is enclosed
  5. μ0(3.0)(0.30)=1.1×10−6\mu_0 (3.0)(0.30) = 1.1 \times 10^{-6} T·m
Explanation: When you encounter an infinite current sheet problem, you're dealing with a classic application of Ampère's law where the key insight is understanding which dimension of your Amperian loop actually matters for the magnetic field circulation. For an infinite current sheet, the magnetic field runs parallel to the sheet's surface and has the same magnitude on both sides. When you draw a rectangular Amperian loop that crosses the sheet, the magnetic field contributes to the line integral only along the sides parallel to the sheet - the perpendicular sides contribute zero because B⃗⊥dl⃗\vec{B} \perp d\vec{l} there. Since the field has equal magnitude but opposite directions on each side of the sheet, only one side of your rectangle (the one in the direction of B⃗\vec{B}) contributes positively to ∮B⃗⋅dl⃗\oint \vec{B} \cdot d\vec{l}. This means the circulation equals BB times the length of the side parallel to the sheet. For this loop, that's the 0.20 m dimension, giving us ∮B⃗⋅dl⃗=μ0K×0.20=μ0(3.0)(0.20)\oint \vec{B} \cdot d\vec{l} = \mu_0 K \times 0.20 = \mu_0(3.0)(0.20), which matches answer A. Answer B incorrectly uses the 0.10 m dimension - this would be right if that were the side parallel to the sheet. Answer C uses the area (0.02 m²), which is never correct for Ampère's law calculations. Answer D misunderstands current enclosure - the surface current density KK represents current per unit length flowing through the sheet itself. Remember: for current sheets, only the dimension of your loop that's parallel to the sheet (and perpendicular to the current flow) determines the magnetic circulation.

Question 20

Three long parallel wires carry currents I1=4.0I_1 = 4.0 A (into page), I2=6.0I_2 = 6.0 A (out of page), and I3=2.0I_3 = 2.0 A (into page). An Amperian loop encloses wires 1 and 3 but not wire 2. What is the net current enclosed by the loop?

  1. 12.0 A
  2. 6.0 A
  3. -6.0 A (correct answer)
  4. 2.0 A
  5. -2.0 A
Explanation: When you encounter Ampère's law problems involving multiple current-carrying wires, focus on one key principle: only the currents that pass through your Amperian loop contribute to the net enclosed current. The direction of current determines the sign. To find the net enclosed current, you need to identify which currents pass through the loop and apply the right-hand rule for sign convention. Since the loop encloses wires 1 and 3 but not wire 2, only I1I_1 and I3I_3 matter for this calculation. Wire 2's current is irrelevant regardless of its magnitude. Using the standard sign convention, currents going into the page are negative and currents coming out of the page are positive. Wire 1 carries 4.0 A into the page (−4.0-4.0 A) and wire 3 carries 2.0 A into the page (−2.0-2.0 A). The net enclosed current is (−4.0)+(−2.0)=−6.0(-4.0) + (-2.0) = -6.0 A. Looking at the wrong answers: A) 12.0 A incorrectly adds all three currents without considering signs or which wires are enclosed. B) 6.0 A either ignores the sign convention or incorrectly includes wire 2's current. D) 2.0 A suggests only considering wire 3's contribution while ignoring wire 1. Remember this strategy: for Amperian loop problems, first identify which currents are actually enclosed by your loop, then carefully apply sign conventions based on current direction. Currents outside the loop don't contribute to Ampère's law calculations, no matter how large they are.