Ampère's Law - College Physics

1. An infinite current sheet carries surface current density $K = 3.0$ 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 $\oint \vec{B} \cdot d\vec{l}$?
   Answer: $\mu_0 (3.0)(0.20) = 7.5 \times 10^{-7}$ T·m 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 $\vec{B} \perp d\vec{l}$ 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 $\vec{B}$) contributes positively to $\oint \vec{B} \cdot d\vec{l}$. This means the circulation equals $B$ times the length of the side parallel to the sheet. For this loop, that's the 0.20 m dimension, giving us $\oint \vec{B} \cdot d\vec{l} = \mu_0 K \times 0.20 = \mu_0(3.0)(0.20)$, 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 $K$ 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.
2. 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?
   Answer: $2.0 \times 10^{-3}$ T 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: $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}$. Since the magnetic field is tangent to the circular path and has constant magnitude, this becomes $B \cdot 2\pi r = \mu_0 N I$, where N is the total number of turns and I is the current. Solving for B: $B = \frac{\mu_0 N I}{2\pi r} = \frac{(4\pi \times 10^{-7})(400)(2.5)}{2\pi(0.10)} = \frac{4 \times 10^{-4}}{0.20} = 2.0 \times 10^{-3}$ T. Answer A ($2.0 \times 10^{-3}$ T) is correct. Answer B ($1.0 \times 10^{-3}$ T) results from incorrectly doubling the radius or halving another parameter. Answer C ($4.0 \times 10^{-3}$ T) occurs if you forget to include the factor of 2π in the denominator. Answer D ($5.0 \times 10^{-4}$ 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.
3. 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? ($\mu_0 = 4\pi \times 10^{-7}$ T·m/A)
   Answer: $8.0 \times 10^{-5}$ T 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 $B = \frac{\mu_0 I}{2\pi r}$, where $I$ is the current and $r$ is the distance from the wire. Substituting the given values: $I = 8.0$ A, $r = 0.02$ m, and $\mu_0 = 4\pi \times 10^{-7}$ T·m/A. $B = \frac{(4\pi \times 10^{-7})(8.0)}{2\pi(0.02)} = \frac{32\pi \times 10^{-7}}{0.04\pi} = \frac{32 \times 10^{-7}}{0.04} = 8.0 \times 10^{-5}$ T This confirms answer A is correct. Answer B ($1.6 \times 10^{-4}$ T) is exactly double the correct value, suggesting you might have forgotten the factor of 2 in the denominator. Answer C ($3.2 \times 10^{-5}$ T) could result from using the wrong distance or making an arithmetic error in the calculation. Answer D ($6.4 \times 10^{-5}$ T) is close but likely comes from incorrectly manipulating the $2\pi$ factor. Remember that Ampère's law problems often involve careful attention to the geometric factor (like $2\pi$ for straight wires). Always double-check your formula setup before plugging in numbers, and convert distances to meters before calculating.
4. 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 $\oint \vec{B} \cdot d\vec{l}$ around this Amperian path?
   Answer: $\mu_0 (3.0) = 3.8 \times 10^{-6}$ T·m When you encounter Ampère's law problems, remember that $\oint \vec{B} \cdot d\vec{l}$ depends only on the current enclosed by your Amperian loop, not the loop's size or shape. Ampère's law states that $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}$. 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: $\oint \vec{B} \cdot d\vec{l} = \mu_0 (3.0) = 3.8 \times 10^{-6} \text{ T}\cdot\text{m}$ 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 $(12.0)^2/(8.0)^2$, 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 $\mu_0$ 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.
5. 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?
   Answer: It remains the same because turn density is unchanged 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 $B = \mu_0 n I$, where $n$ is turns per unit length and $I$ 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 $n$ and current $I$ 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.
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?
   Answer: Ampère's law cannot be used because the current path lacks sufficient symmetry 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 $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed}$, 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 $\vec{B}$ 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.
7. Two parallel infinite wires separated by distance $d = 8.0 \text{ cm}$ carry currents $I_1 = 6.0 \text{ A}$ and $I_2 = 3.0 \text{ A}$ in opposite directions. A student applies Ampère's Law using a circular Amperian loop of radius $r = 12.0 \text{ cm}$ centered at the midpoint between the wires. What is the value of $\oint \vec{B} \cdot d\vec{l}$ for this loop?
   Answer: $\mu_0 (6.0 - 3.0) = \mu_0 \times 3.0 \text{ A}$ Ampère's Law states $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}$, where $I_{\text{enc}}$ is the net current enclosed by the loop, taking into account the direction. Since the circular loop of radius 12.0 cm is centered at the midpoint and the wires are only 4.0 cm from the center (half of 8.0 cm separation), both wires are enclosed by the loop. Using the right-hand rule consistently, currents in opposite directions contribute with opposite signs to the net enclosed current: $I_{\text{enc}} = 6.0 - 3.0 = 3.0 \text{ A}$. Choice A incorrectly adds the currents without considering direction. Choice C ignores one current entirely. Choice D incorrectly assumes opposite currents always cancel the line integral, confusing this with the case where equal and opposite currents would give zero net enclosed current.
8. 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 $B = \mu_0 nI$ and gets $3.8 \times 10^{-3}$ T. The actual field at the center of this finite solenoid will be:
   Answer: Less than $3.8 \times 10^{-3}$ T because of finite-length end effects When analyzing solenoid magnetic fields, you need to distinguish between the idealized infinite solenoid and real finite solenoids. The formula $B = \mu_0 nI$ 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: $n = 400/0.20 = 2000 \text{ turns/m}$, so $B = (4\pi \times 10^{-7})(2000)(1.5) = 3.77 \times 10^{-3} \text{ T}$, which rounds to $3.8 \times 10^{-3} \text{ T}$. Answer A is wrong because while Ampère's law is exact, the formula $B = \mu_0 nI$ 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 $3.8 \times 10^{-3} \text{ T}$, 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.
9. 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?
   Answer: $1.0 \times 10^{-5}$ T 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 $\pi(0.04^2 - 0.02^2) = \pi(0.0016 - 0.0004) = 0.0012\pi \text{ m}^2$ The current density is $J = \frac{6.0 \text{ A}}{0.0012\pi \text{ m}^2} = \frac{5000}{\pi} \text{ A/m}^2$ The enclosed current at radius 3.0 cm includes only the annular region from 2.0 cm to 3.0 cm: $I_{enc} = J \times \pi(0.03^2 - 0.02^2) = \frac{5000}{\pi} \times \pi(0.0009 - 0.0004) = 5000 \times 0.0005 = 2.5 \text{ A}$ Using Ampère's law: $B(2\pi r) = \mu_0 I_{enc}$ $B = \frac{\mu_0 I_{enc}}{2\pi r} = \frac{(4\pi \times 10^{-7})(2.5)}{2\pi(0.03)} = 1.0 \times 10^{-5} \text{ T}$ Choice A ($2.0 \times 10^{-5}$ T) uses the wrong enclosed current calculation. Choice C ($4.0 \times 10^{-5}$ T) likely uses the full 6.0 A current incorrectly. Choice D ($3.0 \times 10^{-5}$ 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.
10. 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 $4.0 \times 10^{-5}$ T, what current flows through the wire?
    Answer: 10 A 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 $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}$. 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 $B \cdot 2\pi r = \mu_0 I$, where $r$ is the loop radius. Solving for current: $I = \frac{B \cdot 2\pi r}{\mu_0}$ Substituting the given values: $r = 0.050 \text{ m}$, $B = 4.0 \times 10^{-5} \text{ T}$, and $\mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A}$: $I = \frac{(4.0 \times 10^{-5})(2\pi)(0.050)}{4\pi \times 10^{-7}} = \frac{4.0 \times 10^{-6}}{4\pi \times 10^{-7}} = 10 \text{ A}$ This confirms answer A is correct. Answer B (5.0 A) results from forgetting the factor of $2\pi$ in the circumference. Answer C (2.5 A) comes from both missing $2\pi$ 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 $2\pi r$.
11. Two infinite parallel wires are separated by distance $d = 6.0 \text{ cm}$. Wire 1 carries current $I_1 = 4.0 \text{ A}$ upward, and wire 2 carries current $I_2 = 4.0 \text{ A}$ downward. A student chooses a rectangular Amperian loop with dimensions $8.0 \text{ cm} \times 10.0 \text{ cm}$ that encloses both wires. If the student then moves wire 2 to a position $12.0 \text{ cm}$ away from wire 1 (outside the original loop), how does $\oint \vec{B} \cdot d\vec{l}$ change?
    Answer: Changes from $\mu_0 \times 0 \text{ A}$ to $\mu_0 \times 4.0 \text{ A}$ When you encounter Ampère's law problems involving multiple wires, focus on which currents are enclosed by your loop and their directions. Ampère's law states that $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}$, where $I_{\text{enc}}$ is the net current enclosed by the loop. Initially, both wires are inside the rectangular loop. Wire 1 carries 4.0 A upward and wire 2 carries 4.0 A downward. Using the right-hand rule consistently around your loop, these currents contribute with opposite signs to the enclosed current. Therefore, $I_{\text{enc}} = 4.0 - 4.0 = 0$ A, making $\oint \vec{B} \cdot d\vec{l} = \mu_0 \times 0$. After moving wire 2 outside the loop (12.0 cm from wire 1), only wire 1 remains enclosed. Now $I_{\text{enc}} = 4.0$ A, so $\oint \vec{B} \cdot d\vec{l} = \mu_0 \times 4.0$ A. Answer A incorrectly suggests the initial value was $\mu_0 \times 8.0$ A, which would mean adding the currents instead of considering their directions. Answer B reverses the initial and final values entirely. Answer D fails to recognize that moving wire 2 outside changes the enclosed current from zero to 4.0 A. Study tip: Always carefully track which currents are enclosed and their relative directions. Ampère's law depends only on enclosed current—the exact positions of wires inside the loop don't matter, but whether they're inside or outside is crucial.
12. Two concentric circular Amperian loops of radii 3.0 cm and 6.0 cm surround a straight wire carrying current I. If $\oint \vec{B} \cdot d\vec{l} = 2.4 \times 10^{-6}$ T·m for the inner loop, what is $\oint \vec{B} \cdot d\vec{l}$ for the outer loop?
    Answer: $2.4 \times 10^{-6}$ T·m When you encounter Amperian loops and line integrals of magnetic fields, you're working with Ampère's Law, which states that $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}$, where $I_{enc}$ 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 $\oint \vec{B} \cdot d\vec{l}$ 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 $\oint \vec{B} \cdot d\vec{l} = 2.4 \times 10^{-6}$ T·m for the inner loop, it must also equal $2.4 \times 10^{-6}$ T·m for the outer loop. This confirms answer C. Answer A ($1.2 \times 10^{-6}$ T·m) incorrectly assumes the integral decreases with radius. Answer B ($4.8 \times 10^{-6}$ T·m) wrongly suggests the integral doubles with doubled radius. Answer D ($6.0 \times 10^{-6}$ 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 $\oint \vec{B} \cdot d\vec{l}$ remains constant regardless of loop radius.
13. 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?
    Answer: $1.9 \times 10^{-3}$ T 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 $B = \mu_0 n I$, where $\mu_0 = 4\pi \times 10^{-7}$ T·m/A is the permeability of free space, $n$ is the number of turns per unit length, and $I$ is the current. Substituting the given values: $n = 500$ turns/m and $I = 3.0$ A: $B = (4\pi \times 10^{-7})(500)(3.0) = 6\pi \times 10^{-4} \approx 1.9 \times 10^{-3}$ T This confirms answer A is correct. Looking at the wrong answers: B ($9.4 \times 10^{-4}$ T) appears to result from using $3\pi \times 10^{-4}$ instead of $6\pi \times 10^{-4}$, possibly from forgetting to multiply by the current or making an arithmetic error. C ($3.8 \times 10^{-3}$ T) is roughly double the correct answer, suggesting someone might have made an error with the $\pi$ factor or doubled something incorrectly. D ($6.3 \times 10^{-4}$ T) is close to $2\pi \times 10^{-4}$, 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 $4\pi \times 10^{-7}$ for $\mu_0$ and calculating $n \times I$ correctly.
14. 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:
    Answer: Equal to the field of a single solenoid by continuity 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 $B = \mu_0 n I$, where $n$ is the turn density and $I$ 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.
15. 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?
    Answer: The magnetic field decreases by a factor of 4 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 $B = \frac{\mu_0 I}{2R}$, 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: $B_{new} = \frac{\mu_0 (I/2)}{2(2R)} = \frac{\mu_0 I}{8R}$ Comparing to the original field $B_{original} = \frac{\mu_0 I}{2R}$, we get: $\frac{B_{new}}{B_{original}} = \frac{1/8R}{1/2R} = \frac{1}{4}$ 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.
16. Three long parallel wires carry currents $I_1 = 4.0$ A (into page), $I_2 = 6.0$ A (out of page), and $I_3 = 2.0$ A (into page). An Amperian loop encloses wires 1 and 3 but not wire 2. What is the net current enclosed by the loop?
    Answer: -6.0 A 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 $I_1$ and $I_3$ 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 ($-4.0$ A) and wire 3 carries 2.0 A into the page ($-2.0$ A). The net enclosed current is $(-4.0) + (-2.0) = -6.0$ 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.
17. 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)?
    Answer: $5.0 \times 10^{-4}$ T 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 $J = \frac{I_{total}}{\pi R^2}$, where $R$ is the conductor's radius. At distance $r$ from the center (where $r < R$), the enclosed current is $I_{enc} = J \cdot \pi r^2 = I_{total} \cdot \frac{r^2}{R^2}$. Using Ampère's law with a circular path of radius $r = 1.0$ mm: $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}$ Since the magnetic field is symmetric, $B \cdot 2\pi r = \mu_0 I_{total} \frac{r^2}{R^2}$ Solving for $B$: $B = \frac{\mu_0 I_{total} r}{2\pi R^2}$ Substituting values: $B = \frac{(4\pi \times 10^{-7})(8.0)(1.0 \times 10^{-3})}{2\pi (2.0 \times 10^{-3})^2} = 5.0 \times 10^{-4}$ T This confirms answer B is correct. A ($1.0 \times 10^{-3}$ T) likely results from using the full current instead of the enclosed current. C ($2.0 \times 10^{-3}$ T) might come from incorrectly applying the formula for points outside the conductor. D ($8.0 \times 10^{-4}$ 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.
18. 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 $\oint \vec{B} \cdot d\vec{l}$ around this loop?
    Answer: $\mu_0 (12 \text{ A}) = 1.5 \times 10^{-5}$ T·m When you encounter Ampère's law problems, remember that the law states $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}$, 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 $\mu_0$ times the enclosed current. Since your rectangular loop completely encloses the 12 A wire, the result is simply $\mu_0 (12 \text{ A}) = 1.5 \times 10^{-5}$ 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: $\mu_0 (12 \text{ A}) = 1.5 \times 10^{-5}$ 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 ($3.0 \times 10^{-5}$ T·m) suggests you might have doubled the current or the permeability constant—a common arithmetic error. Answer C ($6.0 \times 10^{-6}$ 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 $\mu_0$. 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.
19. A hollow cylindrical conductor has inner radius $a = 2.0 \text{ cm}$ and outer radius $b = 5.0 \text{ cm}$. It carries a total current $I = 15.0 \text{ A}$ uniformly distributed over its cross-sectional area. Using a circular Amperian loop of radius $r = 3.5 \text{ cm}$, what current is enclosed by this loop?
    Answer: $\frac{(3.5)^2 - (2.0)^2}{(5.0)^2 - (2.0)^2} \times 15.0 \text{ A} = \frac{8.25}{21} \times 15.0 \text{ A}$ When you encounter problems involving current through hollow conductors, you need to carefully consider which portion of the conductor actually carries current and apply the concept of current density. In a hollow cylindrical conductor, current flows only through the conducting material between the inner radius $a$ and outer radius $b$. The current density is uniform throughout this annular cross-section. To find the enclosed current, you must determine what fraction of the total conducting area lies within your Amperian loop. The total conducting area is $\pi(b^2 - a^2) = \pi(5.0^2 - 2.0^2) = 21\pi \text{ cm}^2$. Your Amperian loop at $r = 3.5 \text{ cm}$ encloses the area from the inner radius to the loop radius: $\pi(3.5^2 - 2.0^2) = 8.25\pi \text{ cm}^2$. The enclosed current is therefore $\frac{8.25}{21} \times 15.0 \text{ A}$, making choice A correct. Choice B incorrectly assumes the loop encloses the entire conductor, but at $r = 3.5 \text{ cm}$, you're still inside the outer radius. Choice C makes the error of calculating the fraction using the entire circular area $\pi r^2$ rather than just the conducting annular region. Choice D misunderstands the geometry—the loop at $3.5 \text{ cm}$ is outside the inner radius, so it does enclose current-carrying material. Study tip: For hollow conductors, always subtract the inner area when calculating current fractions. The key formula is: enclosed current = total current × (enclosed conducting area)/(total conducting area).
20. A solenoid with $n = 2000$ turns per meter carries a current $I = 1.5 \text{ A}$. A student chooses an Amperian loop that is a rectangle with one side of length $L = 0.15 \text{ m}$ inside the solenoid (parallel to the axis), one side of the same length outside the solenoid, and two sides of length $w = 0.08 \text{ m}$ connecting them (perpendicular to the axis). How many current-carrying wires are enclosed by this Amperian loop?
    Answer: $300$ wires, since $nL = 2000 \times 0.15 = 300$ The Amperian loop is a rectangle that cuts through the solenoid. The number of current-carrying wires (turns) enclosed by the loop is determined by how many turns the loop encircles as it passes through the solenoid. Since the solenoid has 2000 turns per meter and the loop extends 0.15 m along the solenoid's axis (length L), the number of enclosed turns is $n \times L = 2000 \times 0.15 = 300$. The width $w = 0.08$ m is perpendicular to the axis and doesn't affect the count of enclosed turns. Choice B incorrectly uses the perpendicular dimension. Choice C incorrectly multiplies by both dimensions as if counting turns in an area. Choice D ignores the actual length over which turns are enclosed.

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