Physics 2 Quiz: Amperes Law

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Amperes LawQuestion 1 of 8

An infinitely long solid cylindrical conductor of radius $R$ carries a total current $I$ . However, rather than being uniform, the current density varies with radial distance $r$ from the axis as $J(r) = J_0 \left(1 - \dfrac{r}{R}\right)$ , where $J_0$ is a constant and $0 \leq r \leq R$ .

Which expression correctly gives the magnetic field magnitude at a point inside the conductor at radius $r < R$ ?

B = \dfrac{\mu_0 J_0}{2}\left(r - \dfrac{r^2}{R}\right)

, obtained by evaluating the current density at the outer boundary

r

and multiplying by the enclosed cross-sectional area, rather than integrating.

B = \mu_0 J_0\left(\dfrac{r}{2} - \dfrac{r^2}{3R}\right)

, obtained by integrating

J(r')

over the disk of radius

r

to find

I_{\text{enc}}

, then applying

B = \mu_0 I_{\text{enc}}/(2\pi r)

B = \dfrac{\mu_0 J_0 r}{2}\left(1 - \dfrac{r}{R}\right)

, obtained by treating the current density as uniform at its local value

J(r)

and using the formula for a uniform cylinder.

B = \dfrac{\mu_0 J_0}{2\pi}\left(\dfrac{r}{2} - \dfrac{r^2}{3R}\right)

, obtained by correctly integrating

J(r')

but omitting the

2\pi

factor from the Amperian loop circumference on the left-hand side.

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Physics 2 Quiz

Physics 2 Quiz: Amperes Law

Practice Amperes Law in Physics 2 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 Physics 2.

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

Which expression correctly gives the magnetic field magnitude at a point inside the conductor at radius $r < R$ ?

$B = \dfrac{\mu_0 J_0}{2}\left(r - \dfrac{r^2}{R}\right)$ , obtained by evaluating the current density at the outer boundary $r$ and multiplying by the enclosed cross-sectional area, rather than integrating.
$B = \mu_0 J_0\left(\dfrac{r}{2} - \dfrac{r^2}{3R}\right)$ , obtained by integrating $J(r')$ over the disk of radius $r$ to find $I_{\text{enc}}$ , then applying $B = \mu_0 I_{\text{enc}}/(2\pi r)$ . (correct answer)
$B = \dfrac{\mu_0 J_0 r}{2}\left(1 - \dfrac{r}{R}\right)$ , obtained by treating the current density as uniform at its local value $J(r)$ and using the formula for a uniform cylinder.
$B = \dfrac{\mu_0 J_0}{2\pi}\left(\dfrac{r}{2} - \dfrac{r^2}{3R}\right)$ , obtained by correctly integrating $J(r')$ but omitting the $2\pi$ factor from the Amperian loop circumference on the left-hand side.

Explanation: Whenever you see a non-uniform current density inside a conductor, Ampère's Law still applies — but you must integrate

J(r')

over the enclosed area rather than plugging in a single value of

J

. The key relationship is

\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}

. For a circular Amperian loop of radius

r

inside the conductor, the left side gives

B(2\pi r)

. The right side requires computing

I_{\text{enc}} = \int_0^r J(r')\, 2\pi r'\, dr'

. Substituting

J(r') = J_0(1 - r'/R)

I_{\text{enc}} = 2\pi J_0 \int_0^r \left(r' - \frac{r'^2}{R}\right)dr' = 2\pi J_0\left(\frac{r^2}{2} - \frac{r^3}{3R}\right)

Setting

B(2\pi r) = \mu_0 I_{\text{enc}}

and dividing by

2\pi r

B = \mu_0 J_0\left(\frac{r}{2} - \frac{r^2}{3R}\right)

This confirms B is correct. Choice A multiplies

J(r)

— the local current density at radius

r

— by the area

\pi r^2

, as if the density were uniform throughout. This ignores how

J

actually varies from the axis outward. Choice C makes a similar conceptual error: it treats

J(r)

as a uniform density filling the whole cylinder of radius

r

, producing the wrong enclosed current. Choice D performs the integration correctly but forgets to divide by

2\pi r

from the left-hand side of Ampère's Law, leaving an extra

2\pi

in the denominator. Study tip: On non-uniform current problems, always set up the integral

\int J(r')\, 2\pi r'\, dr'

explicitly — treating

J

as constant is the most common trap, and it appears in two different disguises here (choices A and C).

Question 2

A hollow cylindrical conductor of inner radius $a$ and outer radius $b$ carries a total current $I$ distributed non-uniformly with current density $J(r) = \dfrac{C}{r}$ , where $C$ is a constant and $r$ is the radial distance from the axis ( $a \leq r \leq b$ ).

What is the magnetic field magnitude at a radial distance $r$ with $a < r < b$ ?

$B = \dfrac{\mu_0 C}{r}\ln\!\left(\dfrac{b}{a}\right)$ , obtained by computing the total current through the entire shell from $a$ to $b$ and using this full current in Ampère's Law regardless of the field point location inside the conductor.
$B = \dfrac{\mu_0 C \ln(r/a)}{r}$ , obtained by integrating $J(r') = C/r'$ but omitting the factor of $r'$ from the area element $dA$ , effectively computing $\int_a^r (C/r')\,dr'$ instead of $\int_a^r (C/r')\,2\pi r'\,dr'$ .
$B = \dfrac{\mu_0 C(r-a)}{2\pi r^2}$ , obtained by correctly computing $I_{\text{enc}} = 2\pi C(r-a)$ but then dividing by an extra factor of $2\pi r$ on the left-hand side of Ampère's Law rather than just $r$ .
$B = \dfrac{\mu_0 C(r-a)}{r}$ , obtained by integrating $J(r') = C/r'$ correctly over the annular area from $a$ to $r$ using $dA = 2\pi r'\,dr'$ , which causes the $r'$ factors to cancel and gives $I_{\text{enc}} = 2\pi C(r-a)$ . (correct answer)

Explanation: Whenever you see a non-uniform current density inside a conductor, Ampère's Law still applies — but you must carefully identify only the current enclosed within your Amperian loop, not the total current through the entire conductor. For a circular Amperian loop of radius

r

(with

a < r < b

), Ampère's Law gives

B(2\pi r) = \mu_0 I_{\text{enc}}

. The enclosed current requires integrating

J(r') = C/r'

over the annular area from

a

r

, using the area element

dA = 2\pi r'\,dr'

I_{\text{enc}} = \int_a^r \frac{C}{r'}\cdot 2\pi r'\,dr' = 2\pi C\int_a^r dr' = 2\pi C(r-a)

Notice the

r'

in the denominator of

J

cancels with the

r'

from

dA

, leaving a clean integral. Substituting into Ampère's Law:

B = \dfrac{\mu_0 \cdot 2\pi C(r-a)}{2\pi r} = \dfrac{\mu_0 C(r-a)}{r}

. This confirms D is correct. A uses the total current through the full shell (integrating from

a

b

) instead of only the enclosed portion up to

r

— a classic Ampère's Law misapplication. B drops the

r'

factor from

dA

, treating the area element as simply

dr'

rather than

2\pi r'\,dr'

, which is a geometry error. C computes

I_{\text{enc}}

correctly but then divides by

2\pi r \cdot 2\pi r

instead of

2\pi r

, introducing a phantom extra factor. Study tip: In Ampère's Law problems, always write out

dA = 2\pi r'\,dr'

explicitly before integrating — many errors come from forgetting this cylindrical area element entirely.

Question 3

A student is asked whether Ampère's Law $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}$ can be used to determine the magnetic field at a specific point on an Amperian loop, given a current configuration that does not possess the symmetry required to extract $B$ from the integral.

Which statement most precisely characterizes the relationship between Ampère's Law and symmetry in such a case?

Without symmetry, Ampère's Law still yields the correct field if one chooses an Amperian loop that passes through the point of interest perpendicularly, because perpendicular intersection ensures $\vec{B} \cdot d\vec{l} = 0$ everywhere except at the target point.
Ampère's Law is only valid when the current distribution possesses cylindrical, planar, or toroidal symmetry; in asymmetric configurations, the law itself breaks down and gives incorrect results for the line integral.
Ampère's Law remains a valid, exact statement about the line integral of $\vec{B}$ around any closed path, but without sufficient symmetry, one cannot algebraically extract the value of $B$ at a specific point from the integral — the law is true but not directly useful for field calculation at a point. (correct answer)
In the absence of symmetry, Ampère's Law can still be applied by replacing $\vec{B}$ with its magnitude averaged over the Amperian loop, which equals $\mu_0 I_{\text{enc}} / (2\pi r)$ for any loop of radius $r$ , regardless of the current geometry.

Explanation: Whenever you encounter a question about Ampère's Law and symmetry, the key distinction to keep in mind is the difference between a law being valid and a law being useful for extracting a specific quantity. These are not the same thing. Ampère's Law,

\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}

, is a fundamental law of electromagnetism — it holds exactly for any closed Amperian loop, regardless of the current geometry. The line integral on the left always equals

\mu_0 I_{\text{enc}}

. The problem arises when you try to solve for

B

at a point: to pull

B

outside the integral, you need symmetry to guarantee that

|\vec{B}|

is constant and

\vec{B} \parallel d\vec{l}

along the loop. Without that, the integral is correct but you have one equation containing infinitely many unknown values of

\vec{B}

, making it algebraically unsolvable for a specific point. This is exactly what C describes — the law remains exact, but it cannot directly yield

B

at a point. A is wrong because no loop geometry isolates a single point. Even if

\vec{B} \perp d\vec{l}

almost everywhere, the integral still mixes contributions from every point on the path, and you cannot isolate one location. B is wrong because it confuses applicability with utility. Ampère's Law never "breaks down" — it's always valid. Symmetry is required for calculation, not for the law's truth. D is wrong because averaging

\vec{B}

over an arbitrary loop does not equal

\mu_0 I_{\text{enc}}/(2\pi r)

in general; that formula only applies when full cylindrical symmetry is already established. A reliable study tip: on any Ampère's Law problem, first ask yourself two questions — is the law valid here? (always yes) and can I extract

B

from the integral? (only with symmetry). Keeping those questions separate will protect you from every distractor in this question family.

Question 4

A long coaxial cable consists of an inner solid cylindrical conductor of radius $a$ carrying a uniformly distributed current $I$ directed out of the page, surrounded by a thin outer cylindrical shell of radius $b$ (where $b > a$ ) carrying a current $2I$ directed into the page. The region between the conductors is vacuum.

At a radial distance $r$ such that $a < r < b$ , what is the magnitude of the magnetic field, and what is its direction (taking 'counterclockwise' as the positive sense when viewed from the front)?

$B = \dfrac{\mu_0 I}{2\pi r}$ , directed counterclockwise, because only the inner conductor's current is enclosed by the Amperian loop at this radius. (correct answer)
$B = \dfrac{\mu_0 I}{2\pi r}$ , directed clockwise, because the net enclosed current at this radius is $I$ directed out of the page, which by the right-hand rule produces a clockwise field when viewed from the front.
$B = \dfrac{3\mu_0 I}{2\pi r}$ , directed clockwise, because the Amperian loop at this radius encloses both the inner current $I$ (out) and must account for the outer shell current $2I$ (in) by superposition, giving a net magnitude of $3I$ .
$B = \dfrac{\mu_0 I}{\pi r}$ , directed counterclockwise, because the field from the inner conductor must be doubled to account for the image current induced in the outer conductor at radius $b$ .

Explanation: Whenever you see a magnetic field question involving a coaxial cable, your go-to tool is Ampère's Law:

\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}

. The key principle is that only the current enclosed by your chosen Amperian loop contributes to the field at that location. For a circular Amperian loop of radius

r

where

a < r < b

, the loop sits entirely in the vacuum gap between the two conductors. The only current threading through this loop is the inner conductor's current

I

, directed out of the page. The outer shell at radius

b

lies outside your loop, so its current

2I

contributes nothing. Applying Ampère's Law:

B(2\pi r) = \mu_0 I

, giving

B = \dfrac{\mu_0 I}{2\pi r}

. Using the right-hand rule with current pointing out of the page, curl your fingers around the wire — the field circles counterclockwise. That's answer A, the correct choice. Answer B reaches the right magnitude but gets the direction wrong. A current coming out of the page produces a counterclockwise field, not clockwise — B has the right-hand rule backwards. Answer C incorrectly includes the outer shell current in the enclosed current. Since the shell at radius

b > r

lies outside the Amperian loop, it is simply not enclosed and cannot be added by superposition in this context. Answer D invents a concept — "image current doubling" — that has no basis in magnetostatics for this configuration. Study tip: Always draw your Amperian loop first and ask: "What current passes through this loop?" Currents outside the loop are completely irrelevant to Ampère's Law, regardless of how large they are.

Question 5

Consider an infinite slab of conducting material of thickness $2d$ (extending from $z = -d$ to $z = +d$ ), carrying a uniform current density $\vec{J} = J_0 \hat{x}$ . The slab is infinite in the $x$ and $y$ directions.

A student wants to use Ampère's Law to find the magnetic field at a point outside the slab at height $z > d$ . Which Amperian loop geometry is valid and yields the correct result, and what is the magnitude of $\vec{B}$ at that exterior point?

A rectangular loop in the $yz$ -plane with sides at $+z$ and $-z$ is the correct geometry, and the field at height $z$ outside the slab is $B = \mu_0 J_0 z$ , because the enclosed current grows as $z$ increases beyond $d$ .
A rectangular loop in the $yz$ -plane with one long side at $z > d$ and one long side at $z = 0$ (the midplane) is the valid choice. The enclosed current is $J_0 d L$ , and the field at height $z$ is $B = \mu_0 J_0 d / 2$ , independent of $z$ .
A circular Amperian loop of radius $r$ in the $xy$ -plane centered on the $z$ -axis is the valid choice because it best exploits the cylindrical symmetry of the current distribution, giving $B = \mu_0 J_0 d / (\pi r)$ .
A rectangular loop in the $yz$ -plane with sides of length $L$ along $\hat{y}$ placed symmetrically at $+z$ and $-z$ (both outside the slab) is valid. By antisymmetry, $\vec{B}$ is in the $\pm\hat{y}$ direction and both long sides contribute $BL$ in the same sense, while the short sides contribute zero. This gives $B = \mu_0 J_0 d$ , independent of $z$ . (correct answer)

Explanation: When applying Ampère's Law to a symmetric current distribution, your first job is identifying the correct loop geometry — one that matches the symmetry of the field so that

\oint \vec{B} \cdot d\vec{\ell}

simplifies cleanly. For an infinite current slab with

\vec{J} = J_0\hat{x}

, the magnetic field must point in the

\pm\hat{y}

direction by symmetry (perpendicular to both

\hat{x}

and

\hat{z}

), and its magnitude depends only on

z

. Crucially, above and below the slab, the field points in opposite directions:

\vec{B} = +\mu_0 J_0 d\,\hat{y}

for

z > d

and

\vec{B} = -\mu_0 J_0 d\,\hat{y}

for

z < -d

. This antisymmetry is the key to choosing the right loop. Choice D places a rectangular loop symmetrically at

+z

and

-z

, both outside the slab. Along both long sides (length

L

, oriented along

\hat{y}

), the field is parallel to the path and contributes

BL

each. Because the field reverses direction on opposite sides but the loop traverses them in opposite senses, both sides add constructively:

\oint \vec{B}\cdot d\vec{\ell} = 2BL

. The enclosed current is

I_{enc} = J_0 \cdot (2d) \cdot L

. Setting

2BL = \mu_0 J_0 (2d)L

gives

B = \mu_0 J_0 d

, constant outside. Choice A is wrong because the enclosed current doesn't grow once you're fully outside the slab — all

2d

of current is already enclosed for any

z > d

. Choice B uses a loop with one side at

z = 0

where

\vec{B} = 0

by symmetry, which works but gives only half the circulation (one contributing side), leading to an error by a factor of 2. Choice C fails because the slab has no cylindrical symmetry — a circular loop in the

xy

-plane doesn't align with the field direction at all. Your strategy: always match your Amperian loop to the actual symmetry of

\vec{B}

, and check which sides of the loop give nonzero (and additive) contributions before applying the law.

Question 6

A toroidal solenoid has $N$ total turns of wire, a mean radius $R_{\text{mean}}$ , and carries current $I$ . The inner radius of the torus is $R_1$ and the outer radius is $R_2$ , with $R_1 < R_{\text{mean}} < R_2$ . A student applies Ampère's Law using three different circular Amperian loops — Loop 1 at radius $r_1 < R_1$ , Loop 2 at radius $r_2$ with $R_1 < r_2 < R_2$ , and Loop 3 at radius $r_3 > R_2$ — all centered on the toroid's axis.

Which statement correctly describes the results for all three loops and identifies a key assumption required for Ampère's Law to yield $B = 0$ on Loop 3?

Loop 1 gives $B = 0$ , Loop 2 gives $B = \mu_0 N I / (2\pi r_2)$ , and Loop 3 gives $B = \mu_0 N I / (2\pi r_3)$ because all $N$ turns thread through Loop 3 in the same sense as they do through Loop 2, just at a larger radius.
Loop 1 and Loop 3 both give $B = 0$ exactly, while Loop 2 gives $B = \mu_0 N I / (2\pi r_2)$ . For Loop 3, $B = 0$ requires that the toroid be ideal (all flux confined inside the torus), which means the net current threading Loop 3 is zero because each turn's outgoing and returning current cancel when $N$ full turns are enclosed. (correct answer)
Loop 1 gives $B = 0$ , Loop 2 gives $B = \mu_0 N I / (2\pi R_{\text{mean}})$ (using the mean radius regardless of loop position), and Loop 3 gives $B = 0$ because the field must vanish outside any closed current distribution by Gauss's Law for magnetism.
All three loops give $B = 0$ because the toroid's symmetry means the magnetic field is purely azimuthal and the line integral of a purely azimuthal field around any circle centered on the axis is identically zero by symmetry.

Explanation: Whenever you encounter an Ampère's Law problem involving a toroid, your first move should be identifying how much net current threads each loop — because that's the only thing that determines the magnetic field along a symmetric path. For a toroid with

N

turns carrying current

I

, think carefully about what each loop encloses. Loop 1 (inside the torus hole,

r_1 < R_1

) encloses zero current, so

B = 0

. Loop 2 (inside the windings,

R_1 < r_2 < R_2

) encloses all

N

turns going in one direction, giving

B = \mu_0 N I / (2\pi r_2)

. Loop 3 (outside the torus,

r_3 > R_2

) encloses all

N

turns going forward and all

N

return paths — net current is zero, so

B = 0

. This is exactly what answer B describes, and it correctly names the key assumption: the toroid must be ideal, meaning the wire wraps perfectly so every outgoing current is paired with a returning current when the full loop is enclosed. A is wrong because it assumes all

N

turns thread Loop 3 in only one sense — ignoring that each turn's return path also passes through the loop, canceling the contribution. C is wrong on two counts: it incorrectly uses

R_\text{mean}

instead of

r_2

in Loop 2's result, and it invokes Gauss's Law for magnetism (which governs flux, not circulation) to explain why

B = 0

outside. D is wrong because a purely azimuthal

\vec{B}

field does not automatically produce a zero line integral — in fact, Loop 2 has a nonzero result precisely because the field is azimuthal there. Your study tip: always count net enclosed current by tracking both the "go" and "return" segments of every wire through your Amperian loop. Sign mistakes here are the most common trap on toroid problems.

Question 7

A long straight wire of radius $R$ carries a current $I$ uniformly distributed over its cross-section. A second, identical wire runs parallel to the first at a center-to-center separation of $d \gg R$ , carrying the same current $I$ in the opposite direction. A student wishes to use Ampère's Law to find the magnetic field midway between the two wires.

Which statement correctly explains why Ampère's Law alone is insufficient to directly calculate the magnetic field at the midpoint between the two wires, and what method must instead be used?

Ampère's Law is insufficient because the two-wire system lacks the continuous rotational or translational symmetry needed to extract a single $B$ value from the line integral. The Biot-Savart Law (or superposition of the individual wire fields using Ampère's Law on each wire separately) must be applied instead. (correct answer)
Ampère's Law is insufficient because the midpoint lies on neither wire's axis, so the enclosed current is zero for any Amperian loop centered at the midpoint, making the law trivially true but uninformative regardless of the field value.
Ampère's Law is insufficient at the midpoint because the opposing currents create a net zero enclosed current for any symmetric Amperian loop, which by Ampère's Law requires $B = 0$ at the midpoint — but this conclusion is actually correct, so the law is both valid and sufficient here.
Ampère's Law cannot be applied to systems of multiple conductors because it is derived only for single-conductor configurations; for two or more wires, only the Biot-Savart Law is valid as a fundamental law.

Explanation: Whenever you see a question about Ampère's Law, ask yourself one critical question first: does this configuration have sufficient symmetry to make the line integral useful? Ampère's Law,

\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}

, is always mathematically true — but it only solves for

B

when you can argue that

|\vec{B}|

is constant along your chosen Amperian loop, letting you pull it out of the integral. A single infinite straight wire has perfect cylindrical symmetry, so a circular Amperian loop centered on it gives

B(2\pi r) = \mu_0 I_{\text{enc}}

cleanly. The two-wire system destroys that symmetry. No matter what loop shape you draw through the midpoint, the field magnitude varies along it — you cannot factor out a single

B

. The law gives you a true but unsolvable equation. The correct approach, as answer A states, is superposition: use Ampère's Law on each wire individually to find each wire's field at the midpoint, then vector-add the results. Answer B contains a subtle error: it claims the enclosed current is zero for any loop centered at the midpoint. A loop enclosing only one wire has nonzero

I_{\text{enc}}

, so "any loop" is false. The real problem is symmetry, not enclosed current. Answer C is dangerously wrong. A net-zero enclosed current in

\oint \vec{B} \cdot d\vec{l} = 0

does not require

B = 0

everywhere — it only means the integral cancels. The field at the midpoint is actually nonzero. Answer D is simply false. Ampère's Law is a fundamental law valid for any current configuration; it has no single-conductor restriction. Study tip: On exam questions about Ampère's Law, always check for symmetry before setting up a loop — if the geometry breaks cylindrical, planar, or solenoid-like symmetry, default to Biot-Savart plus superposition.

Question 8

An infinite solenoid of radius $R$ has $n$ turns per unit length and carries current $I$ . A student attempts to find the field inside by choosing a rectangular Amperian loop with both sides parallel to the solenoid axis — one side at radius $r_1 < R$ (inside) and one side at radius $r_2 > R$ (outside). The loop has length $L$ along the axis.

In applying Ampère's Law with this loop, which of the following correctly identifies all contributions to $\oint \vec{B} \cdot d\vec{l}$ and gives the correct enclosed current?

Both axial sides contribute — the inner side gives $+BL$ and the outer side gives $+B_{\text{fringe}}L$ from the non-zero fringe field outside the solenoid. The two perpendicular sides are zero. The enclosed current is $2nLI$ because the loop's two short sides each cross a set of windings, giving $B = \mu_0 nI$ coincidentally but for incorrect reasons.
All four sides contribute to the line integral because $\vec{B}$ has both axial and radial components inside a solenoid, arising from the helical winding geometry. The enclosed current is $nLI$ , but the radial contributions from the two perpendicular sides must be subtracted, reducing the result to $B = \mu_0 nI/2$ .
The two sides perpendicular to the axis contribute zero because $\vec{B}$ is axial while $d\vec{l}$ on those sides is radial. The outer axial side contributes zero because $B = 0$ outside an ideal solenoid. Only the inner axial side contributes $BL$ . The enclosed current is $nLI$ , giving $B = \mu_0 nI$ . (correct answer)
The outer axial side contributes $-BL$ (opposite in sense to the inner side) by Lenz's Law, since the outer field opposes any change in flux. The perpendicular sides contribute zero. The net line integral is therefore zero, consistent with zero net enclosed current, which would imply $B$ is the same inside and outside the solenoid.

Explanation: Whenever you apply Ampère's Law, your first job is to carefully evaluate

\oint \vec{B} \cdot d\vec{l}

by analyzing each segment of the loop separately — asking whether

\vec{B}

is parallel, perpendicular, or zero along that segment. For an ideal infinite solenoid, the magnetic field is purely axial (along the axis) inside and exactly zero outside. With a rectangular Amperian loop straddling the solenoid wall, you have four segments to consider. The two sides perpendicular to the axis run radially — since

d\vec{l}

is radial but

\vec{B}

is axial, the dot product

\vec{B} \cdot d\vec{l} = 0

on both. The outer axial side lies outside the solenoid where

B = 0

, contributing nothing. Only the inner axial side contributes, giving

+BL

. The enclosed current is the number of turns threading the loop, which is

nL

turns each carrying

I

, so

I_{\text{enc}} = nLI

. Ampère's Law then gives

BL = \mu_0 nLI

, confirming

B = \mu_0 nI

. This is answer C. Choice A is wrong because the field outside an ideal solenoid is strictly zero — there is no fringe field to contribute, and the enclosed current count of

2nLI

is invented. Choice B invents radial field components that don't exist in an ideal solenoid; the field is purely axial, so there are no radial contributions to subtract. Choice D misapplies Lenz's Law, which governs induced EMF, not static magnetic field direction — it has no role here, and the outer field is zero, not

-B

. Your study tip: when evaluating Amperian loop integrals, always check each side independently — is

\vec{B}

parallel or perpendicular to

d\vec{l}

, and is

|\vec{B}|

even nonzero there? Skipping this step is the root of every wrong answer here.