Physics 2 Quiz: Magnetic Field Solenoid

14 questions · exam conditions

0:00

Magnetic Field SolenoidQuestion 1 of 14

A toroidal solenoid (toroid) has $N$ total turns uniformly wound on a torus with inner radius $a$ and outer radius $b$ . A student claims that for the purposes of computing the magnetic field inside the windings (i.e., at radius $r$ where $a < r < b$ ), the toroid can be treated as an equivalent straight ideal solenoid with $n = N/(2\pi r)$ turns per unit length at that radius.

Which of the following statements best evaluates the student's claim and identifies the correct expression for the magnetic field inside the toroid at radius $r$ (where $a < r < b$ )?

The claim is correct and complete. The field is

B = \mu_0 n I = \mu_0 N I / (2\pi r)

, and this value is uniform throughout the interior of the toroid because the average radius

\bar{r} = (a+b)/2

is used consistently.

The claim is correct as a computational shortcut but physically incomplete: the field is

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

, and it varies with $r$ inside the winding region, being strongest near

r = a

and weakest near

r = b

, unlike a straight solenoid where

B

is uniform.

The claim is incorrect because a toroid cannot be modeled as a straight solenoid. The correct field inside the winding is

B = \mu_0 N I / (2\pi) \cdot \ln(b/a)

, which accounts for the varying path length around the torus.

The claim is incorrect because the Amperian loop for a toroid does not enclose the current in the same way as for a straight solenoid. The correct interior field is

B = \mu_0 N I / [\pi(a+b)]

, which uses the mean circumference of the torus as the effective path length.

← Back to quizzes

Physics 2 Quiz

Physics 2 Quiz: Magnetic Field Solenoid

Practice Magnetic Field Solenoid 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 Magnetic Field Solenoid, 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 of the following statements best evaluates the student's claim and identifies the correct expression for the magnetic field inside the toroid at radius $r$ (where $a < r < b$ )?

The claim is correct and complete. The field is $B = \mu_0 n I = \mu_0 N I / (2\pi r)$ , and this value is uniform throughout the interior of the toroid because the average radius $\bar{r} = (a+b)/2$ is used consistently.
The claim is correct as a computational shortcut but physically incomplete: the field is $B = \mu_0 N I / (2\pi r)$ , and it varies with $r$ inside the winding region, being strongest near $r = a$ and weakest near $r = b$ , unlike a straight solenoid where $B$ is uniform. (correct answer)
The claim is incorrect because a toroid cannot be modeled as a straight solenoid. The correct field inside the winding is $B = \mu_0 N I / (2\pi) \cdot \ln(b/a)$ , which accounts for the varying path length around the torus.
The claim is incorrect because the Amperian loop for a toroid does not enclose the current in the same way as for a straight solenoid. The correct interior field is $B = \mu_0 N I / [\pi(a+b)]$ , which uses the mean circumference of the torus as the effective path length.

Explanation: Whenever you encounter a question about magnetic fields in a toroid, your anchor concept should be Ampère's Law:

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

. The key insight is that you choose a circular Amperian loop of radius

r

centered on the toroid's axis, and symmetry guarantees

B

is constant along that loop. For a circular path at radius

r

(where

a < r < b

), the loop encloses all

N

turns carrying current

I

, so

I_{enc} = NI

. Applying Ampère's Law:

B(2\pi r) = \mu_0 NI

, which gives

B = \mu_0 NI / (2\pi r)

. The student's shortcut of writing

n = N/(2\pi r)

as a local turns-per-unit-length is mathematically equivalent and gives the right formula — but it obscures a critical physical truth: because

r

appears in the denominator, the field is not uniform inside the winding. It is strongest near the inner radius

a

and weakest near the outer radius

b

. This makes B the correct answer. Choice A is wrong because it claims the field is uniform inside the toroid — it is not. Uniformity holds for an ideal straight solenoid, not a toroid, where the varying path circumference creates a radial dependence. Choice C is wrong because

\mu_0 NI \ln(b/a)/(2\pi)

is actually the formula for the flux through the toroid's cross-section, not the field at a point. Choice D is wrong because using the mean circumference

\pi(a+b)

gives only an average approximation, not the exact field at a specific

r

. Your study tip: always distinguish between the field at a point (found via Ampère's Law with a loop through that point) and derived integrated quantities like flux. Confusing these is one of the most common traps in magnetostatics problems.

Question 2

An ideal solenoid of length $L$ , radius $R$ , total turns $N$ , and current $I$ is oriented along the $z$ -axis. A long straight wire carrying current $I_0$ is placed along the central axis of the solenoid. Which of the following correctly describes the magnetic field inside the solenoid far from its ends?

The field inside is $B = \mu_0 n I + \mu_0 I_0 / (2\pi r)$ , directed azimuthally near the axis and axially far from it, because the solenoid field is axial while the wire's field curls around the $z$ -axis at radius $r$ .
The field inside is purely axial with magnitude $B = \mu_0 n I$ , unchanged from the solenoid alone, because the long straight wire on the axis produces a field that is entirely azimuthal ( $\hat{\phi}$ direction) and therefore orthogonal to and independent of the axial solenoid field.
The field inside is purely axial with magnitude $B = \mu_0 n I + \mu_0 I_0 / (2\pi R)$ , where $R$ is the solenoid radius, because the wire's field must be evaluated at the solenoid wall where it is largest and adds to the interior field.
The field inside is $\sqrt{(\mu_0 n I)^2 + (\mu_0 I_0 / 2\pi r)^2}$ , directed at an angle to the axis that depends on $r$ , and the net axial component equals $\mu_0 n I$ while the net azimuthal component equals $\mu_0 I_0/(2\pi r)$ at every interior point. (correct answer)

Explanation: When multiple current sources exist in the same region, you must apply superposition carefully — not just add magnitudes, but recognize that fields pointing in different directions combine as vectors, not scalars. Here, the solenoid produces a uniform axial field

B_{sol} = \mu_0 n I

in the

\hat{z}

direction throughout its interior (where

n = N/L

). The central wire produces a field

B_{wire} = \mu_0 I_0 / (2\pi r)

that curls azimuthally in the

\hat{\phi}

direction at every interior point a distance

r

from the axis. These two fields are perpendicular to each other at every interior point — one axial, one azimuthal — so you cannot simply add their magnitudes. Instead, the net field magnitude is the vector sum:

B_{net} = \sqrt{(\mu_0 n I)^2 + (\mu_0 I_0 / 2\pi r)^2}

, and its direction tilts at an angle that depends on

r

. This makes D correct. A is wrong because it claims the field is "directed azimuthally near the axis and axially far from it," which mischaracterizes the geometry — both components exist simultaneously at every interior point, not in separate regions. B is the most tempting trap: it correctly identifies that the wire's field is azimuthal and the solenoid's is axial, but then wrongly concludes the wire's field can simply be ignored. Orthogonality means you add them as perpendicular vectors — the wire's field is real and measurable inside. C incorrectly evaluates the wire's field at

r = R

(the wall) and treats it as a uniform axial addition, which is wrong on both counts — the wire field is azimuthal and varies with

r

. Study tip: Whenever fields from two sources are perpendicular, always use

B = \sqrt{B_1^2 + B_2^2}

— never

B_1 + B_2

. Orthogonal doesn't mean ignorable.

Question 3

A finite solenoid of length $L$ and radius $R$ has $n$ turns per unit length and carries current $I$ . An ideal (infinite) solenoid with the same $n$ and $I$ would have field $B_0 = \mu_0 n I$ everywhere inside. For the finite solenoid, the field on the central axis at one end (i.e., at the rim of the solenoid, on-axis) is exactly $B_0/2$ .

A student uses the known end-field result to argue: 'If I place two identical finite solenoids end-to-end, aligned coaxially with currents in the same direction, the field at the junction between them equals $B_0/2 + B_0/2 = B_0$ .' Under what condition is this argument most physically meaningful, and what subtlety limits its practical usefulness?

The argument is always valid regardless of solenoid dimensions. Each solenoid independently contributes $B_0/2$ at its own end, and since the junction is simultaneously the end of both solenoids, superposition gives exactly $B_0$ under all circumstances with no limitations.
The argument is valid only when $L \ll R$ (short solenoid limit), because only then does each solenoid behave like a flat coil whose axial field at the edge equals half its central value. For long solenoids, the $B_0/2$ end-field result overestimates the actual junction contribution.
The argument is most useful when $L \gg R$ (long solenoid limit). In this limit, $B_0 = \mu_0 n I$ accurately describes the nearly uniform interior field, so the junction field approaching $B_0$ is physically meaningful. For short solenoids ( $L \sim R$ ), the interior field is non-uniform and $B_0$ no longer represents a well-defined interior value, so the superposition result, while mathematically correct, loses its practical significance. (correct answer)
The argument is valid only for the axial component of the field. At the junction, radial fringe-field components from each solenoid point in opposite directions and cancel by symmetry, but they induce a back-reaction that reduces the net axial field below $B_0$ by an amount proportional to $(R/L)^2$ .

Explanation: Whenever you see a superposition argument applied to solenoids, ask yourself: what physical regime makes the underlying formula meaningful? The end-field result

B_{end} = B_0/2

is mathematically exact for any finite solenoid, but

B_0 = \mu_0 n I

only represents a well-defined, uniform interior field when

L \gg R

. C is correct because the student's superposition argument is mathematically valid in all cases — placing two solenoids end-to-end genuinely produces

B_0/2 + B_0/2 = B_0

at the junction. The real question is whether

B_0

is a meaningful benchmark. In the long solenoid limit (

L \gg R

), the interior field is nearly uniform and well-approximated by

\mu_0 n I

, so saying the junction recovers

B_0

is physically informative: the seam essentially "heals" the edge effect. For short solenoids (

L \sim R

), the field is non-uniform throughout and

B_0

doesn't represent any characteristic interior value — so the result, while correct, tells you little. A is wrong because it ignores that "valid" and "physically meaningful" are different things. Superposition always works, but the significance of the result depends on context. B is wrong and reverses the physics entirely. The

B_0/2

end-field result holds for any aspect ratio; short solenoids (

L \ll R

) are precisely where

B_0

becomes meaningless, not where the formula is uniquely valid. D is wrong because it invents a fictitious "back-reaction" mechanism. Radial fringe fields do exist but don't reduce the axial superposition — that's not how magnetostatics works. Study tip: On questions mixing superposition with solenoid physics, always separate mathematical correctness from physical significance — examiners love to test whether you know the difference.

Question 4

An ideal solenoid with $n$ turns per unit length, radius $R$ , and current $I$ has a magnetic field $B = \mu_0 n I$ inside and zero outside. A student claims that if the solenoid is bent into a complete torus (a toroid) without changing $n$ , $R$ , or $I$ , the field inside the toroid will still equal $\mu_0 n I$ everywhere, since the local winding geometry is unchanged. Which response best evaluates this claim?

The claim is correct. Bending the solenoid into a toroid preserves the local winding geometry, so by Ampere's law applied locally, the field at every interior point remains $\mu_0 n I$ . The toroidal curvature has no effect on the local field magnitude.
The claim is incorrect. Bending the solenoid into a toroid changes the effective turns density seen by an Amperian loop: inner turns are closer together (higher local $n$ ) and outer turns are farther apart (lower local $n$ ). The field varies as $B(r) = \mu_0 N I/(2\pi r)$ , where $n$ in the original solenoid is the turns density at the mean radius, and the field is not uniform. (correct answer)
The claim is incorrect. The toroidal geometry completely confines the field inside the winding, which reduces the effective field below $\mu_0 n I$ everywhere because the field lines must curve to follow the torus, requiring additional energy that reduces the magnitude.
The claim is partially correct. The field inside the toroid equals $\mu_0 n I$ only at the mean radius $r = (a+b)/2$ , where $a$ and $b$ are the inner and outer radii. At all other interior points, the field is higher (for $r < (a+b)/2$ ) or lower (for $r > (a+b)/2$ ) due to the curvature, but the average field over the cross-section equals $\mu_0 n I$ exactly.

Explanation: Whenever you see a question comparing a solenoid to a toroid, the key tool is Ampere's law applied globally, not just a local geometric argument. The law states

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

, and the shape of your Amperian loop determines everything. For a toroid with

N

total turns, choose a circular Amperian loop of radius

r

centered on the toroid's axis. The enclosed current is

N I

, and by symmetry

B

is constant along the loop, giving

B(2\pi r) = \mu_0 N I

, so

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

. This field clearly depends on

r

: it is stronger near the inner radius and weaker near the outer radius. The original solenoid's uniform

n

becomes a non-uniform effective turns-per-length around the torus — inner windings are compressed (higher local density) and outer windings are stretched (lower local density) — making the field nonuniform. Answer B captures this exactly. Answer A is the trap the question is built around. Local winding geometry looks unchanged, but Ampere's law is a global statement. The curvature of the path means the same

N

turns contribute differently at different radii, and you cannot ignore that by reasoning "locally." Answer C invents a fictional energy-based reduction mechanism. Toroidal geometry confines the field (which is actually an advantage), but confinement does not reduce field magnitude. Answer D is a plausible-sounding compromise, but it's wrong:

\mu_0 n I

does not hold at the mean radius specifically, nor does the cross-section average equal

\mu_0 n I

in general. Study tip: When geometry changes, always re-apply Ampere's law from scratch — local intuition can mislead you when path curvature matters.

Question 5

An ideal solenoid is stretched to twice its length with same total turns and current. What happens to B?

Quartered
Unchanged
Doubled
Halved (correct answer)

Explanation: The magnetic field inside a solenoid is B = mu0 times (turns per length) times current. Stretching it to twice the length with the same total turns cuts the turns per length in half, so B is halved. The tempting wrong answer is unchanged, but the total number of turns doesn't set B; the turns per unit length does.

Question 6

Solenoid 1 has n turns/m and current I; solenoid 2 has 2n turns/m and current I/3. What is $B_2/B_1$ ?

2/3 (correct answer)
3/2
6
1/6

Explanation: Solenoid field equals mu_0 times turns per meter times current. Solenoid 2 gives mu_0 times 2n times I/3, which is (2/3)mu_0 nI, so B2/B1 = 2/3. A tempting mistake is 3/2, but that flips the ratio and ignores that the one-third current outweighs the doubled turns.

Question 7

An ideal solenoid of radius r has interior B. If radius doubles but N, L, I are fixed, what happens to B?

Doubles
Halves
The same (correct answer)
4 times

Explanation: The field inside an ideal solenoid is B = mu0 N I / L, which depends only on the number of turns, current, and length, not on the radius. So doubling r changes nothing about B. The tempting mistake is to think the field spreads out over a larger area and weakens, but B is set by the winding density and current, not the cross-sectional area.

Question 8

An ideal solenoid is 5.0 cm long, has 200 turns, and carries 2.0 A. What is B inside?

0.10 mT
1.0 mT
10 mT (correct answer)
100 mT

Explanation: The field inside a solenoid is mu_0 n I. Convert length to meters: n = 200 / 0.050 = 4,000 turns/m. Then B = (4 pi x 10^-7)(4,000)(2.0) ≈ 0.010 T = 10 mT. A tempting wrong answer is 0.10 mT, which comes from leaving length in cm as turns/cm instead of converting to turns/m.

Question 9

An ideal solenoid has 500 turns/m and current 2.0 A. Viewed from the left, current is clockwise. What is B inside?

1.3 mT, leftward
1.3 mT, rightward (correct answer)
0.63 mT, rightward
0.63 mT, leftward

Explanation: Inside a solenoid, B = mu0 n I = (4 pi x 10^-7)(500)(2) = 1.3 mT. For clockwise current viewed from the left, your right hand curls clockwise and your thumb points away from you, so the field inside points rightward. The tempting wrong choice is 1.3 mT leftward: the magnitude is right, but that direction would mean counterclockwise when viewed from the left.

Question 10

A solenoid is wound with two separate layers of wire. The first layer has $n_1 = 1500 \text{ turns/m}$ and carries current $I_1 = 2.0 \text{ A}$ . The second layer, wound directly on top of the first, has $n_2 = 1000 \text{ turns/m}$ and carries current $I_2 = 3.0 \text{ A}$ . The currents in the two layers circulate in the same direction when viewed from the same end of the solenoid. The two layers have different wire gauges but the same total length $L$ .

What is the magnitude of the magnetic field deep inside this double-wound solenoid, and which single-layer solenoid configuration is equivalent to it?

$B = \mu_0(n_1 I_1 + n_2 I_2) = \mu_0(3000 + 3000) = 6000\mu_0 \approx 7.54 \text{ mT}$ ; equivalent only to a single solenoid with $n_1 + n_2 = 2500 \text{ turns/m}$ carrying the total current $I_1 + I_2 = 5.0 \text{ A}$ , since both the turns density and current must be combined.
$B = \mu_0(n_1 + n_2)\bar{I} = \mu_0(2500)(2.5) = 6250\mu_0 \approx 7.85 \text{ mT}$ ; equivalent to a solenoid with combined turns density $n_1 + n_2 = 2500 \text{ turns/m}$ carrying the mean current $\bar{I} = (I_1 + I_2)/2 = 2.5 \text{ A}$ .
$B = \mu_0(n_1 I_1 + n_2 I_2) = \mu_0(3000 + 3000) = 6000\mu_0 \approx 7.54 \text{ mT}$ ; equivalent to a single solenoid with any $n$ and $I$ satisfying $nI = 6000 \text{ A/m}$ , such as $n = 3000 \text{ turns/m}$ and $I = 2.0 \text{ A}$ . (correct answer)
$B = \mu_0 n_1 I_1 = 3000\mu_0 \approx 3.77 \text{ mT}$ ; equivalent to a single solenoid with $n_1 = 1500 \text{ turns/m}$ and $I_1 = 2.0 \text{ A}$ , because only the innermost winding layer contributes to the field at the center — the outer layer's field is screened by the inner layer.

Explanation: When dealing with a multi-layer solenoid, the key principle is superposition: each layer independently creates its own magnetic field deep inside, and you simply add them. For an ideal solenoid, the field is

B = \mu_0 n I

, where

n

is turns per meter and

I

is the current. Because both layers' currents circulate in the same direction, their fields point the same way and add constructively. Applying superposition here gives

B = \mu_0 n_1 I_1 + \mu_0 n_2 I_2 = \mu_0(1500)(2.0) + \mu_0(1000)(3.0) = \mu_0(3000 + 3000) = 6000\mu_0 \approx 7.54 \text{ mT}

. Any single-layer solenoid that produces this same field only needs to satisfy

nI = 6000 \text{ A/m}

— for instance,

n = 3000 \text{ turns/m}

with

I = 2.0 \text{ A}

, or countless other combinations. That's exactly what C states, making it correct. A gets the field calculation right but claims the only equivalent solenoid must use

n_1 + n_2

with

I_1 + I_2

. This is false — any

(n, I)

pair satisfying

nI = 6000

works; there's no unique equivalent configuration. B incorrectly uses the mean current rather than each layer's actual current. Averaging the currents has no physical basis here — you need to weight each layer's contribution by its own

n

and

I

. D introduces a fictitious "shielding" effect. Solenoid layers don't screen each other magnetically; both layers contribute fully to the interior field. Study tip: Whenever you see multiple current-carrying layers in a solenoid, immediately think superposition — calculate

\mu_0 n_i I_i

for each layer separately and sum them. There's no shielding, no averaging, just addition.

Question 11

An ideal solenoid has $N$ turns, length $L$ , cross-sectional radius $R$ , and carries current $I$ . The wire is now replaced with a superconducting wire of the same gauge, and the solenoid geometry is unchanged. The current is then slowly increased to $2I$ while the solenoid is maintained below its critical temperature. Simultaneously, the solenoid length is doubled to $2L$ by adding an equal number of turns (so total turns become $2N$ ) while keeping $R$ constant. What is the ratio of the new magnetic field to the original magnetic field?

$B_{new}/B_{old} = 2$ , because although the length doubles, the number of turns doubles as well, keeping $n$ constant, while the current doubles; these two effects multiply to give a net factor of 2. (correct answer)
$B_{new}/B_{old} = 4$ , because both the turns density $n = N/L$ and the current $I$ double independently, and since $B = \mu_0 n I$ , the field scales as the product of both factors, yielding a factor of $2 \times 2 = 4$ .
$B_{new}/B_{old} = 1$ , because adding turns to double the length keeps $n = N/L$ unchanged, and the superconducting property forces the total magnetic flux through the solenoid to remain constant regardless of changes in current.
$B_{new}/B_{old} = 2\sqrt{2}$ , because the field is proportional to $\sqrt{N \cdot I}$ for a solenoid, and doubling both $N$ and $I$ gives a factor of $\sqrt{2}\cdot\sqrt{2} = 2$ from turns and $\sqrt{2}$ from current, yielding $2\sqrt{2}$ .

Explanation: Whenever you see a solenoid problem with multiple simultaneous changes, your anchor should be the field formula

B = \mu_0 n I

, where

n = N/L

is the turn density (turns per unit length). Apply each change to this formula systematically before combining them. Here, the length doubles to

2L

and the number of turns doubles to

2N

, so the new turn density is

n_{new} = 2N/2L = N/L

. The turn density is unchanged. However, the current doubles from

I

2I

. Plugging into the formula:

B_{new} = \mu_0 n_{new}(2I) = \mu_0 (N/L)(2I) = 2\mu_0 nI = 2B_{old}

. The ratio is exactly 2, confirming answer A. Answer B claims the turn density doubles, but this is wrong — adding turns proportionally to the added length keeps

n

constant. Doubling

N

alone would double

n

, but doubling both

N

and

L

together cancels out, leaving

n

the same. The factor-of-4 result follows from a false premise. Answer C invokes flux conservation in a superconductor, but this only applies to a closed superconducting loop carrying a fixed, externally-unchanged current. Here, the current is being driven externally from

I

2I

, so flux conservation does not freeze the field. Answer D uses a fabricated formula

B \propto \sqrt{N \cdot I}

, which has no physical basis. The correct dependence is linear in both

n

and

I

. Study tip: When a solenoid problem changes both

N

and

L

, always compute

n = N/L

first — it often simplifies dramatically before you even touch the current.

Question 12

Two solenoids share the same axis. Solenoid 1 has $n_1 = 1000 \text{ turns/m}$ and carries current $I_1 = 3.0 \text{ A}$ . Solenoid 2 has $n_2 = 2000 \text{ turns/m}$ and carries current $I_2 = 1.0 \text{ A}$ . The solenoids are wound in opposite senses (i.e., viewed from the same end, the currents circulate in opposite directions). Solenoid 2 is physically located entirely inside Solenoid 1.

What is the magnitude of the net magnetic field in the interior region where both solenoids overlap, far from any end effects?

$B = \mu_0 |n_1 I_1 - n_2 I_2| = \mu_0 |3000 - 2000| = 1000\mu_0 \approx 1.26 \text{ mT}$ , because the fields are antiparallel and the net field is the difference of $n_1 I_1$ and $n_2 I_2$ . (correct answer)
$B = \mu_0 (n_1 I_1 + n_2 I_2) = \mu_0 (3000 + 2000) = 5000\mu_0 \approx 6.28 \text{ mT}$ , because both solenoids contribute positive flux in the overlapping region regardless of winding direction.
$B = \mu_0 n_2 I_2 = 2000\mu_0 \approx 2.51 \text{ mT}$ , because Solenoid 2, being inside Solenoid 1, shields its interior from the outer solenoid's field by Lenz's law, so only Solenoid 2's field persists in the overlap region.
$B = \mu_0 (n_1 I_1 - n_2 I_2)/2 = \mu_0(3000-2000)/2 = 500\mu_0 \approx 0.63 \text{ mT}$ , because the net field in the overlap region is the arithmetic mean of the two individual fields, weighted by their respective volumes.

Explanation: When two solenoids share an axis, you treat each one as an independent source of magnetic field and apply superposition — the net field is the vector sum of the individual fields. The key detail here is the winding direction, which determines whether each solenoid's field points along the same axis direction or opposite directions. For a single ideal solenoid, the interior field is

B = \mu_0 n I

, directed along the axis according to the right-hand rule. Here, Solenoid 1 produces

B_1 = \mu_0 (1000)(3.0) = 3000\mu_0

and Solenoid 2 produces

B_2 = \mu_0 (2000)(1.0) = 2000\mu_0

. Because the solenoids are wound in opposite senses, their fields point in opposite directions inside the overlap region. Superposition then gives a net magnitude of

|B_1 - B_2| = \mu_0|3000 - 2000| = 1000\mu_0 \approx 1.26 \text{ mT}

, confirming that A is correct. B is wrong because it ignores winding direction entirely — adding the magnitudes only applies when fields are parallel (same winding sense). C misapplies Lenz's law: that principle governs induced currents from changing flux, not the static superposition of steady fields. A steady solenoid field cannot "shield" anything by Lenz's law. D invents an averaging rule that has no physical basis; superposition adds fields at each point in space, not averages them by volume. A useful habit: whenever a problem mentions winding direction or current circulation, immediately assign a directional sign (

+

-

) to each field before doing any arithmetic. Opposite windings mean subtraction; same windings mean addition.

Question 13

An ideal solenoid has $n$ turns per unit length, radius $R$ , and carries current $I$ . A student argues that if the solenoid is stretched uniformly so that its length doubles while the wire is re-wound to maintain the same total number of turns $N$ , the magnetic field inside doubles because the solenoid is now longer. A second student argues the field is halved. A third student claims the field is unchanged. Which student is correct, and what is the correct field after stretching?

The first student is correct. Stretching the solenoid doubles its length, which increases the path length for Ampere's law integration, effectively doubling the enclosed current linkage per unit length and therefore doubling $B$ .
The second student is correct. Stretching doubles the length while $N$ is fixed, so the turns density drops to $n' = N/(2L) = n/2$ . Applying $B = \mu_0 n' I$ gives $B' = \mu_0(n/2)I = B/2$ , so the field is halved. (correct answer)
The third student is correct. For an ideal solenoid, the field depends only on the current and the total number of turns, not on how those turns are distributed along the length, so $B = \mu_0 N I$ remains unchanged after stretching.
The third student is correct. Stretching the solenoid reduces the turns density but simultaneously reduces the flux per turn by the same factor, leaving the total field $B = \mu_0 n I$ invariant because $n$ and the cross-sectional flux compensate each other exactly.

Explanation: Whenever you see a solenoid problem involving physical changes, your first instinct should be to return to the fundamental formula and track exactly which quantities change:

B = \mu_0 n I

, where

n

is the turns per unit length, not the total number of turns. Here's the key insight: when the solenoid is stretched so its length doubles from

L

2L

, the total number of turns

N

stays fixed (the wire is re-wound to preserve

N

), but those turns are now spread over twice the length. The new turns density becomes

n' = N/(2L) = n/2

. Plugging into Ampere's law, which for an ideal solenoid gives

B = \mu_0 n I

, yields

B' = \mu_0 (n/2) I = B/2

. The field is halved — the second student is correct, making B the right answer. Choice A contains a fundamental misconception: Ampere's law doesn't reward longer path length with a stronger field. A longer Amperian loop encloses more length but proportionally fewer turns per unit length, so the enclosed current per unit length actually decreases. Choice C is wrong because

B

depends on

n

(turns per unit length), not on

N

alone —

\mu_0 N I

isn't even dimensionally the correct formula for field. Choice D is a fabricated compensation argument; flux per turn and turns density are not related in a way that holds

B

constant, and

n

itself changes when you stretch the solenoid. Your study tip: always distinguish between

N

(total turns) and

n

(turns per unit length). Solenoid problems frequently exploit this difference — if the geometry changes, recompute

n

before applying

B = \mu_0 n I

Question 14

An ideal solenoid with $n = 5000 \text{ turns/m}$ and radius $R = 2.0 \text{ cm}$ carries a current that varies in time as $I(t) = I_0 \sin(\omega t)$ where $I_0 = 4.0 \text{ A}$ and $\omega = 100\pi \text{ rad/s}$ . At time $t = 1/(200) \text{ s}$ , what is the magnitude of the magnetic field inside the solenoid?

$B = \mu_0 n I_0 = (4\pi\times10^{-7})(5000)(4.0) \approx 25.1 \text{ mT}$ , because $t = 1/200 \text{ s}$ corresponds to $\omega t = \pi/2$ , where $\sin(\pi/2) = 1$ , so the current is at its maximum value $I_0$ . (correct answer)
$B = \mu_0 n I_0 \cos(\omega t)|_{t=1/200} = \mu_0 n I_0 \cos(\pi/2) = 0$ , because the magnetic field inside a solenoid with time-varying current is proportional to $dI/dt$ , not to $I(t)$ itself, and $dI/dt = I_0\omega\cos(\omega t)$ , which is zero at $\omega t = \pi/2$ .
$B = \mu_0 n I_0 \omega = (4\pi\times10^{-7})(5000)(4.0)(100\pi) \approx 7.90 \text{ T}$ , because the time-varying current induces an additional EMF-driven field proportional to $\omega$ , which must be added to the quasi-static solenoid field.
$B = \mu_0 n I_0 / \sqrt{2} \approx 17.8 \text{ mT}$ , because for a sinusoidally varying current, the magnetic field inside a solenoid is given by the RMS value of the current, which equals $I_0/\sqrt{2}$ , and the instantaneous field is always equal to this RMS value.

Explanation: When a solenoid carries a time-varying current, the magnetic field inside it is still given by the same static formula —

B = \mu_0 n I

— evaluated at the instantaneous current. This is the quasi-static approximation, valid when the current changes slowly enough that we don't need to account for radiation effects (which is the case here). So the approach is straightforward: find

I

t = 1/200

s, then plug into

B = \mu_0 n I

. Evaluating the argument:

\omega t = 100\pi \cdot \frac{1}{200} = \frac{\pi}{2}

. Therefore

I = I_0 \sin(\pi/2) = I_0 = 4.0

A. The field is

B = \mu_0 n I_0 = (4\pi \times 10^{-7})(5000)(4.0) \approx 25.1 \text{ mT}

, confirming A is correct. Choice B confuses the magnetic field with the induced EMF. The EMF around a loop does depend on

dB/dt \propto dI/dt \propto \cos(\omega t)

, but the field itself tracks

I(t)

, not its derivative. Don't mix up Faraday's law with the field formula. Choice C invents a fictitious "EMF-driven field" proportional to

\omega

. No such term exists in the quasi-static solenoid field. The field depends only on the instantaneous current, not on how fast it's changing. Choice D misapplies RMS. RMS current is useful for computing average power, not for finding the instantaneous magnetic field. The field at any moment equals

\mu_0 n I(t)

, period. Study tip: Whenever you see a time-varying current in a solenoid, remember that

B = \mu_0 n I(t)

— just substitute the instantaneous current. Reserve RMS for power calculations and

dI/dt

for EMF calculations.