Physics 2 Quiz: Magnetic Field Straight Wire

6 questions · exam conditions

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Magnetic Field Straight WireQuestion 1 of 6

Two long parallel wires separated by distance $d$ carry equal currents $I$ in the same direction. A third long wire carrying current $3I$ is placed parallel to the first two, at distance $d$ from Wire 1 and distance $2d$ from Wire 2 (i.e., all three wires are collinear in cross-section). The net magnetic force per unit length on the third wire is measured. If the current in Wire 1 is reversed (now anti-parallel to Wire 3), while everything else remains unchanged, by what factor and in what direction does the net force on Wire 3 change?

The magnitude of the net force doubles and its direction reverses, because reversing Wire 1's current converts its attractive force on Wire 3 to a repulsive force of equal magnitude

\frac{3\mu_0 I^2}{2\pi d}

per unit length. Since Wire 2's attractive contribution

\frac{3\mu_0 I^2}{4\pi d}

per unit length remains unchanged, the original net force toward Wire 1 becomes a larger net force away from Wire 1.

The magnitude of the net force decreases by a factor of 3 and its direction reverses. Originally, Wire 1 attracts Wire 3 with force

\frac{3\mu_0 I^2}{2\pi d}

per unit length, and Wire 2 attracts Wire 3 with

\frac{3\mu_0 I^2}{4\pi d}

per unit length, both directed toward Wire 2's side. After reversal, Wire 1 repels Wire 3, so the net force becomes

\frac{3\mu_0 I^2}{4\pi d} - \frac{3\mu_0 I^2}{2\pi d}

per unit length, directed toward Wire 1.

The magnitude of the net force decreases by a factor of 3 and its direction reverses. Originally, both wires attract Wire 3: Wire 1 contributes

\frac{3\mu_0 I^2}{2\pi d}

and Wire 2 contributes

\frac{3\mu_0 I^2}{4\pi d}

per unit length, both toward Wire 1, giving a net of

\frac{9\mu_0 I^2}{4\pi d}

toward Wire 1. After reversal, Wire 1 repels Wire 3 with

\frac{3\mu_0 I^2}{2\pi d}

while Wire 2 still attracts with

\frac{3\mu_0 I^2}{4\pi d}

, giving a net of

\frac{3\mu_0 I^2}{4\pi d}

directed away from Wire 1.

The magnitude of the net force remains unchanged but its direction reverses, because reversing Wire 1's current changes its contribution from attractive to repulsive by exactly the same magnitude

\frac{3\mu_0 I^2}{2\pi d}

per unit length. The asymmetry introduced by the different distances to Wire 1 and Wire 2 is exactly compensated, preserving the total force magnitude while flipping its direction.

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Physics 2 Quiz

Physics 2 Quiz: Magnetic Field Straight Wire

Practice Magnetic Field Straight Wire 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 Straight Wire, 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

The magnitude of the net force doubles and its direction reverses, because reversing Wire 1's current converts its attractive force on Wire 3 to a repulsive force of equal magnitude $\frac{3\mu_0 I^2}{2\pi d}$ per unit length. Since Wire 2's attractive contribution $\frac{3\mu_0 I^2}{4\pi d}$ per unit length remains unchanged, the original net force toward Wire 1 becomes a larger net force away from Wire 1.
The magnitude of the net force decreases by a factor of 3 and its direction reverses. Originally, Wire 1 attracts Wire 3 with force $\frac{3\mu_0 I^2}{2\pi d}$ per unit length, and Wire 2 attracts Wire 3 with $\frac{3\mu_0 I^2}{4\pi d}$ per unit length, both directed toward Wire 2's side. After reversal, Wire 1 repels Wire 3, so the net force becomes $\frac{3\mu_0 I^2}{4\pi d} - \frac{3\mu_0 I^2}{2\pi d}$ per unit length, directed toward Wire 1.
The magnitude of the net force decreases by a factor of 3 and its direction reverses. Originally, both wires attract Wire 3: Wire 1 contributes $\frac{3\mu_0 I^2}{2\pi d}$ and Wire 2 contributes $\frac{3\mu_0 I^2}{4\pi d}$ per unit length, both toward Wire 1, giving a net of $\frac{9\mu_0 I^2}{4\pi d}$ toward Wire 1. After reversal, Wire 1 repels Wire 3 with $\frac{3\mu_0 I^2}{2\pi d}$ while Wire 2 still attracts with $\frac{3\mu_0 I^2}{4\pi d}$ , giving a net of $\frac{3\mu_0 I^2}{4\pi d}$ directed away from Wire 1. (correct answer)
The magnitude of the net force remains unchanged but its direction reverses, because reversing Wire 1's current changes its contribution from attractive to repulsive by exactly the same magnitude $\frac{3\mu_0 I^2}{2\pi d}$ per unit length. The asymmetry introduced by the different distances to Wire 1 and Wire 2 is exactly compensated, preserving the total force magnitude while flipping its direction.

Explanation: When parallel wires carry currents in the same direction, they attract; opposite directions, they repel. The force per unit length between two wires is

\frac{\mu_0 I_1 I_2}{2\pi r}

. Keep this formula and these sign rules at the forefront whenever you see multi-wire force problems. Here's how to work through the original configuration. Wire 3 (current

3I

) sits at distance

d

from Wire 1 and

2d

from Wire 2, with all currents initially parallel. Wire 1 attracts Wire 3 with

\frac{\mu_0(I)(3I)}{2\pi d} = \frac{3\mu_0 I^2}{2\pi d}

per unit length, directed toward Wire 1. Wire 2 attracts Wire 3 with

\frac{\mu_0(I)(3I)}{2\pi(2d)} = \frac{3\mu_0 I^2}{4\pi d}

per unit length, also directed toward Wire 1 (since Wire 2 is on the far side). The net original force is

\frac{3\mu_0 I^2}{2\pi d} + \frac{3\mu_0 I^2}{4\pi d} = \frac{9\mu_0 I^2}{4\pi d}

toward Wire 1. After reversing Wire 1's current, it now repels Wire 3:

-\frac{3\mu_0 I^2}{2\pi d}

. Wire 2 still attracts:

+\frac{3\mu_0 I^2}{4\pi d}

. The new net force is

\frac{3\mu_0 I^2}{4\pi d} - \frac{3\mu_0 I^2}{2\pi d} = -\frac{3\mu_0 I^2}{4\pi d}

, meaning magnitude

\frac{3\mu_0 I^2}{4\pi d}

directed away from Wire 1. The ratio of new to old magnitude is

\frac{1}{3}

, confirming answer C. Choice A incorrectly claims the magnitude doubles — it ignores that Wire 2's attraction now partially cancels Wire 1's repulsion rather than reinforcing it. Choice B gets the correct formula for the new force but incorrectly states both original forces point toward Wire 2's side; they both actually point toward Wire 1, so the original net force is

\frac{9\mu_0 I^2}{4\pi d}

, not

\frac{3\mu_0 I^2}{4\pi d}

. Choice D claims the magnitude is unchanged, which would only be true if Wire 2 didn't exist — it completely ignores Wire 2's continued attractive contribution that prevents perfect cancellation. Your strategy: always draw a clear diagram, assign directions explicitly (toward which wire?), and sum forces with signs before comparing scenarios. Forgetting to account for all wires — or misidentifying their directions — is the most common trap in these multi-wire problems.

Question 2

A long straight wire carries a linearly increasing current $I(t) = \alpha t$ , where $\alpha$ is a positive constant. A student argues that the magnetic field at perpendicular distance $r$ from the wire is $B(r,t) = \frac{\mu_0 \alpha t}{2\pi r}$ at every instant, obtained by substituting $I(t)$ into the magnetostatic formula. Which of the following best evaluates the validity and limitations of this claim?

The claim is exactly correct at all distances and all times, because the Ampère–Maxwell law automatically accounts for time-varying currents through the displacement current term, and the cylindrical symmetry of the geometry preserves the $1/r$ dependence regardless of how rapidly the current changes.
The claim is valid only if the wire has negligible resistance, because Ohmic dissipation in a resistive wire causes Joule heating that raises the wire's temperature and alters its resistivity over time, thereby changing the current distribution across the wire's cross-section and violating the uniform-current assumption underlying the $1/r$ formula.
The claim is invalid at all distances, because a time-varying current necessarily produces a time-varying magnetic field, and by Faraday's law this induces an electric field, which in turn modifies the magnetic field through the Ampère–Maxwell law — a self-reinforcing cycle that fundamentally alters the spatial dependence of $B$ and renders the static $1/r$ formula inapplicable under any conditions.
The claim is an approximation that is valid only when $r \ll ct$ , i.e., when the observation point is close enough that the electromagnetic travel time $r/c$ is negligible compared to the timescale over which the current changes appreciably. At larger distances, the finite speed of light means the field has not yet adjusted to the instantaneous current value, and radiation effects become significant. (correct answer)

Explanation: When you see a question about time-varying currents and magnetic fields, ask yourself: does the field update instantaneously everywhere, or does information travel at finite speed? This is the key to the quasi-static approximation in electrodynamics. The magnetostatic formula

B = \frac{\mu_0 I(t)}{2\pi r}

treats the field as responding instantaneously to the current. In reality, electromagnetic disturbances propagate at

c

, so an observation point at distance

r

only "knows" about the current state from a time

r/c

ago. When

r \ll ct

— meaning the travel time

r/c

is tiny compared to the timescale over which the current has been building — the field hasn't had time to deviate significantly from the instantaneous formula, and the approximation holds well. At large distances, however, the field still reflects an earlier (smaller) current value, retardation effects dominate, and radiation fields with different spatial dependence (

\sim 1/r

for radiation, but with phase structure) become significant. So D correctly identifies both the regime of validity and the physical reason for breakdown. A is wrong because the Ampère–Maxwell law does not make the static formula exact — it is the full wave equation that governs the fields, and exact solutions require retarded potentials, not instantaneous substitution. B introduces a real but entirely irrelevant complication; Ohmic heating doesn't determine whether the quasi-static formula is valid. C overcorrects — the self-reinforcing coupling between fields is real, but it doesn't invalidate the formula at all distances; near the wire, quasi-static results are excellent. Remember this rule of thumb: quasi-static approximations are valid when

r \ll \lambda

, where

\lambda = c/f

is the characteristic wavelength of the time variation — equivalent to the condition in D.

Question 3

A long straight wire of circular cross-section has radius $R$ and carries a total current $I$ with a non-uniform current density $J(r) = J_0 \frac{r}{R}$ , where $r$ is the distance from the wire's axis. Which expression correctly gives the magnetic field at a point outside the wire at distance $r > R$ from the axis?

$B = \frac{\mu_0 J_0 R}{3r}$ , because the current density integrated over the cross-section gives a total current proportional to $J_0 R$ , and applying Ampère's law with this enclosed current directly yields a field that scales as $R/r$ .
$B = \frac{\mu_0 J_0 R^2}{4r}$ , because the effective total current is found using the average current density $\langle J \rangle = J_0/2$ over the cross-section, giving $I = \frac{1}{2}J_0 \pi R^2$ , and substituting into Ampère's law yields $B = \frac{\mu_0 J_0 R^2}{4r}$ .
$B = \frac{\mu_0 J_0 r}{2R}$ , because outside the wire the non-uniform current distribution causes the field to grow linearly with $r$ , similar to the behavior inside a wire with uniform current density, scaled by the ratio $r/R$ .
$B = \frac{\mu_0 J_0 R^2}{3r}$ , because integrating the current density over the wire's cross-section gives $I = \int_0^R J_0 \frac{r'}{R}(2\pi r')\,dr' = \frac{2\pi J_0 R^2}{3}$ , and Ampère's law then gives $B(2\pi r) = \mu_0 I$ , yielding $B = \frac{\mu_0 J_0 R^2}{3r}$ . (correct answer)

Explanation: When a wire carries a non-uniform current density, Ampère's law still works beautifully outside the wire — but you must carefully compute the total enclosed current by integrating

J(r)

over the cross-section. That integral is the heart of this problem. For

J(r') = J_0 \frac{r'}{R}

, the total current is found by integrating over thin annular rings of area

dA = 2\pi r'\,dr'

I = \int_0^R J_0\frac{r'}{R}(2\pi r')\,dr' = \frac{2\pi J_0}{R}\int_0^R r'^2\,dr' = \frac{2\pi J_0}{R}\cdot\frac{R^3}{3} = \frac{2\pi J_0 R^2}{3}

Applying Ampère's law at

r > R

B(2\pi r) = \mu_0 I

, so:

B = \frac{\mu_0 J_0 R^2}{3r}

This confirms D is correct. A reaches the right

1/r

dependence but botches the integral, dropping an extra factor of

R

— likely by treating the area element as

2\pi R\,dr'

instead of

2\pi r'\,dr'

. B uses an "average" current density shortcut:

\langle J \rangle = J_0/2

seems intuitive, but averaging a non-uniform

J

over area requires weighting by

r'

, not taking a simple midpoint value. The correct area-weighted average gives

\langle J \rangle = 2J_0/3

, not

J_0/2

. C is fundamentally wrong in concept — outside a wire, the magnetic field always decreases as

1/r

, never grows linearly; linear growth only occurs inside the wire. Study tip: On any problem with non-uniform current density, immediately set up

I_{enc} = \int J(r')\,2\pi r'\,dr'

before touching Ampère's law — the area element

2\pi r'\,dr'

is non-negotiable and the most common place students lose points.

Question 4

A long straight wire lies along the x-axis and carries a current $I$ in the $+x$ direction. A second long straight wire lies along the z-axis and carries a current $2I$ in the $+z$ direction. At the point $(0, d, 0)$ , what is the direction of the net magnetic field produced by both wires?

The net field points in the $+z$ direction, because the wire along the x-axis dominates at this point and produces a field entirely in the z-direction, while the contribution from the z-axis wire vanishes at this location.
The net field points in a direction that is a combination of $+z$ and $-x$ , because the wire along the x-axis produces a field in the $+z$ direction at $(0,d,0)$ and the wire along the z-axis produces a field in the $-x$ direction at $(0,d,0)$ . (correct answer)
The net field points in the $-x$ direction, because the wire along the z-axis produces the dominant contribution at this point and its field is entirely in the $-x$ direction, while the x-axis wire contributes negligibly.
The net field points in the $+y$ direction, because both wires produce field components that combine along the y-axis at the point $(0,d,0)$ , with the z-axis wire contributing in the $+y$ direction and the x-axis wire reinforcing it.

Explanation: Whenever you see a magnetic field direction problem, your go-to tool is the Biot-Savart law combined with the right-hand rule: for a long straight wire, the field curls around the wire, and its direction at any point is tangent to a circle centered on the wire. Start with the wire along the x-axis (current in

+x

). At point

(0, d, 0)

, this point sits directly above the wire along the y-axis. Using the right-hand rule — curl your fingers from the current direction

+\hat{x}

around toward the point — the field at

(0, d, 0)

points in the

+z

direction. Formally:

\hat{I} \times \hat{r} = \hat{x} \times \hat{y} = +\hat{z}

. Now for the wire along the z-axis (current in

+z

). The point

(0, d, 0)

lies in the y-direction from this wire. The field curls around the z-axis:

\hat{z} \times \hat{y} = -\hat{x}

, so this wire produces a field in the

-x

direction at that point. The net field is a superposition of both contributions:

+z

from the x-axis wire and

-x

from the z-axis wire, confirming answer B. A is wrong because it ignores the z-axis wire's nonzero contribution — the point

(0, d, 0)

is at perpendicular distance

d

from the z-axis, so that wire absolutely contributes. C makes the same mistake in reverse, ignoring the x-axis wire. D is wrong because neither wire produces a y-component at this point — you can verify with cross products that no

\hat{y}

terms appear. Study tip: Always identify the perpendicular distance from each wire to your field point separately, then apply

\hat{I} \times \hat{r}

for each — never assume one wire "dominates" without checking geometry first.

Question 5

A long coaxial cable consists of an inner solid cylindrical conductor of radius $a$ carrying current $I$ in the $+z$ direction, and an outer thin cylindrical shell of radius $b > a$ carrying current $I$ in the $-z$ direction. Both conductors have uniform current distributions.

A physicist claims: 'At any point outside the outer conductor ( $r > b$ ), the magnetic field is exactly zero. At any point between the conductors ( $a < r < b$ ), the field is identical to that of just the inner wire alone.' Which of the following correctly evaluates this claim?

Both parts of the claim are correct. Outside ( $r>b$ ), the total enclosed current is $I - I = 0$ , giving $B = 0$ . Between the conductors ( $a < r < b$ ), only the inner wire contributes to $I_{enc}$ , and the outer shell contributes nothing since it lies entirely outside the Amperian loop, giving $B = \frac{\mu_0 I}{2\pi r}$ , identical to an isolated wire. (correct answer)
The first part is correct but the second is incomplete. Outside ( $r>b$ ), $B=0$ by Ampère's law. Between the conductors, the field is $B = \frac{\mu_0 I}{2\pi r}$ , which equals the inner wire alone only in magnitude — the direction is modified by the presence of the outer conductor, which acts as a magnetic shield and partially rotates the field direction inward.
The first part is correct, but the second part is only approximately correct. Between the conductors, the field equals the inner wire's contribution as a zeroth-order approximation, but fringe fields from the finite edges of the outer conductor modify the field slightly in a realistic (non-infinite) cable.
Both parts of the claim are correct only if the currents are steady (DC). For alternating currents, the skin effect causes current to redistribute toward the conductor surfaces, invalidating the uniform current density assumption and therefore the application of Ampère's law as stated.

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

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

. The key insight is that only current enclosed by your chosen Amperian loop matters — current outside the loop contributes zero to

I_{enc}

. For

r > b

, your Amperian loop encloses both conductors:

I_{enc} = I + (-I) = 0

, so

B(2\pi r) = 0

, giving

B = 0

exactly. For

a < r < b

, your loop encloses only the inner conductor, so

I_{enc} = I

, giving

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

— exactly what an isolated wire produces. Both parts of the physicist's claim are fully correct, making A the right answer. Choice B invents a fictional "field rotation" effect from the outer conductor. Ampère's Law is exact for infinite symmetrical geometries — the outer shell causes no directional distortion between the conductors; it simply isn't enclosed by the loop. Choice C introduces fringe-field corrections for finite cables. While true in a pedantic real-world sense, this question explicitly states a long coaxial cable, the standard idealization where Ampère's Law applies exactly. The claim isn't "approximately" correct — it's exactly correct under these stated conditions. Choice D correctly identifies that AC skin effects redistribute current, but this doesn't make Ampère's Law invalid — it just changes

I_{enc}

. More importantly, the problem specifies uniform current distributions, so this caveat is irrelevant here. Study tip: On Ampère's Law problems, always ask yourself what current is inside my loop — current outside the loop is completely irrelevant, no matter how close it is.

Question 6

Two long straight wires are oriented perpendicular to each other but do not intersect. Wire 1 lies along the x-axis carrying current $I_1$ in the $+x$ direction, and Wire 2 lies parallel to the z-axis but shifted to position $(0, d, 0)$ , carrying current $I_2$ in the $+z$ direction. What is the net magnetic force per unit length that Wire 2 exerts on Wire 1?

Zero, because the magnetic field produced by Wire 2 at any point on Wire 1 is parallel to Wire 1's current direction, so the cross product $I_1 d\vec{\ell} \times \vec{B}_2$ vanishes everywhere along Wire 1.
$\frac{\mu_0 I_1 I_2}{2\pi d}$ per unit length, directed in the $-y$ direction, because the geometry of perpendicular wires produces a net attractive force similar to parallel wires, reduced by the sine of the angle between them.
Zero, because for every element of Wire 1 at position $(x, 0, 0)$ , the force contribution is nonzero but the forces on the positive- $x$ and negative- $x$ halves of Wire 1 point in opposite directions and cancel when integrated over the full infinite wire. (correct answer)
$\frac{\mu_0 I_1 I_2}{4\pi d}$ per unit length, directed in the $+z$ direction, because the perpendicular orientation reduces the effective current interaction by a factor of two compared to parallel wires, and the force direction is set by the cross product of the two current directions.

Explanation: When tackling magnetic force problems between wires, your first instinct should be to carefully evaluate both the field produced by one wire and the resulting force element

d\vec{F} = I_1 d\vec{\ell} \times \vec{B}_2

before drawing conclusions — and then consider whether integration over an infinite wire might cause cancellation. Wire 2 (at position

(0, d, 0)

, carrying current in

+\hat{z}

) produces a magnetic field that circles around it. At a point

(x, 0, 0)

on Wire 1, the displacement from Wire 2 is

(x, -d, 0)

, and the field

\vec{B}_2

points in a direction that varies with

x

. Crucially, this field has both

\hat{x}

and

\hat{y}

components that depend on

x

. The force element on Wire 1 is

I_1 dx\,\hat{x} \times \vec{B}_2

. The

\hat{x}

-component of

\vec{B}_2

contributes nothing (since

\hat{x} \times \hat{x} = 0

), and the

\hat{y}

-component contributes a force in

\hat{z}

. When you integrate this

\hat{z}

-directed force from

-\infty

+\infty

, symmetry causes equal and opposite contributions to cancel exactly. The net force is zero, making C correct. A is wrong because it misidentifies the reason —

\vec{B}_2

is not parallel to Wire 1 everywhere; the cross product is nonzero element-by-element, but the integral vanishes. B is wrong because perpendicular infinite wires don't behave like parallel wires — there's no simple angular reduction factor, and the net force is actually zero. D is wrong on both grounds: the force direction it claims and the nonzero magnitude are both incorrect; the integration yields zero, not a reduced version of the parallel-wire formula. Strategy tip: For infinite-wire force problems with unusual geometries, always check whether integration symmetry causes cancellation before assuming a nonzero result — this is a classic trap.