Magnetic Field: Straight Wire - Physics 2

1. 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?
   Answer: $B = \frac{\mu_0 J_0 R^2}{3r}$, 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 $B = \frac{\mu_0 J_0 R^2}{3r}$. 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.
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?
   Answer: 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. 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.
3. 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?
   Answer: 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 $(0,d,0)$. 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.
4. 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?
   Answer: 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. 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 $-\infty$ to $+\infty$, 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.
5. 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?
   Answer: 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. 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.
6. 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?
   Answer: 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. 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.

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