Difficulty: Intermediate | Prerequisites: Ampere's law basics, right-hand rule, trigonometric components of vectors.
Coaxial conductors are one of the most practical applications of Ampere's law. The geometry, two concentric cylinders carrying current in opposite directions, appears in coaxial cables, certain types of power transmission lines, and laboratory apparatus. This problem combines the core skill of computing enclosed current with an extra step: resolving the magnetic field into Cartesian components using the angle between the position vector and the axes. It is a common exam format because it tests Ampere's law, the area-ratio technique, the right-hand rule, and vector decomposition all in one question.
For coaxial cylinders, apply Ampere's law with a circular loop between or outside the conductors. When the loop passes through a conductor, compute the fractional enclosed current using the area ratio. Once you have the magnitude of B, use the geometry (the angle the position vector makes with the x-axis) to extract the x- or y-component.
Coaxial conductors
Two concentric cylindrical conductors sharing the same axis, typically carrying currents in opposite directions. Think of a standard coaxial cable: a central wire surrounded by a cylindrical shield, with current flowing one way in the core and returning through the shield.
Cross-sectional area ratio
The technique for finding the enclosed current when an Amperian loop passes through a uniformly distributed current: I_enc = I_total × (area enclosed) / (total area of conductor). For a solid inner cylinder, this is r²/a². For a hollow outer cylinder, this is (r² − b²) / (c² − b²).
Vector decomposition of B
The magnetic field from Ampere's law points tangentially (perpendicular to the radial direction). To find Bₓ or Bᵧ, you project B onto the x- or y-axis using the angle θ that the position vector makes with the x-axis. In simple terms, if the field is tangential and you know the angle, Bₓ = −|B| sin θ and Bᵧ = |B| cos θ (or similar, depending on the current direction).
Inner solid cylinder: radius a = 2 cm, carries I₁ = 1.2 A in the +z direction (out of the page).
Outer hollow cylinder: inner radius b = 4 cm, outer radius c = 6 cm, carries I₂ = 2.4 A in the −z direction (into the page).
Both currents are uniformly distributed over their respective cross-sections.
Point P is at distance r = 5 cm from the axis, at an angle of 30° above the x-axis.
r = 5 cm falls between b = 4 cm and c = 6 cm, so the Amperian loop passes through the outer conductor.
The loop encloses all of I₁ and a fraction of I₂.
The outer conductor occupies the annular region from b to c.
At radius r, the enclosed portion of the outer conductor's cross-section is π(r² − b²).
The total cross-section of the outer conductor is π(c² − b²).
Fractional enclosed current from the outer conductor:
I₂_enc = I₂ × (r² − b²) / (c² − b²)
I₂_enc = 2.4 × (5² − 4²) / (6² − 4²) = 2.4 × 9/20 = 1.08 A
This current is in the −z direction, so it contributes negatively: −1.08 A.
Total enclosed current: I_enc = I₁ − I₂_enc = 1.2 − 1.08 = 0.12 A (net in the +z direction).
Wait: the solution actually uses I_enc = I₁ − I₂_enc in the signed sense. Let us be careful:
I₁ = +1.2 A (out of page)
I₂_enc = −1.08 A (into page)
Net I_enc = 1.2 − 1.08 = 0.12 A? No. Looking at the worked solution more carefully: B(2πr) = μ₀(I₁ − I₂ × (r² − b²)/(c² − b²)), and the magnitude comes out to |B| = 4.8 × 10⁻⁷ T.
|B| = μ₀|I_enc| / (2πr) = (4π × 10⁻⁷)|I_enc| / (2π × 0.05)
The field B is tangential to the circle at point P. The right-hand rule (for net current in the +z direction) shows B points in the tangential direction at P.
Point P is at angle θ = 30° from the x-axis.
The tangential direction at angle θ from the x-axis has components: (−sin θ, cos θ).
The right-hand rule, applied carefully to the actual net enclosed current direction, gives the sign.
In this problem, the result is Bₓ = −|B| cos 60° = −4.8 × 10⁻⁷ × 0.5 = −2.4 × 10⁻⁷ T.
The key step is recognising that if B is tangential and the radial direction makes 30° with the x-axis, then the tangential direction makes 60° with the x-axis (since tangential is perpendicular to radial). So Bₓ = −|B| cos 60°.
Ampere's law: ∮ B · dl = μ₀ I_enc → B = μ₀ I_enc / (2πr)
Enclosed current for a solid cylinder (r < a): I_enc = I × r² / a²
Enclosed current for a hollow cylinder (b < r < c): I_enc = I × (r² − b²) / (c² − b²)
Cartesian components from tangential field: If the position vector is at angle θ from the +x axis, the tangential unit vector is (−sin θ, cos θ, 0). Multiply by |B| and the appropriate sign from the right-hand rule.
This is literally the physics of a coaxial cable. In a well-designed coax, I₁ = I₂ (equal and opposite), so the field outside the outer conductor is exactly zero. This means no electromagnetic interference leaks out or couples in, which is why coaxial cables are used for TV signals, internet connections, and laboratory instruments where signal integrity matters.
Students often use the wrong area ratio. For the outer hollow conductor, the denominator is (c² − b²), not c² or b². The current occupies only the annular region.
A common error is forgetting that the field is tangential, not radial. When asked for Bₓ, you cannot just write B; you need to decompose using the angle.
Students mix up the angle: if point P is at 30° from the x-axis, the tangential direction is at 90° + 30° = 120° from the x-axis (or equivalently, 60° from the x-axis measured differently). Draw a sketch.
The sign from the right-hand rule is often lost. If the net enclosed current is in the +z direction, B curls counterclockwise when viewed from above. At a specific point, this determines whether Bₓ is positive or negative.
⚠️ The Cartesian decomposition step is where most marks are lost. Practise converting from tangential magnitude to Bₓ and Bᵧ using the angle.
⚠️ Know both area-ratio formulas cold: r²/a² for a solid cylinder and (r² − b²)/(c² − b²) for a hollow one.
⚠️ Problems often give the angle in one convention (from the x-axis) but the decomposition requires a different angle (between B and the x-axis). Always sketch the geometry.
True or false: For a coaxial cable with equal and opposite currents, the magnetic field is zero everywhere outside the outer conductor. True, provided the currents are exactly equal in magnitude.
Fill in the blank: For a hollow cylindrical conductor with inner radius b and outer radius c, the fraction of current enclosed at radius r is ____. (r² − b²) / (c² − b²)
True or false: The magnetic field from a long straight conductor points radially outward. False. It points tangentially (circles the conductor).
Fill in the blank: If a point is at angle 45° from the x-axis and the tangential magnetic field has magnitude B, then |Bₓ| = ____. B sin 45° = B / √2 (because the tangential direction is perpendicular to the radial, so it makes a 45° angle with the x-axis as well in this special case).
Q: A solid wire of radius 1 cm carries 5 A. What is B at r = 0.5 cm from the centre?
A: Inside the wire, I_enc = 5 × (0.5²/1²) = 1.25 A. B = μ₀(1.25) / (2π × 0.005) = 5 × 10⁻⁵ T.
Q: A coaxial cable has inner conductor radius a = 1 cm (carrying 3 A out) and outer conductor from b = 2 cm to c = 3 cm (carrying 3 A in). What is B at r = 2.5 cm?
A: I₂_enc = 3 × (2.5² − 2²)/(3² − 2²) = 3 × (2.25/5) = 1.35 A. Net I_enc = 3 − 1.35 = 1.65 A. B = μ₀(1.65) / (2π × 0.025) = 1.32 × 10⁻⁵ T.
Q: In the above problem, what is B at r = 4 cm (outside everything)?
A: Net I_enc = 3 − 3 = 0 A. B = 0.
Q: Point P is at (3 cm, 4 cm) from the axis. What angle does the position vector make with the x-axis, and how would you find Bₓ?
A: r = 5 cm, θ = arctan(4/3) ≈ 53.1°. The tangential direction at this point is (−sin 53.1°, cos 53.1°) = (−0.8, 0.6). So Bₓ = −|B| × 0.8 (with sign adjusted for the right-hand rule based on current direction).
This connects back to the cylindrical shell problem in the first set of notes: the area-ratio technique is identical. It also previews Faraday's law: if the current in a coaxial cable changes with time, the changing magnetic flux between the conductors induces an EMF, which is the basis for understanding inductance per unit length of coaxial cables.
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