Difficulty: Intermediate | Prerequisites: Biot-Savart law basics, right-hand rule, vector components of magnetic fields.
Ampere's law is one of the core tools for finding magnetic fields in situations with high symmetry. This set of problems deals with a cylindrical conducting shell carrying a distributed current, combined with an infinite straight wire along the axis. It is the kind of geometry that shows up repeatedly in electromagnetism courses because it tests whether you can identify the right Amperian loop, correctly determine the enclosed current, and apply symmetry arguments to simplify the integral. If you are comfortable with Gauss's law for electric fields, the logic here is very similar, just applied to magnetic fields.
Use Ampere's law with circular Amperian loops centred on the axis to find the magnetic field at various distances from a current-carrying cylindrical shell and wire. The field depends on how much current is enclosed by the loop, and for points inside the shell material you need a fractional enclosed-current calculation based on area ratios.
Ampere's law
The line integral of the magnetic field around a closed loop equals μ₀ times the total current enclosed by that loop: ∮ B · dl = μ₀ I_enc. Think of it as the magnetic-field equivalent of Gauss's law: it lets you find B without integrating contributions from every piece of current, provided the geometry is symmetric enough.
Amperian loop
A closed mathematical path you choose to exploit symmetry when applying Ampere's law. It is not a physical object. In simple terms, pick a loop where B is either constant along the path or perpendicular to it, so the integral becomes trivial.
Enclosed current (I_enc)
The net current passing through the surface bounded by your Amperian loop. Only current that actually threads through the loop contributes; current outside the loop does not appear in Ampere's law.
Right-hand rule (for Ampere's law)
Curl the fingers of your right hand in the direction you traverse the Amperian loop. Your thumb points in the direction that defines positive current through the loop. In simple terms, it tells you whether a given current contributes positively or negatively to I_enc.
Uniformly distributed current
Current spread evenly over the cross-sectional area of a conductor. To find how much current passes through a sub-region, you use the ratio of areas: I_enc = I_total × (A_sub / A_total).
A solid cylindrical conducting shell has inner radius a = 5.6 cm and outer radius b = 7.1 cm, with its axis along the z-axis.
The shell carries a uniformly distributed current I₂ = 5.1 A in the positive z-direction.
An infinite wire runs along the z-axis carrying current I₁ = 3 A in the negative z-direction.
At point P, located at (d, 0) with d = 32 cm, you are outside both the wire and the shell.
Draw a circular Amperian loop of radius d centred on the z-axis.
By symmetry, B is tangential and constant in magnitude around this loop, so ∮ B · dl = B(2πd).
The enclosed current is the algebraic sum: I_enc = I₁ + I₂ = (−3) + 5.1 = 2.1 A.
I₁ is negative because it flows in the −z direction.
Solving: B = μ₀(I₁ + I₂) / (2πd).
At point P, the right-hand rule shows B points in the +y direction, so B_y(P) = 1.31 × 10⁻⁶ T.
The integral ∫ B · dl from P to R to S (along the dotted path shown) equals a fraction of the full loop integral.
Section R to S contributes zero because B · dl = 0 along a radial segment (B is perpendicular to dl).
Section P to R is 1/8 of the full circular loop (the path subtends 1/8 of the circle).
Result: ∫ₚˢ B · dl = (1/8) × μ₀(I₁ + I₂) = 3.30 × 10⁻⁷ T·m.
The integral from S back to P along a straight line has the same magnitude but opposite sign: −3.30 × 10⁻⁷ T·m. This follows because the endpoints are the same as the P-to-S path, and the full closed-loop integral fixes the relationship.
At point T, located at (−6 cm, 0), the distance from the axis is r = 6 cm, which falls between a = 5.6 cm and b = 7.1 cm.
The Amperian loop at radius r encloses all of I₁ but only a fraction of I₂.
The fraction of I₂ enclosed is determined by the area ratio:
I₂_enc = I₂ × π(r² − a²) / π(b² − a²)
I₂_enc = 5.1 × (6² − 5.6²) / (7.1² − 5.6²) = 1.24 A
The total enclosed current: I_enc = I₁ + I₂_enc = (−3) + 1.24 = −1.76 A.
The right-hand rule shows that at point T (on the −x side), the positive-z current from the shell produces B in the +y direction, while the negative-z wire current produces B in the −y direction. The net B_y(T) = +5.86 × 10⁻⁶ T.
At (2.8 cm, 0), the point is inside the hollow region (r < a = 5.6 cm).
The Amperian loop at this radius encloses only I₁ (the wire). None of the shell's current is enclosed because the shell starts at r = a.
Doubling I₂ has no effect on the field here. The answer is: B(2.8 cm, 0) remains the same.
Ampere's law (integral form): ∮ B · dl = μ₀ I_enc
Field from Ampere's law with circular symmetry: B(2πr) = μ₀ I_enc → B = μ₀ I_enc / (2πr)
Fractional enclosed current for uniform distribution in a shell: I_enc(shell) = I_total × (r² − a²) / (b² − a²), valid for a ≤ r ≤ b
Permeability of free space: μ₀ = 4π × 10⁻⁷ T·m/A
Coaxial cables are essentially this geometry: a central conductor carrying current in one direction, surrounded by a cylindrical shell carrying current in the opposite direction. Ampere's law explains why the magnetic field outside an ideal coaxial cable is zero (when the currents are equal and opposite), which is exactly why coax is used to shield signals from external interference.
Students often forget to assign a sign to each current based on the right-hand rule and the chosen loop direction. The net enclosed current is an algebraic sum, not the sum of magnitudes.
When the point of interest is inside the shell material, students sometimes use the full shell current instead of computing the fractional enclosed current via the area ratio.
A common error is thinking that doubling a current outside the Amperian loop changes the field inside. If the current is not enclosed, it does not appear in Ampere's law.
Students sometimes try to use Ampere's law along non-circular or arbitrary paths without realising that the simplification B(2πr) only works when B is constant and tangential along the loop.
⚠️ You will almost certainly be asked to find the enclosed current for a point inside the shell material. Practise the area-ratio formula until it is automatic.
⚠️ Line integrals along partial paths (not full loops) appear frequently. Remember that radial segments contribute zero, and arc segments contribute a fraction of the full loop result.
⚠️ The "what happens if we change a current" conceptual question tests whether you understand which currents are enclosed. If the current is outside your loop, changing it does nothing.
True or false: Ampere's law only works for infinitely long, straight wires. False. Ampere's law is always valid. The simplification to B = μ₀I/(2πr) requires cylindrical symmetry, but the law itself applies to any closed loop.
Fill in the blank: For a point inside the conducting shell at radius r, the enclosed shell current is I₂ × ____. (r² − a²) / (b² − a²)
True or false: If the Amperian loop encloses zero net current, the magnetic field is zero everywhere on the loop. False. Ampere's law says the line integral of B around the loop is zero, but B itself can be nonzero at individual points.
Fill in the blank: The line integral ∫ B · dl** along a radial path (pointing directly away from the axis) is ____.** Zero, because B is perpendicular to the radial direction.
Q: A cylindrical shell (inner radius 3 cm, outer radius 5 cm) carries 4 A in the +z direction. An axial wire carries 2 A in the −z direction. What is B at r = 10 cm from the axis?
A: All current is enclosed. I_enc = −2 + 4 = 2 A. B = μ₀(2) / (2π × 0.10) = 4 × 10⁻⁶ T.
Q: In the same setup, what is B at r = 1 cm (inside the hollow region)?
A: Only the wire is enclosed. I_enc = −2 A. B = μ₀(2) / (2π × 0.01) = 4 × 10⁻⁵ T, directed according to the right-hand rule for current in the −z direction.
Q: At what radius would the net enclosed current be zero?
A: Set I₁ + I₂ × (r² − a²)/(b² − a²) = 0. Solve for r. This gives the radius inside the shell where the partial shell current exactly cancels the wire current.
Q: If you double the wire current but leave the shell current unchanged, does the field at r = 4 cm (inside the shell) change?
A: Yes. The point at r = 4 cm (assuming a = 3 cm) is inside the shell material, and the Amperian loop encloses all of the wire current plus a fraction of the shell current. Changing the wire current changes I_enc.
This connects directly to the Biot-Savart law: Ampere's law gives the same results but far more efficiently when symmetry is present. It also connects to Faraday's law later in the course, where changing magnetic flux through a loop induces an EMF, and to the design of real devices like solenoids, toroids, and coaxial cables.
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