Questions: Magnetic Fields in Solenoids and Toroids

Question 1

A long solenoid has 1000 turns per meter carrying 2 A. You measure the magnetic field at the central axis and then at a point 3 cm off-axis but still well inside the solenoid. What do you find?

• A. The field is stronger on the axis — it falls off toward the edges like a bar magnet
• B. The field is the same at both points — it is uniform throughout the interior of an ideal solenoid
• C. The field is weaker off-axis because that point is closer to one side of the winding
• D. The field varies sinusoidally from one end of the solenoid to the other

Answer: (B)

The derivation via Ampère's law shows B = μ₀nI inside an ideal solenoid — this result contains no positional dependence. The rectangular Amperian loop argument works regardless of where inside the solenoid the loop is placed, so the field is the same everywhere in the interior. This uniformity is precisely what makes solenoids useful for applications requiring a controlled, homogeneous magnetic field (like MRI machines). Outside the solenoid, the field is essentially zero.

Question 2

An engineer needs a magnetic device that produces effectively zero field outside its volume so it won't interfere with neighboring circuits. Which geometry is preferable?

• A. A solenoid, since its external field is approximately zero for a long solenoid
• B. A toroid, since the geometry completely confines the magnetic field within the toroidal volume
• C. Both confine the field equally well — the toroid is chosen only for its compact shape
• D. A longer solenoid, which progressively reduces fringe fields at the ends

Answer: (B)

A toroid confines the field exactly to zero outside: any Amperian circle outside the toroid encloses equal amounts of current going in opposite directions, so the net enclosed current is zero and B = 0 exactly. A solenoid is only approximately zero outside (ideal solenoid: zero; real solenoid: small fringe fields at the ends). For applications where stray fields would cause interference — like inductors in electronic circuits or transformer cores — the toroid's exact confinement makes it the correct choice.

Question 3

Doubling the current through a solenoid while keeping the number of turns per unit length constant will double the magnetic field inside the solenoid.

• T. True
• F. False

Answer: True

B = μ₀nI — the field is directly proportional to both n (turns per unit length) and I (current). All other factors held constant, doubling I doubles B. This linear relationship is what makes solenoids precisely controllable: you can dial in a specific target field by adjusting current. The same linear scaling holds for changing n, allowing engineers to tune either the winding density or the current to achieve a desired field strength.

Question 4

The magnetic field inside a toroid is uniform throughout the toroidal cross-section, just as the field inside an ideal solenoid is uniform along its length.

• T. True
• F. False

Answer: False

The field inside a toroid is NOT uniform — it varies as 1/r, where r is the distance from the central axis. B = μ₀NI/(2πr), so the field is strongest at the inner radius of the torus and weakest at the outer radius. In contrast, the solenoid's field is uniform everywhere inside. This non-uniformity in toroids is a practical consideration: for applications requiring a truly uniform field, a solenoid is better; for applications requiring complete external confinement, a toroid is better.

Question 5

Why does a solenoid produce a uniform magnetic field inside while a toroid does not? What about their geometries accounts for this difference?

Model answer: A solenoid is geometrically uniform along its axis: the Amperian rectangle argument gives B = μ₀nI with no dependence on position within the solenoid. A toroid is a solenoid bent into a ring, so the Amperian loop is a circle of radius r. Since circumference = 2πr, the Ampère's law equation B·(2πr) = μ₀NI gives B = μ₀NI/(2πr), which decreases with radius. The inner edge of the toroid is at a shorter radius than the outer edge, so the field is stronger at the inner edge.

The key is that symmetry determines what the Amperian loop integral looks like. For the solenoid, a rectangular loop gives B·L = μ₀·(nL)·I regardless of where L is placed — no r-dependence. For the toroid, a circular loop of different radii encloses the same total current NI but has different circumferences, so B must vary with r to satisfy Ampère's law.