Two solenoids have the same number of turns per unit length and carry the same current, but one has twice the radius of the other. How do their interior magnetic fields compare?
AThe wider solenoid has twice the field, since it encloses more area
BThe narrower solenoid has twice the field, since the field lines are more concentrated
CBoth solenoids have the same field — B = μ₀nI depends only on turn density and current, not on radius
DThe fields cannot be compared without knowing the solenoids' total lengths
B = μ₀nI for an ideal solenoid, where n is turns per unit length and I is current. Radius does not appear in the formula — the Ampère's law derivation shows the interior field is uniform and independent of the solenoid's radius. What matters is how densely turns are packed (n) and how much current flows, not the solenoid's diameter. Both solenoids with the same n and I have identical interior fields.
Question 2 Multiple Choice
A solenoid carries current I and has n turns per unit length. The current is doubled while n is halved (turns are spread more loosely). What happens to the interior field?
AThe field doubles, since more current creates more field
BThe field is halved, since fewer turns per unit length means less field generation
CThe field is unchanged — B = μ₀nI, and (n/2)(2I) = nI, so the product nI is unchanged
DThe field changes in a way that depends on the solenoid's total length
B = μ₀nI depends only on the product nI. If n is halved and I is doubled, nI → (n/2)(2I) = nI — unchanged. The field is the same. What matters physically is the current-per-unit-length (sometimes written as the surface current density K = nI): the total ampere-turns packed into each unit of length. The same field can be achieved with few dense turns carrying high current, or many loose turns carrying low current, as long as the product nI is equal.
Question 3 True / False
For an ideal (infinitely long) solenoid, the magnetic field outside is exactly zero — all magnetic flux is confined to the interior.
TTrue
FFalse
Answer: True
True. This follows from the Ampère's law derivation: a rectangular Amperian loop with one side outside and one inside has its outer contribution equal to zero (the external field is zero) because neighboring turns produce external fields in opposite directions that cancel exactly for an infinite solenoid. The interior leg contributes B·L = μ₀nLI. In reality, finite solenoids produce fringe fields at their ends, and the exterior field is not exactly zero — but it becomes negligible far from the ends compared to the strong uniform interior field.
Question 4 True / False
A solenoid's magnetic field B = μ₀nI is proportional to the current squared, which is why the energy stored in a solenoid scales as I².
TTrue
FFalse
Answer: False
False — with an important distinction. The magnetic field B = μ₀nI is proportional to I (linear, not squared). However, the energy stored in the magnetic field is proportional to B², which means energy ∝ (μ₀nI)² ∝ I². So energy scales as I², not the field itself. Inductance L is defined such that energy = ½LI², where L ∝ n². The statement conflates the field (linear in I) with the stored energy (quadratic in I).
Question 5 Short Answer
Explain why the magnetic field inside a solenoid is approximately uniform, and why this is surprising given that the solenoid is made of discrete wire loops.
Think about your answer, then reveal below.
Model answer: Each circular loop produces a dipole-like field that is strongest near that loop's center and falls off with distance. When many loops are stacked tightly, their interior contributions all point in the same axial direction and add constructively regardless of position inside the stack. Meanwhile, their exterior fields cancel destructively, because adjacent loops generate opposing external fields at any exterior point. The result is a uniform axial interior field B = μ₀nI independent of position — an emergent collective property of the ensemble, not a property of any individual loop.
The Ampère's law derivation confirms this: no matter where inside the solenoid you place the interior leg of the Amperian loop, the same nL turns thread through it and the same B·L appears in the line integral. The field cannot depend on radial position inside because the Amperian analysis gives the same answer regardless of where the interior leg is placed. This is the power of Ampère's law for symmetric configurations — it extracts the field from global symmetry without having to sum contributions from each individual current element.