Solenoid Field: Centre Versus End
A student winds 40 turns of insulated copper wire in a single uniform layer along a 25 cm length of cardboard tube of diameter 3.0 cm, then connects the coil to a 2.0 A dc supply. At the centre, a Hall probe reads 3.8 × 10−4 T; at one end the same probe reads 2.0 × 10−4 T. Using B = μ0nI, the student calculates 4.0 × 10−4 T for the centre and 4.0 × 10−4 T for the end, because the solenoid looks symmetric from either side. He concludes that the Hall probe is faulty and that the end reading in particular is impossible. The teacher notes that the probe has an active area 8 mm wide, that the winding is slightly non-uniform near one end, and that the tube is not infinitely long. The supply leads are twisted and kept away from the probe.
1. Calculate n and the ideal central field. Is the centre reading of 3.8 × 10−4 T strong enough evidence that the probe is faulty?
Read the worked answer
The turn density is n = 40/0.25 = 160 m−1, so the long-solenoid estimate is B = μ0nI = (4π × 10−7)(160)(2.0) = 4.02 × 10−4 T. The measured centre value is 3.8 × 10−4 T, about 5.5% low. A finite-length correction at the centre is L/√(L2 + 4R2) = 0.25/√(0.252 + 4 × 0.0152) ≈ 0.993, which predicts 3.99 × 10−4 T, only 0.7% lower. The remaining few percent can reasonably come from the finite wire thickness changing the effective turn spacing, the 8 mm probe averaging over a non-uniform field, calibration tolerance, and slight winding irregularity. A 5% deviation is not by itself proof of a faulty probe; it should be checked against a known field or by reversing the current and checking linearity.
2. Explain why the field at the end is about half the central field, and why the symmetry of the solenoid does not imply equal fields at the centre and the end.
Read the worked answer
For a long solenoid, the axial field at a point is the sum of contributions from all turns. At the centre, turns on both sides contribute nearly equally; at an end, only the turns on one side are present, and the integral of the contributions from that half gives approximately half the central value. More formally, the field of a finite solenoid on its axis falls from μ0nI at the centre to about ½μ0nI at the end when the length is much larger than the radius. Symmetry about the central plane means B(−z) = B(z) measured from the centre, not that the centre and end are equivalent. The ends break translational symmetry along the axis, so the field need not be the same there. The observed 2.0 × 10−4 T is therefore expected, not impossible.
3. If the student doubles the current to 4.0 A, predict the centre and end readings and state two practical checks before trusting the result.
Read the worked answer
An air-core solenoid is linear in current, so doubling I should double B: the centre reading should become about 2 × 3.8 × 10−4 = 7.6 × 10−4 T and the end reading about 2 × 2.0 × 10−4 = 4.0 × 10−4 T. Before trusting this, check that the coil does not overheat and that the supply can maintain 4.0 A without its terminal voltage falling; copper resistance and heating can change the current. Also check linearity at several currents and reverse the current to detect any zero offset or nearby magnetic material. Earth’s field, about 5 × 10−5 T, is a small background but should be considered if high precision is claimed.