Current Balance near the End of a Solenoid
A student builds a current balance to measure the magnetic field inside a long solenoid. The solenoid has 500 turns per metre and carries 2.0 A. A short straight wire of length 5.0 cm is mounted horizontally across the solenoid near one end and carries 1.0 A perpendicular to the solenoid axis. The wire is attached to a balance that reads a force of 4.0×10⁻⁵ N. Using B = μ₀nI, the student predicts 1.26×10⁻³ T and a force of 6.3×10⁻⁵ N, so she suspects the balance or the solenoid. The wire is actually 2 cm from the solenoid end, the solenoid is 30 cm long, and the wire's leads run along the field for part of their length. The balance has a zero error of 0.5×10⁻⁵ N. She can move the wire along the axis, reverse the current, and use a Hall probe.
1. Calculate B at the centre and the expected force, then use the zero error to find the effective field at the wire. Compare the two fields.
Read the worked answer
At the centre, B = μ₀nI = (4π×10⁻⁷)(500)(2.0) = 1.257×10⁻³ T. For a 5.0 cm wire carrying 1.0 A perpendicular to B, F = BIL = (1.257×10⁻³)(1.0)(0.050) = 6.28×10⁻⁵ N, matching the student's prediction. The balance zero error is 0.5×10⁻⁵ N, so the corrected measured force is 4.0×10⁻⁵ − 0.5×10⁻⁵ = 3.5×10⁻⁵ N. The effective field at the wire is B_eff = F/(IL) = 3.5×10⁻⁵/(1.0×0.050) = 7.0×10⁻⁴ T. This is 0.56 of the centre field. A 30 cm solenoid is not very long compared with its diameter, and 2 cm from the end is in the fringing region, where the field is expected to be well below the centre value. The difference is therefore not necessarily a faulty balance.
2. Why can the current in the leads change the measured force even if the wire segment is perpendicular to B? What checks would separate this from the end effect?
Read the worked answer
Only the part of a lead that is perpendicular to B experiences a force; a lead running parallel to the field contributes no magnetic force. If the return lead or a twisted pair crosses the field at an angle, it adds an upward or downward force and changes the balance reading. Move the leads outside the solenoid or twist them together so their forces cancel, and repeat. Reverse the current: the force on the perpendicular wire must reverse, while a zero error does not. Move the wire to the centre: if the reading rises toward 6.3×10⁻⁵ N, the original low value was mainly the end field; if not, lead forces or alignment are suspect. A Hall probe map along the axis gives an independent check.
3. Plan a measurement of B along the solenoid axis and state what graph would show whether the long-solenoid approximation is valid.
Read the worked answer
Mount a short wire, or a Hall probe, on a scale along the axis and keep the solenoid current and wire current constant. At each position record the force, correct the zero error, and compute B_eff = F/(IL). Reverse the current at each point and average the magnitudes to reduce offset and alignment errors. Plot B_eff against distance from the centre. A long solenoid should show a broad flat plateau in the middle and a rapid fall near the ends. If the plateau is short compared with the solenoid length, or if the centre value is not close to μ₀nI, the long-solenoid approximation is not valid and the centre value should be used rather than a value near the end. The graph also shows the fringing length, which is comparable to the radius.