Magnetic Field Inside a Solenoid
1 · Predict
A solenoid (a tightly wound coil) carries current through its many turns. If you increase the current, does the interior field strength scale in direct proportion?
- Yes — the interior field is directly proportional to the current.
- The field grows faster than the current, since each turn's contribution compounds.
- The interior field doesn't depend on current.
2 · Set Up
- Open the solenoid-field preset and press Reset. The solenoid has 400 turns over a 0.45 m length.
- Enable the interior field-strength readout.
- Set the coil current for each trial and record the interior field strength.
3 · Collect Data
| Coil current I (A) | Interior field B (mT) |
|---|---|
| 1 | |
| 2 | |
| 3 |
Plot field strength B (y-axis) against current I (x-axis) for your three trials. Is the line straight through the origin?
4 · Analyze
- For one trial, compute B = μ₀NI/L using μ₀ = 4π×10⁻⁷ H/m, N = 400, L = 0.45 m. Compare to the table.
- Explain why the interior field depends on the number of turns per unit length (N/L), not on the turns and length separately — a longer solenoid with proportionally more turns gives the same field.
5 · Extend
- An electromagnet (like in a junkyard crane) is essentially a solenoid wrapped around an iron core, which can multiply the field by a factor of hundreds or thousands (the core's high permeability). Explain why a plain air-core solenoid like this one produces a much weaker field for the same current and turns.
- Outside an ideal solenoid, the magnetic field is close to zero, unlike a single wire's field which extends outward from it. Explain, using the many individual turns' fields, why the fields from opposite sides of the coil largely cancel outside but reinforce inside.
The Physics Behind This Experiment
Solenoid Interior Field
A long, tightly wound solenoid produces a nearly uniform interior magnetic field set by its turns density and current: B = μ₀NI/L, where N is the total number of turns and L is the solenoid's length.