Solenoid Field and Self-Inductance
For a long ideal solenoid B = mu n I. Self-inductance L is flux linkage per current for a linear magnetic system, and its induced emf is proportional to current's rate of change.
Why this shows up in the exam
Electromagnets · Chokes and relays · Switch-mode power supplies
Learn the idea
A solenoid creates a nearly uniform field and opposes changes in its own current. Current in a long solenoid creates an internal magnetic field. If that current changes, its own magnetic flux changes and produces a back emf that resists the change.
🧠 Memory hook: Solenoid field follows current; inductance fights current change.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- B = mu n I — field inside a long ideal solenoid
- N Phi = L I — definition for a linear inductor
- emf_L = -L dI/dt — self-induced back emf
- L_solenoid = mu N² A/l — long ideal solenoid
How to approach it
- 1Decide whether the question asks for field or inductance
- 2Use B = mu n I for the field
- 3Use -L dI/dt only when current changes
Common slip-ups that cost marks
- •Using total turns instead of turns per metre in B = mu n I
- •Treating an inductor as resistance
- •Using final current in place of dI/dt
- •Forgetting the N-squared dependence of a solenoid
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A conducting rod of length 2 m moves at 3 m/s perpendicular to a 4 T magnetic field. Find the motional emf.
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