Home › Karnataka PUC II › Physics › Electromagnetic Induction › Self-inductance of a long solenoid of length l, …
Self-inductance of a long solenoid of length l, cross-section A, with n turns per unit length is
AL = μ₀ A / (n² l)
BL = μ₀ n² A l
CL = μ₀ n A / l
DL = μ₀ n l / A
Answer & Solution
Correct answer: B. L = μ₀ n² A l
B = μ₀nI inside; Φ_B per turn = μ₀nI·A; total linkage N Φ = (nl)(μ₀nIA) = μ₀n²AlI. So L = μ₀n²Al.
Related questions
The coil of an AC generator connects to the outside circuit through:A coil is labelled 2 H. The H stands for the unit called the:A magnet dropped through an aluminium pipe, compared with a PVC pipe, takes:Some trains brake with electromagnets held above the rails. The braking feels:Cutting slots in the swinging copper plate makes it:A copper plate swings between the poles of a magnet and soon stops. The cause is:A loop moves while lying fully inside a uniform magnetic field. The induced current is:Where does the work you do pushing the magnet into the coil end up?