Question

(a) How is self-inductance defined and what physical principle allows us to define it ?

(b) In discussing LRC circuits we found that comparison with simple oscillating mechanical systems such as a mass on a spring with friction was very useful. Using this analogy, what are the mechanical equivalents of: charge, current, inductance, capacitance, and magnetic flux ?

(c) Derive from basic principles the self-inductance of a solenoid having cross sectional area A, length L and N windings.

Answer 1

a)The property of self-inductance is a particular form of electromagnetic induction. Self inductance is defined as the induction of a voltage in a current-carrying wire when the current in the wire itself is changing. In the case of self-inductance, the magnetic field created by a changing current in the circuit itself induces a voltage in the same circuit. Therefore, the voltage is self-induced.

Consider a single conducting circuit around which a current I is flowing. This current generates a magnetic field B which gives rise to a magnetic flux ϕ linking the circuit. We expect the flux ϕ to be directly proportional to the current I, given the linear nature of the laws of magnetostatics, and the definition of magnetic flux. Thus, we can write  

ϕ= L*I

where the constant of proportionality L is called the self inductance of the circuit.

Quantity	mechanical equivalents
Charge	Momentum
Current	Force
Inductance	Compliance
Capacitance	Mass
magnetic flux	Momentum

We use Lenz law to define self inductance.

Lenz's law states that an induced current has a direction such that its magnetic field opposes the change in magnetic field that induced the current.

b)

c) Consider a solenoid of length L and cross-sectional area A. Suppose that the solenoid has N turns. When a current I flows in the solenoid, a uniform axial field of magnitude

B= μ0NI/L  -------------------(1)

is generated in the core of the solenoid. The field-strength outside the core is negligible. The magnetic flux linking a single turn of the solenoid is ϕ= BA. Thus, the magnetic flux linking all N turns of the solenoid is  

ϕ= NBA =μ0N2AI/L   -----------------(2)

According to Eq.(1), the self inductance of the solenoid is given by ϕ= L*I , which reduces to  

L =μ0N2A/L

Note that L is positive. Furthermore, L is a geometric quantity depending only on the dimensions of the solenoid, and the number of turns in the solenoid.