(a) State Lanz's law. Give one example to illustrate this law. "The Lenz's law is a consequence of the principle of conservation of energy." Justify this statement.
(b) Deduce a expression for the mutual inductance of two long coaxial solenoids but having different radii and different number of turns.

Explanation

(a)Lenz's law:It states that the direction of induced current or `emf` in a circuit is always such that it opposes the cause which produces it.
It gives the direction of current or `emf` induced in a circuit.
Lenz's law is in accordance with the principles of conservation of energy.In electromagnetic induction, the electrical energy (in the form of induced current or induced `e.m.f`) is obtained at the expense of mechanical energy.
(b)Coefficient of mutual induction:The coefficient of mutual induction or mutual inductance of two coils is numerically equal to the amount of magnetic flux linked with one coil when unit current flows through the neighbouring coil.
It is denoted by `M`.Its `SI` unit is weber/ampere.
Mutual inductance of two long coaxial solenoids:Consider the following fig.Which shows two long coaxial solenoids each of length `l` let the radius of the inner solenoid `S_(1)` be `r_(1)` and the number of turns per unit length be `n_(1)`
The corresponding quantities for the outer solenoid `S_(2)` are `r_(2)` and `n_(2)` respectively. Let `N_(1)` and `N_(2)`, be the total number of turns of coils `S_(1)` and `S_(2)` respectively.
`N_(1)phi_(1)=M_(12)I_(2)` ..(1)
where `M_(12)` is called the mutual inductance of solenoid `S_(1)` with respect to solenoid `S_(2)`.The magnetic field due to the current `I_(2)` in `S_(2)` is `mu_(0)n_(2)l_(2)`.The resulting flux linkage with coil `S_(1)` is `N_(1)phi_(1)=(n_(1)l)(pir^(2))(mu_(0)n_(2)I_(2))=mu_(0)n_(1)n_(2)pir_(1)^(2)lI_(2)`..(2)
where `n_(1)l` is the total number of turns in solenoid `S_(1)` From eq.(1) and (2) we get
`M_(12)I_(2)=mu_(0)n_(1)n_(2)pir_(1)^(2)lI_(2)`
or `M_(12)=mu_(0)n_(1)n_(2)pir_(1)^(2)l`...(3)
Now consider the reverse case a current `I_(1)` is passed through the solenoid `S_(1)` and the flux linkage with coil `S_(2)` is
`N_(2)phi_(2)=M_(21)I_(1)` ..(4)
Where `M_(21)`,is called the mutual inductance of solenoid `S_(2)` with respect to solenoid `S_(1)`.The flux due to the current `I_(1)` in `S_(1)` can be assumed to be confined only inside `S_(1)`, Since the solenoids are very long.Thus, flux linkage with solenoid `S_(2)` is
`N_(2)phi_(2)=(n_(2)l)(pir_(1)^(2))(mu_(0)nl_(1)^(2))`...(5)
Where `n_(2)l` is the total number of turns of `S_(2)`.
From eq.(4) and (5), we get
`M_(21)=mu_(0)n_(1)n_(2)pir_(1)^(2)l`...(6)
From eq (3) and (4), we get
`M_(12)=M_(21)=M`(say)
Therefore `M=mu_(0)n_(1)n_(2)pir^(2)l`