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(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. |
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Answer» Solution :(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 produced at the expense of MECHANICAL energy. (b) Mutual Inductance : Suppose there are two coils `C_(1)` and `C_(2)`. The current `I_(1)` is flowing in Primary coil `C_(1)`, due to which an effective magnetic flux `phi_(2)` is linked with secondary coil `C_(2)`. By experiments. `phi_(2) prop I_(1) or phi_(2)=MI_(1)` Where, M is a CONSTANT, and is called the coefficient of mutual inductance. From `M=f_(2)/I_(1)`. If `I_(1)=1` ampere, `M=phi_(2)` i.e. the mutual inductance between two coils is numerically equal to the effective flux linkage with secondary coil, when current flowing in primary coil is 1 ampere. Mutual Inductance of Two co-axial Solenoids : CONSIDER two long co-axial solenoid each of length `l` with secondary coil, when current flowing in primary coil is N ampere. Mutual Unductance of two co-axial Solenoids : Consider two long co-axial solenoid each of length `l` with number of turns `N_(1)` and `N_(2)` wound one over the other. Number of turns per unit length in order (primary) solenoid, `n=N_(1)/l`. If `I_(1)` is the current flowing in primary solenoid, the magnetic field produced within this solenoid, `B_(1)=(mu_(0) N_(1) I_(1))/l` The flux linked with each turn of inner solenoid coil is `phi_(2)=B_(1)A_(2)`, where `A_(2)` is the cross-sectional area of inner solenoid. The total flux linkage with innercoil of `N_(2)`-turns. `phi_(2)=N_(2)phi_(2)=N_(2)B_(1)A_(2)` `=N_(2) ((mu_(0) N_(1)I_(1))/l)A_(2)` `=(mu_(0) N_(1)N_(2))/l A_(2)l_(1)` By definition, Mutual Inductance, `M= phi_(i)/I_(1)=(mu_(0)N_(1)N_(2)A_(2))/l` If `n_(1)` is number of turns per unit length of outer solenoid and `r_(2)` is radius of inner solenoid, then `M=mu_(0) n_(1)N_(2) pi r_(2)^(2)` 
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