Could you please derive the mutual induction and self induction for a conductor?
Let us deduce the same for conductor in the form of a solenoid:
Self-Inductance of a Long Solenoid
A long solenoid is one whose length is very large as compared to its radius of cross-section. The magnetic field B at any point inside such a solenoid is practically constant and is given by,
μ0 = Absolute magnetic permeability of free space
N = Total number of turns in the solenoid
∴ Magnetic flux through each turn of the solenoid coil = B × Area of each turn
A = Area of each turn of the solenoid
Total magnetic flux linked with the solenoid
If L is coefficient of self-inductance of the solenoid, then
∴Φ = LI … (iii)
Equating (ii) and (iii),
If core is of any other magnetic material, μ0 is replaced by
The phenomenon according to which an opposing emf is produced in a coil as a result of change in current, hence, the magnetic flux linked with a neighbouring coil is called mutual induction.
Coefficient of mutual induction − Consider two coils P and S. Suppose that a current I is flowing through the coil P at any instant i.e.,
Φ ∝ I
Φ = MI … (i)
Where, M is called coefficient of mutual induction
If ‘e’ is the induced emf produced in the S-coil, then
Mutual Inductance of Two Long Solenoids
Consider two long solenoids S1 and S2 of same length l, such that solenoid S2 surrounds solenoid S1completely.
n1 − Number of turns per unit length of S1
n2 − Number of turns per unit length of S2
I1 − Current passed through solenoid S1
Φ21 − Flux linked with S2 due to current flowing through S1
Φ21 ∝ I1
Φ21 = M21I1
Where, M21 is the coefficient of mutual induction of the two solenoids
When current is passed through solenoid S1, an emf is induced in solenoid S2.
Magnetic field produced inside solenoid S1 on passing current through it,
B1 = μ0n1I1
Magnetic flux linked with each turn of solenoid S2 will be equal to B1 times the area of cross-section of solenoid S1.
Magnetic flux linked with each turn of the solenoid S2 = B1A
Therefore, total magnetic flux linked with the solenoid S2,
Φ21 = B1A × n2l = μ0n1I1 × A× n2l
Φ21 = μ0n1n2AI1
∴ M21 = μ0n1n2Al
Similarly, the mutual inductance between the two solenoids, when current is passed through solenoid S2 and induced emf is produced in solenoid S1, is given by
M12 = μ0n1n2Al
∴M12 = M21 = M (say)
Hence, coefficient of mutual induction between the two long solenoids
Φis directly proportional to i its means if flux changes then current will also changes.
therefore, Φ is equal to ki where k called coefficient of self induction we can also write L in the place of k
now induced e.m.f is E=-dΦ/dt
now it will written as -d/dt(L/i)
at last we will get required equation for self induction that is
the direction of induced e.m.f will be according to lenz law that is it will oppose the cause for its induction.
now Derivation of mutual induction
1 and 2 are two coils placed parrallel to each other .current in 1 changes and hence its magnetic flux change and this process will be same for the 2nd coil ,
now change in flux will be Φ2 is directly proportional to 1 coil
therefore fi 2 =M 1
where M is coefficient of mutual induction
induced e.m.f in 2 coil will be E2 =-dfi2/dt =-d/dt(M1)
now E2=-M d1/dt is required equation for mutual induction .Dear friend let me trully tell u dat i have nt defined d self n mutual induction as well so this will be ur task to do this. i have written fi symbol for magnetic flux 1 or 2 times i think in the ABOVE DERIVATION so plz correct it wherever symbols will be necessarily.due to shortage of time plz see it n if any concern regarding this derivation then u will ask me
ASHISH MISHRA (RAO MOHAR SINGH SCHOOL)