|Class 12 Physics Electromagnetic Induction||Mutual Inductance|
In mutual inductance there are 2 coils, current is passed in one coil, as current increases there is change in the flux, and as aresult current is induced in the second coil.
Consider coil 1 connected to battery and coil 2 is connected to the galvanometer.
Mutual Inductance between long co-axial solenoids
Problem:- A solenoid of length 50cm with 20 turns per cm and area of cross-section 40cm2 completely surrounds another co-axial solenoid of the same length, area of cross section 25cm2 with 25 turns per cm. Calculate the mutual inductance of the system?
Answer:- Let the outer solenoid be 1 and inner solenoid be 2.
Length l1 =50cm=50x10-2m
n1 =2000 turns /m
A1 = 40 x10-4 m2
n2 =2500 turns/m
M12= (μ0N1N2A2)/length =μ0n1n2A2length
Problem:- (a) A toroidal solenoid with an air core has an average radius of 15cm, area of cross-section 2cm2 and 1200 turns. Obtain the self-inductance of the toroid. Ignore field variations across the cross-section of toroid.
(b) A second coil of 300 turns is wound closely on the toroid above. If the current in the primary coil is increased from 0 to 2A in 0.05s.Obtain the induced emf in the second coil.
Answer:- (a) Radius r1=15x10-2m
B=μ0nI= (μ0N1I)/ (2πr1)
Total flux: -N1xBxA1=N1x(μ0xN1I/2πr1) x A1
Also Total flux =LI
LI =μ0xA1x N12I/ (2πr1)
= (4πx10-7x (1200)2x 2x10-4)/ (2πx15x10-2)
The self-inductance of the toroid is 2.304x10-4 H.
(dI1/ dt) = (If –Ii)/dt = (2-0)/ (0.05) =40A/s
Total flux in 2:- Φ2 =N2xB1xA1
=N2(μ0n1I1)/A1 equation (1)
Also,Φ2 =MI1 equation (2)
e2 =M (dI1/dt) =μ0N2n1A1x40
=4πx10-7x 300x (N1/2πr1) x2x10-4x40
The emf induced in the second coil is 0.023V.