Question

121. A solenoid of length 20 cm, area of cross-
to 40 cm. and having 4000 turns is placed
inside another solenoid of 2000 turns having a
cross sectional area 8.0 cm' and length 10 cm.
The mutual inductance between the solenoids is??​

Answer 1

Mutual inductance

M=μ0N1N2πr21l

=4π×10−7×4×103×2×103

×4×10−4×10×10−2

=0.04×10−2H.

a_1 = 4 cm^2, a_2 = 8 cm^2n_1 = 4000//0.2 m,n_2 = 2000//0.1 ml_1 = 20 cm = 0.20 m,l_2 = 10 cm = 0.1 mB = mu_(0) n_(2) i[Letthecurrentthroughouterso≤noidbei]phi = n_(1) B.A = n_(1) n_2) mu_(0) i xx a_(1)= 2000 xx (2000)/(0.1) xx 4pi xx10^(-7) xx i xx 4 xx 10^(-4)E = (dphi)/(dt) = 64pi xx 10^(-4) xx (dl)/(dt)NowM = E/lt= 6 xx 10^(-4) H = 2 xx 10^(-2) H`.

Answer 2

SOLUTION :

Solenoid I : a1 = 4cm2 ;

n1 = 4000/0.2m;

l1= 20cm = 0.20m

Solenoid II : a2 = 8cm2;

n2 = 2000/0.1m ;

l2 = 10cm = 0.10m

B = μ0n2i

Let the current through outer solenoid be

i. ϕ = n1B.A = n1n2μ0i x a1

This is the best explanation I can give you