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Numerical · Q23

Q.Two long coaxial solenoids, each of length 0.5 m0.5\ \text{m}, are wound one inside the other: the inner solenoid has radius 2 cm2\ \text{cm} and 10001000 turns per metre, and the outer solenoid has 40004000 turns per metre. Calculate the mutual inductance of the pair.

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Given: l=0.5 ml=0.5\ \text{m} (both solenoids), inner radius r1=2 cm=0.02 mr_1=2\ \text{cm}=0.02\ \text{m}, n1=1000 turns/mn_1=1000\ \text{turns/m} (inner), n2=4000 turns/mn_2=4000\ \text{turns/m} (outer).

M=μ0n1n2πr12lM = \mu_0 n_1 n_2 \pi r_1^2 l

Step 1 -- n1n2n_1 n_2:

n1n2=1000×4000=4×106n_1n_2 = 1000\times4000 = 4\times10^{6}

Step 2 -- πr12\pi r_1^2:

πr12=π×(0.02)2=π×4×10−4≈1.2566×10−3 m2\pi r_1^2 = \pi\times(0.02)^2 = \pi\times4\times10^{-4} \approx 1.2566\times10^{-3}\ \text{m}^2

Step 3 -- combine with μ0\mu_0, ll:

M=(1.2566×10−6)×(4×106)×(1.2566×10−3)×0.5M = (1.2566\times10^{-6})\times(4\times10^{6})\times(1.2566\times10^{-3})\times0.5

1.2566×10−6×4×106=5.02651.2566\times10^{-6}\times4\times10^{6} = 5.0265 …

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