Question

Two air-cored coils are disposed relatively to each other so that 75% of the flux in one coil links all the turns of the other coil. Each coil has a cross-sectional mean diameter of 2cm and a length of 50 cm. If there are 2000 turns of wire on one coil, calculate the number of turns required on the other coil to give a mutual inductance of 20 mH.

Answer 1

Answer:

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Answer 2

16887 Turns

Given:

To air-cored coils are disposed relatively to each other so that 75% of the flux in one coil links all the turns of the other coil.

Cross-sectional mean diameter = 2cm

Length = 50cm

No. of turns of wire on one coil = 2000

mutual inductance = 20mH

To find:

Number of turns required on the other coil

Solution:

S = $\frac{1}{\frac{\pi }{4}*(0.02)^{2}*4\pi *10^{-7}*1 }$ = $1.2665 * 10^{-9} AT/W_{b}$

Now,

⇒ L1 = $N_{1} ^{2} /S$ and L2 = $N_{2} ^{2} /S$

∴ $\sqrt{L1L2}$ = $N_{1}N_{2}/S$

Also, M = $K\sqrt{L1L2}$ = K * $N_{1}N_{2}/S$

It is given that, M = $20 * 10^{-3} H$

$N_{1} = 2000$

K = 0.75

∴ $20 * 10^{-3} = \frac{0.75 * 2000* N_{2} }{1.2665 * 10^{9} }$

⇒ $N_{2} = \frac{20 * 10^{-3}*1.2665 * 10^{9} }{0.75* 2000}$

⇒ 16887 turns

Hence, the number of turns on the other coil is 16887 turns.

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