Question

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Answer 1

AnswEr:

GivEn:

• Radius of wire = R
• Turns (Revolution) = 200
• Self-Inductance = 108 mH

To Find:

• Find Self-Inductance of coil turn 500 ?

SolutiOn:

$\\ {\sf{ L_{1} = \dfrac{ \mu_{0} N^2_{1} πr^2}{l} }}$

and,

$\\ {\sf{ L_{2} = \dfrac{ \mu_{0} N^2_{2} πr^2}{l} }}$

$\implies{\sf{ \dfrac{L_{2}}{L_{1}} = ( \dfrac{N_{2}}{N_{1}} )^2 }} \\ \\ \implies{\sf{ L_{2} = L_{1} ( \dfrac{N_{2}}{N_{1}} )^2 }} \\ \\ \implies{\sf{ L_{2} = 108 \times ( \dfrac{500}{200} )^2 }} \\ \\ \implies{\sf{108 \times 6.25 }} \\ \\ \implies{\sf{675 \: mH}}$

Hence,

The Self-Inductance of similar coil of 500 turn will be 675 mH.

Extra Dose:

• $\bullet{\boxed{\sf { V_{L} = N \dfrac{d \phi }{dt} }}}$

Remarks:

• VL = induced voltage in volts
• N = number of turns in the coil
• dø/dt = rate of change of magnetic flux in webers/second