Question

A current of 1.5 amperes in a 800 turns coil produces
1.5 x 10-5 weber magnetized flux in each turn. Find out the
self-inductance of coil.​

Answer 1

SOLUTION

GIVEN

A current of 1.5 amperes in a 800 turns coil produces $\sf{1.5 \times {10}^{ - 5} \: \: \: weber}$ magnetized flux in each turn.

TO DETERMINE

The self-inductance of coil.

EVALUATION

Here it is given that a current of 1.5 amperes in a 800 turns coil produces $\sf{1.5 \times {10}^{ - 5} \: \: \: weber}$ magnetized flux in each turn.

So by the given condition

Ф = Magnetic Flux $\sf{ = 1.5 \times {10}^{ - 5} \: \: \: weber}$

n = The number of turns = 800

i = The current flow = 1.5 amperes

L = The self inductance of the material = ?

We know that

$\displaystyle \sf{n \phi = L \times i}$

$\displaystyle \sf{ \implies \: L = \frac{n \phi }{i} }$

$\displaystyle \sf{ \implies \: L = \frac{800 \times 1.5 \times {10}^{ - 5} }{1.5} }$

$\displaystyle \sf{ \implies \: L = 800 \times {10}^{ - 5} }$

$\sf{ \implies \: L = 8 \times {10}^{ - 3} \: }$

$\sf{ \implies \: L = 8\: mH}$

FINAL ANSWER

The self-inductance of coil = 8 mH

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