Question

1000 फेरों वाली कुण्डली में 2.5 ऐम्पियर की विद्युत धारा प्रत्येक फेरे में
0.5x10-5 वेबर चुम्बकीय फ्लक्स उत्पन्न करती है। कुण्डली का स्वप्रेरकत्व ज्ञात
कीजिए।​

Answer 1

SOLUTION

GIVEN

A current of 2.5 amperes in a 1000 turns coil produces $\sf{0.5 \times {10}^{ - 5} \: \: \: weber}$ magnetized flux in each turn.

TO DETERMINE

The self-inductance of coil.

EVALUATION

Here it is given that a current of 2.5 amperes in a 1000 turns coil produces $\sf{0.5 \times {10}^{ - 5} \: \: \: weber}$ magnetized flux in each turn.

So by the given condition

Ф = Magnetic Flux $\sf{ = 0.5 \times {10}^{ - 5} \: \: \: weber}$

n = The number of turns = 1000

i = The current flow = 2.5 amperes

L = The self inductance of the material = ?

We know that

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

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

$\displaystyle \sf{ \implies \: L = \frac{1000 \times 0.5 \times {10}^{ - 5} }{2.5} }$

$\displaystyle \sf{ \implies \: L = \frac{1000 \times {10}^{ - 5} }{5} }$

$\displaystyle \sf{ \implies \: L = 200 \times {10}^{ - 5} }$

$\displaystyle \sf{ \implies \: L = 2 \times {10}^{ - 3} }$

$\sf{ \implies \: L = 2\: mH}$

FINAL ANSWER

The self-inductance of coil $\sf{ = 2\: mH}$

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