Question

21
चुम्बकीय फ्लक्स क्या है? 400 फेरों वाली कुण्डली में 1.5 ऐम्पियर की विद्युत धारा
प्रत्येक फेरे में 0.4-10-5 वेबर का चुम्बकीय फ्लक्स उत्पन्न करती है। कुण्डली का
स्वप्रेरकत्व ज्ञात कीजिए।​

Answer 1

SOLUTION

GIVEN

A current of 1.5 amperes in a 400 turns coil produces $\sf{0.4 \times {10}^{ - 5} \: \: \: weber}$ 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 400 turns coil produces $\sf{0.4 \times {10}^{ - 5} \: \: \: weber}$ magnetized flux in each turn.

So by the given condition

Ф = Magnetic Flux $\sf{ = 0.4 \times {10}^{ - 5} \: \: \: weber}$

n = The number of turns = 400

i = The current flow = 1.5 amperes

L = The self inductance of the material = ?

We know that

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

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

$\displaystyle \sf{ \implies \: L = \frac{400 \times 0.4 \times {10}^{ - 5} }{1.5} }$

$\displaystyle \sf{ \implies \: L = \frac{400 \times 4 \times {10}^{ - 5} }{15} }$

$\displaystyle \sf{ \implies \: L = \frac{16 \times {10}^{ - 3} }{15} }$

$\displaystyle\sf{ \implies \: L = \frac{16}{15} \: mH}$

FINAL ANSWER

The self-inductance of coil $\displaystyle\sf{ = \frac{16}{15} \: mH}$

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