Question

उसके प्रति फेरे में चुम्बका
का स्वप्रेरकत्व ज्ञात कीजिए।
(a) What is coefficient of self-induction?
(b) 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.
.
[L
107A से 1010A तरंग-दैर्घ्य परास वाले विद्युत-चुम्बकीय तरंग का नाम लिखिए और
इसके दो महत्वपूर्ण उपयोग लिखिए।
Name the electromagnetic wave of wavelength range 1078 to 10108
and write two important uses of it.
किसी 30 सेमी फोकस दूरी के उत्तल लेंस के संपर्क में रखे 20 सेमी. फोकस दूरी के
अवतल लेंस के संयोजन से बने संयुक्त लेंस की फोकस दूरी क्या है? क्या यह संयोजन
​

Answer 1

Coefficient of self-induction :

Explanation:

1)

Coefficient of self-induction :

• The coefficient of self-inductance is quantitative relation of electromotive force produced in a circuit by self-induction to the rate of change of current producing it.

• Unit of coefficient of self-inductance H.

2)

Given:

I = 1.5 A

N = 800

φ = 1.5  Wb

To find:

L=?

Explanation:

The self-inductance of the coil is defined as the magnetic flux passing through the coil due to current within the coil itself.

It is calculated by the formula

Nφ = LI

where

N = no of turns in the coil

φ - electric flux

L = self-induction of the coil

I = current passing through the coil

Put given values in above eq

800( 1.5  ) = L ( 1.5 )

L = 8 $\times 10^{-3$

L = 8 mH

∴ The self-induction of the coil of the given coil is 8 mH.