Question

An inductor coil having 500 turns, produces
magnetic flux of 10-4 weber per turn when current
through it is 0.2 A. The coefficient of self
inductance of the coil is
(1) 2.5 mH
(2) 25 mH
(3) 250 mH
(4) 2.5 H​

Answer 1

Explanation:

hope this will help you

Answer 2

Answer:

The coefficient of self-inductance of the coil is 250mH.i.e.option(3).

Explanation:

The emf of the coil in terms of magnetic flux is given as,

$e=\frac{d\phi}{dt}$           (1)

And in terms of the induced current is given as,

$e=\frac{LdI}{dt}$        (2)

By equating the equations (1) and (2) we get;

$\frac{d\phi}{dt}=\frac{LdI}{dt}$

$\phi=LI$        (3)

Where,

Ф=total magnetic flux

L=inductance of the coil

I=current through the coil

From the question we have,

Number of turns(N)=500

The magnetic flux per unit turn=10⁻⁴Weber per turn

I=0.2A

So, the total magnetic flux will be given as,

$\phi=500\times 10^-^4=5\times 10^-^2Weber$     (4)

By substituting the value of Ф and I in equation (3) we get;

$5\times 10^-^2=L\times 0.2$

$L=\frac{5\times 10^-2}{0.2}=25\times 10^-^2=250mH$

Hence, the coefficient of self-inductance of the coil is 250mH.option(3).