Question

वीडियोस इन एक्सपेरिमेंट फॉर सेल्फ इंडक्शन ऑफ ए लोंग सोलेनाइड फॉर लैंड एंड हेवी एंड डांस​

Answer 1

Answer:

Consider a current(I) is passed through the inductor (L).

A: Cross-sectional area

N : Number of turns

L : Length

Magnetic field generated at inside the solenoid is ,

B=

L

μ

o

NI

So, the flux through the coil is a obtained:

ϕ

B

=N(

B

.

A

) (

B

and

A

are along same direction)

ϕ

B

=

L

μ

o

N

2

IA

...(1)

Now, flux (ϕ

B

) is related to inductance (L) as

ϕ

B

=LI

⇒L=

I

ϕ

B

From (1) we get,

⇒L=

L

μ

0

N

2

A