Question

On making a coil of copper wire of length and coil radius r, the value of self inductance is obtained as l. If the coil of same wire, but of coil radius r/2, is made, the value of self inductance will be

Answer 1

Hi,

Answer:

We know the formula for self-inductance of the coil “L” is given as,

L = N (φB) / I = [N*B*A] / I = [N* µo*N*I* πr²]/[2r * I] = [N²*µo*π*r]/2 …… (i)

Where

N = no. of turns of coil

I = current  

φB = magnetic flux =  B*A

B = magnetic field = [µo*N*I]/2r  

Case 1:

Initial coil radius = r

Here let the initial self-inductance be denoted as “l₁”.

∴ l₁ =  [N²*µo*π*r]/2 …… [from eq. (i)] ……. (ii)

Case 2:

Here we have the coil of same wire but the coil radius = r/2

Let the self- inductance after the change in the radius of the coil be denoted as “l₂”.

∴ l₂ =  [N²*µo*π*(r/2)]/2 …… [from eq. (i)] ……. (iii)

Now, on dividing the equation (iii) by (ii), we get

l₂/l₁ = [{N²*µo*π*(r/2)}/2] / [{N²*µo*π*r}/2]

⇒ l₂/l₁ = 1/2 …… [cancelling all the similar terms]

⇒ l₂ = l₁/2

Thus, the value of self-inductance after the coil radius becomes r/2 is

l₂ = l₁/2.

Hope this is helpful!!!!!