The ans given is four times, but if we increase the number of turns by two times without change in length of the solenoid, shouldn't the radius also decrease by 2 times? Hence, won't the self inductance become eight times?: 26. When the number of turns in a solenoid are doubled without any change in the length of the solenoid, it self inductance becomes. (1) Half              (2) Double           (3) Four Times              (4) Eight times.

Dear student,
Here it is mentioned that only the turns are increased while keeping other variables as constant.So there will be no difference in the radius as well.

The self inductance of a solenoid is given as

L = μ0N2A / l = L1

where

N is the number of turns of the solenoid

A is the area of each turn of coil

l is the length of the solenoid

and μ0 is the permeability constant

so, if the number of turns was to be doubled the self-inductance would be

L2 =  μ0  (2N)2A / l

or

L2 = 4L1

it would be quadrupled or increase fourfold.

Regards

• 0
I think the fault is in the reasoning itself self inductance L is directly proportional to square no of turns per unit length n^2  and also directly proportional to cross sectional area A and if the length remains unchanged n^2 becomes 4 times and the radius gets halved and cross sectional area gets reduced by four times so inductance will not change at all from its original value and not to 8 times as stated by you
• 1
What are you looking for?