There is one solenoid of length I and self-inductance L. What is the length of wire used to make this solenoid?

Options

A. `sqrt((pilL)/(mu_(0)))`
B. `sqrt((4pilL)/(mu_(0)))`
C. `sqrt((pilL)/(4mu_(0)))`
D. `sqrt((pilL)/(2mu_(0)))`

Answer

B

Explanation

Let n be the number of turns per unit length of solenoid, r be the radius of solenoid then magnetic flux linked with the coil can be written as follows, when current I is flowing through it.
`phi = (mu_(0)nI)(pir^(2))(nl)`
Comparing it with `phi = LI`, we get self-inductance as follows:
`L= mu_(0)pin^(2)r^(2)l`
Total length of wire used to make solenoid can be written as follows:
`L.= (2pir)(nl)= 2pirnl`
Value of nr to be substituted in equation (i) can be taken from equation (i) as follows:
`n^(2)r^(2)= (L)/(mu_(0)pil)`
`nr= sqrt((L)/(mu_(0)pil))`
Substituting in equation (ii) we get length of wire as follows:
`L.= 2pi sqrt((L)/(mu_(0)pil))`
`L.= sqrt((4pilL)/(mu_(0)))`