The length of a thin wire required to manufacture a solenoid of length`l = 100 cm` and inductance `L = 1 mH`, if the solenoid's cross-sectional diameter is considerably less than its length is

Options

A. `1.0` km
B. `0.10` km
C. `0.010` km
D. `10` km

Answer

B

Explanation

Magnetic induction at centre of solenoid is
`B=(mu_(0)N)/(l_(0))I`
then, flux linked with whole solenoid is
`phi=NBA=(mu_(0)N^(2)pir^(2)I)/(i_(0))`
Also, `phi=LI`
:. `L=(mu_(0)N^(2)pir^(2))/(l_(0))` or `L=mu_(0)(N^(2).4pi.pir^(2))/(4piI_(0))=mu_(0)(N^(2).4pi^(2).r^(2))/(4piI_(0))`
`impliesL=mu_(0).((N.2pir)^(2))/(4pii_(0))`
Now, `N.2pir=l=`length of wire used in solenoid
:. `L=(mu_(0).I^(2))/(4piI_(0))`
then, `l=sqrt((4piI_(0)L)/(mu_(0)))` then, `l=0.10` km.