Question

Calculate inductance of solenoid 0.5 m long of cross sectional area 20 and with 500 turns

Answer 1

Answer:

The inductance will be 1.25 mH

Explanation:

According to the problem the solenoid has the length ,l= 0.5 m and the area is 20 having 500 turns

Let the inductance , L

Therefore from the concept of self inductance

L= N∅/I

Now the magnetic produced for solenoid,B= μ0N/l x I [ where N = number of turns and l= length, I = current  and μ0= permittivity ]

From here the flux,∅=  BAcos θ

as θ= 0 therefore,

∅=  BA

Therefore,

L = N  x μ0N I/l x A/I

 =  N^2 x  μ0 x A/l

  = (500)^2 x (4 π x 10^(-7)) x (20 x10^(-4))/0.5

 = 1.25 mH