A long currentcarrying solenoid with an air core has 1800 tu

A long, current-carrying solenoid with an air core has 1800 turns per meter of length and a radius of 0.0185 m. A coil of 240 turns is wrapped tightly around the outside of the solenoid, so it has virtually the same radius as the solenoid. What is the mutual inductance of this system?

Solution

N/L = 1800

r = 0.0185 m

N\' = 240

Mutual inductance is given by :

M = u*(N/L)*N\'*A

where:

A = area of loop = pi*r^2 = pi*0.0185^2

So,

M = 1.26*10^-6*1800*240*(pi*0.0185^2)

= 5.85*10^-4 H <------answer