Question

A long, current-carrying solenoid with an air core has 1550 turns per meter of length and a radius of 0.0240 m. A coil of 200 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

Answer 1

`M=7.05*10^(-4)`

Step-by-step explanation:

From the question we are told that:

Coil one turns N_1=1550 Turns/m

Radius
`r=0.0240m`

Turns 2
`N_2=200N`

Generally the equation for area is mathematically given by

`A=\pi*r^2`

`A=\pi*0.024^2`

`A=\1.81*10^(-3) m^2`

Therefore

The mutual inductance of this system is

`M=\mu*N_1*N_2*A`

`M=(4 \pi*10^(-7))*1550*200*1.81*10^(-3)`

`M=7.05*10^(-4)`