Question

A cylindrical insulated wire of diameter 2.0 mm is tightly wound 200 times around a cylindrical core to form a solenoid with adjacent coils touching each other. When a 0.10 A current is sent through the wire, what is the magnitude of the magnetic field on the axis of the solenoid near its center?

Answer 1

`B=6.3*10^(-5)Teslas`

Step-by-step explanation:

We apply the Ampere's Law to the solenoid:

`BL=uNI\\`

N= number of turns on the solenoid

L=solenoid length

u=relative permeability of air= 4*pi*10^-7 N/A^2

I=current

B=magnetic field

`B=uNI/L\\`

if the coils touching each other:

L=N*d

d=diameter coil

Finally:

`B=uNI/(Nd)=uI/d=4\pi *10^(-7)*0.1/(2*10^(-3))`

`B=6.3*10^(-5)Teslas`