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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 :

jorgezarate

Answer:

[tex]B=6.3*10^{-5}Teslas[/tex]

Explanation:

We apply the Ampere's Law to the solenoid:

[tex]BL=uNI\\[/tex]

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

[tex]B=uNI/L\\[/tex]

if the coils touching each other:

L=N*d

d=diameter coil

Finally:

[tex]B=uNI/(Nd)=uI/d=4\pi *10^{-7}*0.1/(2*10^{-3})[/tex]

[tex]B=6.3*10^{-5}Teslas[/tex]

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