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Compare the magnitude of the magnetic field at the center of a circular current loop of radius 30 mm with the magnitude of the magnetic field at the center of a solenoid of the same radius and with 3.0 turn per millimeter. Assume the current is the same through the current loop and the solenoid.

Respuesta :

Answer:

Ratio of magnetic field will be [tex]\frac{B}{B'}=0.0055[/tex]

Explanation:

We have given radius of the loop r = 30 mm = 0.03 m

We know that magnetic field at the center of the loop is given by

[tex]B=\frac{\mu _0i}{2r}[/tex]---------eqn 1

Number of turns in the solenoid is given as n = 3 turn per mm = 3000 turn per meter

We know that magnetic field due to solenoid is given by

[tex]B'=n\mu _0i[/tex]-------------eqn 2

Now dividing eqn 1 by eqn 2

[tex]\frac{B}{B'}=\frac{\frac{\mu _0i}{2r}}{\mu _0ni}=\frac{1}{2nr}=\frac{1}{2\times 3000\times 0.03}=0.0055[/tex]

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