Question

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.

Answer 1

Ratio of magnetic field will be
`(B)/(B')=0.0055`

Step-by-step 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

`B=(\mu _0i)/(2r)`---------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

`B'=n\mu _0i`-------------eqn 2

Now dividing eqn 1 by eqn 2

`(B)/(B')=((\mu _0i)/(2r))/(\mu _0ni)=(1)/(2nr)=(1)/(2* 3000* 0.03)=0.0055`