What current is required in the windings of a long solenoid that has 420 turns uniformly distributed over a length of 0.575 m in order to produce inside the solenoid a magnetic field of magnitude 4.22×10−5 t? the permeablity of free space is 1.25664 × 10 tm/a.

The current in the solenoid=0.463 Ampere

Explanation:

Given:

Number of turns , N=420

Length of the solenoid=0.575 m

Magnitude of Magnetic Field,

We know that the magnitude of the magnetic Field inside the solenoid is

Where

According to question

Hence the current is calculated.

hi there!

out of the two, (b) precision would be more reasonable.   however, both a and b could be pretty reasonable.

-awesomerepublic   : )

true is the correct answer

hope i