Question

What current is required in the windings of a long solenoid that has 544 turns uniformly distributed over a length of 0.439 m in order to produce inside the solenoid a magnetic field of magnitude 0.0001132 T? The permeablity of free space is 1.25664 × 10−6 T m/A. Answer in units of mA.

Answer 1

I=72.7 mA

Step-by-step explanation:

Given that

N= 544 turns

L= 0.439 m

Number of turn per unit length

n=N/L= 1238.17

B= 0.0001132 T

μo=1.25664 × 10⁻⁶

We know that magnetic filed given as'

B= μo n I

I =Current

`I=(B)/(n\mu _o)`

By putting the values

`I=(0.0001132)/(1238.17*  1.25664* 10^(-6))`

I=0.0727

I=72.7 mA