A long solenoid that has 1 040 turns uniformly distributed over a length of 0.390 m produces a magnetic field of magnitude 1.00 10-4 T at its center. What current is required in the windings for that to occur

This is the complete question,

A long solenoid that has 1 040 turns uniformly distributed over a length of 0.390 m produces a magnetic field of magnitude 1.00 x T at its center. What current is required in the windings for that to occur ? mA

The current required would be 29.84 mA

Explanation:

Given that

The number of turns N = 1040 turns

length of the solenoid L= 0.390 m

the magnetic field B =  1.00 x T

The magnetic field at the center of a solenoid can be obtained with the expression bellow;

B = μ n l ............1

Where μ  is the permeability of free space = 4π x T.m/A

I is the current on the solenoid

n is the turns per unit length = N/L

Substituting the new value of n into equation 1 we have;

B =  μ N l / L

making the current (I) the subject formula, it would be;

I = BL /μ N

I = (1.00 x T x 0.390 m) / ( 4π x T.m/A x 1040)

I = 0.000039  / 0.0013069

I = 0.029841

I = 29.84 x A

I = 29.84 mA

Therefore the current required would be 29.84 mA

it takes about 30 days for it to break down

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.