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

the answer is 46 mls because when you add the speed per second it increses the speed

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2