A solenoid 74.5 cm long has 350 turns and

a radius of 2.26 cm.
If it carries a current of 1.69 A, find the
magnetic field along the axis at its center.
Answer in units of T

Magnetic field = permeability x turn density x current

For a solenoid of length L =0.745m with N =350 turns,
the turn density is n=N/L turns/m.
If the current in the solenoid is I =1.69 amperes
and the relative permeability of the core is k =1( air)
then the magnetic field at the center of the solenoid is

1*(350/.745)*1.69 T

It says its wrong

To find the magnetic field along the axis at the center of a solenoid, we can use the formula for the magnetic field inside a solenoid:

B = μ₀ * n * I

Where:
B = Magnetic field (in tesla, T)
μ₀ = Vacuum permeability (constant, approximately 4π * 10^-7 T m/A)
n = Number of turns per unit length (in turns/m)
I = Current flowing through the solenoid (in amperes, A)

First, let's calculate the number of turns per unit length (n) using the given information:
Length of the solenoid = 74.5 cm = 0.745 m
Radius of the solenoid = 2.26 cm = 0.0226 m
Number of turns = 350

n = Number of turns / Length of the solenoid

n = 350 turns / 0.745 m

n ≈ 470.47 turns/m (rounded to 2 decimal places)

Now that we have the value of n, we can substitute it along with the other given value into the formula to find the magnetic field (B):

B = μ₀ * n * I

B = (4π * 10^-7 T m/A) * 470.47 turns/m * 1.69 A

B ≈ 3.16 x 10^-4 T

Therefore, the magnetic field along the axis at the center of the solenoid is approximately 3.16 x 10^-4 T.