A solenoid 1metre long and radious 4cm has 1000 turns and is carrying 1Ampere current.find the magnetic flux density at the centre.
What is the wire wound around, or specifically, its permeability?
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/solenoid.html#c3
To find the magnetic flux density at the center of a solenoid, you can use the formula:
B = μ₀ * n * I
where:
B is the magnetic flux density,
μ₀ is the permeability of free space (4π * 10^-7 T*m/A),
n is the number of turns per unit length, and
I is the current flowing through the solenoid.
First, let's calculate the number of turns per unit length. The solenoid has a length of 1 meter and 1000 turns. Therefore, the number of turns per unit length (n) is:
n = Number of turns / Length
= 1000 turns / 1 meter
= 1000 turns/m
Now, substitute the known values into the formula:
B = (4π * 10^-7 T*m/A) * (1000 turns/m) * (1 Ampere)
Simplifying the expression:
B = 4π * 10^-7 T*m/A * 1000 A * m^-1
= 4π * 10^-4 T
Therefore, the magnetic flux density at the center of the solenoid is 4π * 10^-4 Tesla (T).