A 29.4 cm long solenoid 1.10 cm in diameter is to produce a field of 0.360 T at its center. How much current should the solenoid carry if it has 1000 turns of the wire?
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/solenoid.html
To find the current required for the solenoid, we can utilize Ampere's Law. Ampere's Law states that the magnetic field inside a solenoid is directly proportional to the product of the number of turns, the current, and the permeability of free space.
The formula for the magnetic field inside a solenoid is given by:
B = (μ₀ * n * I) / L
Where:
B is the magnetic field (0.360 T),
μ₀ is the permeability of free space (4π × 10⁻⁷ T·m/A),
n is the number of turns (1000),
I is the current (to be determined),
L is the length of the solenoid (29.4 cm or 0.294 m).
Rearranging the formula to solve for the current I:
I = (B * L) / (μ₀ * n)
Substituting the given values into the formula:
I = (0.360 T * 0.294 m) / (4π × 10⁻⁷ T·m/A * 1000 turns)
Simplifying the calculation:
I = (0.10584 N/A) / (12.56637 × 10⁻⁷ N/A²)
I = 8.42 A
Therefore, the solenoid should carry a current of approximately 8.42 Amperes to produce a magnetic field of 0.360 Tesla at its center.