A solenoid used to produce magnetic fields for research purposes is 1.9m long, with an inner radius of 26cm and 1500 turns of wire. When running, the solenoid produced a field of 1.1 T in the center.
Given this, how large a current does it carry?
Does the answer have something to do with the formula B = u0*N*I/2R?
Thank you!
Given this, how large a current does it carry?
Does the answer have something to do with the formula B = u0*N*I/2R?
Thank you!
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See the website below for pictures and equations for solenoids.
http://hyperphysics.phy-astr.gsu.edu/hba…
For a solenoid, B = u0 * N/L * I
u0 = 1.26 * 10^-6
B = 1.1 tesla
N/L = number of turns ÷ length of the solenoid = 1500 ÷ 1.9
1.1 = 1.26 * 10^-6 * (1500 ÷ 1.9) * I
I = 1.1 ÷ [1.26 * 10^-6 * (1500 ÷ 1.9)]
I = 1105.8 amps
http://hyperphysics.phy-astr.gsu.edu/hba…
For a solenoid, B = u0 * N/L * I
u0 = 1.26 * 10^-6
B = 1.1 tesla
N/L = number of turns ÷ length of the solenoid = 1500 ÷ 1.9
1.1 = 1.26 * 10^-6 * (1500 ÷ 1.9) * I
I = 1.1 ÷ [1.26 * 10^-6 * (1500 ÷ 1.9)]
I = 1105.8 amps
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B = muzero N I /L for a solenoid.