Incomplete question as we have not told which quantity to find.So the complete question is here
A solenoid used to produce magnetic fields for research purposes is 2.2 mm long, with an inner radius of 25 cmcm and 1300 turns of wire. When running, the solenoid produced a field of 1.5 TT in the center.Given this, how large a current does it carry?
Answer:
[tex]I=2.021A[/tex]
Explanation:
Magnetic field B=1.5 T
Length L=2.2mm =0.0022m
Number of turns N=1300 turns
To find
Current I
Solution
From the magnetic at the center of loop we know that:
[tex]B=uI\frac{N}{L}[/tex]
Substitute the given values
[tex]B=uI\frac{N}{L}\\ as\\u=4\pi *10^{-7} T.A/m\\So\\B=uI\frac{N}{L}\\I=\frac{LB}{Nu}\\ I=\frac{0.0022m(1.5T)}{(1300)(4\pi *10^{-7} T.A/m)}\\I=2.021A[/tex]
