Answer:
35.7 mA
Explanation:
The magnetic field inside a solenoid is given by:
[tex]B=\mu_0 I n[/tex] (1)
where
[tex]\mu_0 = 4\pi \cdot 10^{-7} H/m[/tex] is the vacuum permeability
I is the current
n is the number of turns per unit length
Since we have
N = 870 turns
L = 0.390 (length of the solenoid)
we can calculate n
[tex]n=\frac{N}{L}=\frac{870}{0.390}=2230.8[/tex]
And now we can re-arrange eq.(1) to find the current, I:
[tex]I=\frac{B}{\mu_0 n}=\frac{1.00\cdot 10^{-4} T}{(4\pi\cdot 10^{-7} H/m)(2230.8m^{-1})}=0.0357 A = 35.7 mA[/tex]
