Answer
$35mA$
Work Step by Step
We know that
$I=\frac{BL}{\mu_{\circ}N}$
We plug in the known values to obtain:
$I=\frac{(5.0\times 10^{-5})(0.38)}{4\pi \times 10^{-7}(430)}$
$I=35mA$
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