Answer
$B_{0} = 5.03 \times 10^{5} tesla$
Work Step by Step
Given:
coil wire 0.5 m long
20 turns with current 1.0 A
Required:
flux density within a vacuum
Solution:
Using the combination of Equations 20.1 and 20.3:
$B_{0} = µ_{0}H = \frac{µ_{0}NI}{l} = \frac{(1.257 \times 10^{-6} H/m)(20 turns)(1.0 A)}{(0.5 m)} = 5.03 \times 10^{5}\ tesla$
