Answer
(a) $57A $
(b) $0.15T $
(c) $8.7KJ/m^3$
Work Step by Step
(a) We can find the required amount of current as follows:
$ L=\mu_{\circ}N^2(\frac{\pi d^2}{4})$
We plug in the known values to obtain:
$ L=\frac{(4\pi\times 10^{-7}m/A)(580truns)^2(\pi \frac{(7.2\times 10^{-2}m)^2}{4})}{28\times 10^{-2}m}$
$ L=6.15\times 10^{-3}H $
Now $ U=\frac{1}{2}LI^2$
This can be rearranged as:
$ I=\sqrt{\frac{2U}{L}}$
We plug in the known values to obtain:
$ I=\sqrt{\frac{2(9.9J)}{6.15\times 10^{-3}H}}$
$ I=57A $
(b) As $ B=\mu_{\circ}\frac{N}{l}I $
We plug in the known values to obtain:
$ B=\frac{(4\pi\times 10^{-7}T.m/A)(580 turns)(57A)}{28\times 10^{-2}m}$
$ B=0.15T $
(c) The energy density can be determined as
$ U_B=\frac{B^2}{2\mu_{\circ}}$
We plug in the known values to obtain:
$ U_B=\frac{(0.148T)^2}{2(4\pi\times 10^{-7}T.m/A)}$
$\implies U_B=8.7\times 10^3J/m^3=8.7KJ/m^3$
