Answer
(a) $1.6mT$
(b) The amplitudes will be increased by a factor of $\sqrt 2$.
Work Step by Step
(a) We know that
$B=\sqrt{2u_{\circ}u_{avg}}$
We plug in the known values to obtain:
$B=\sqrt{2(4\pi \times 10^{-7})(1.0)}$
$B=1.6mT$
(b) As the energy density is directly proportional to the square of the field amplitude, the amplitudes will be increased by a factor of $\sqrt 2.$
