Answer
(a) $16.5mT$ to the left.
(b) $17.6mT$ to the right.
Work Step by Step
(a) We know that
$B=\mu_{\circ}n_1I_1(-\hat x)$
We plug in the known values to obtain:
$B=-(4\pi\times 10^{-7}T.m/A)(105cm^{-1}(\frac{100cm}{1m}))(1.25)\hat x$
$B=-1.65\times 10^{-3}T\hat x$
$B=(-1.65mT)\hat x$
The negative sign shows that the magnetic field is directed toward the left.
(b) We know that
$B=\mu_{\circ}(n_2I_2-n_1I_1)\hat x$
We plug in the known values to obtain:
$B=(4\pi \times 10^{-7}T.m/A)[125cm^{-1}(\frac{100cm}{1m})(2.17A)-(105cm^{-1}(\frac{100cm}{1m})(1.25A)]\hat x$
$B=(17.6\times 10^{-3}T)\hat x$
$B=(17.6mT)\hat x$
The positive sign shows that the magnetic filed is directed toward the right.
