c
(c)\({B^2} = B_V^2 + B_H^2\, \Rightarrow \,{B_V} = \sqrt {{B^2} - B_H^2} = \sqrt {{{(0.5)}^2} - {{(0.3)}^2}} = 0.4\)
Now \(\tan \phi = \frac{{{B_V}}}{{{B_H}}} = \frac{{0.4}}{{0.3}} = \frac{4}{3}\)
\(==>\) \(\phi = {\tan ^{ - 1}}\left( {\frac{4}{3}} \right)\).