\(\displaystyle\sum{F}={m}{g}\)

\(\displaystyle{I}{L}{B}{\sin{\theta}}={m}{g}\)

\(\displaystyle{I}={\frac{{{m}{g}}}{{{L}{B}{\sin{\theta}}}}}\)

We need the minimum value of current (I) \(\displaystyle\therefore{L}{B}{\sin{\theta}}\) to be the minimum value.

\(\displaystyle\therefore\theta={90}\)

\(\displaystyle{I}={\frac{{{m}{g}}}{{{L}{B}{\sin{{\left({90}\right)}}}}}}={\frac{{{m}{g}}}{{{L}{B}}}}={\frac{{{\left({0.050}\right)}{\left({9.80}\right)}}}{{{\left({1.00}\right)}{\left({0.10}\right)}}}}={4.9}{A}\)

\(\displaystyle{I}{L}{B}{\sin{\theta}}={m}{g}\)

\(\displaystyle{I}={\frac{{{m}{g}}}{{{L}{B}{\sin{\theta}}}}}\)

We need the minimum value of current (I) \(\displaystyle\therefore{L}{B}{\sin{\theta}}\) to be the minimum value.

\(\displaystyle\therefore\theta={90}\)

\(\displaystyle{I}={\frac{{{m}{g}}}{{{L}{B}{\sin{{\left({90}\right)}}}}}}={\frac{{{m}{g}}}{{{L}{B}}}}={\frac{{{\left({0.050}\right)}{\left({9.80}\right)}}}{{{\left({1.00}\right)}{\left({0.10}\right)}}}}={4.9}{A}\)