Answer:
0.1675 A
Explanation:
Since [tex]BILsin\theta=mg[/tex] then making I the subject [tex]I=\frac {mg}{BLsin\theta}[/tex] where L is the length of the rod, in this case given as 1.0792 m, B is magnetic field which is given as 2.88578 T, m is the mass which is 53.1794 g which is equivalent to 0.0531794 Kg . For minimum current, [tex]sin\theta=1.[/tex]
Substituting the given values then
[tex]I=\frac {0.0531794\times 9.81}{1.0792\times 2.88578}=0.167512525 A\approx 0.1675 A[/tex]
