Question

A thin, horizontal copper rod is 1.0792 m long and has a mass of 53.1794 g. The acceleration of gravity is 9.8 m/s 2 . What is the minimum current in the rod that will cause it to float in a horizontal magnetic field of 2.88578 T

Answer 1

Answer:

0.1675 A

Explanation:

Since $BILsin\theta=mg$ then making I the subject $I=\frac {mg}{BLsin\theta}$ 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, $sin\theta=1.$

Substituting the given values then

$I=\frac {0.0531794\times 9.81}{1.0792\times 2.88578}=0.167512525 A\approx 0.1675 A$