A thin copper rod 1.0 m long has a mass of 0.050 kg and is in a magnetic field of 0.10 T.What minimum current in the rod is needed in order for the magnetic force to balance the weight of the rod?

Question

Quizplus · Accepted Answer

To balance the weight of the rod with the magnetic force, we use the equation for the magnetic force on a current-carrying conductor: $F = BIL$, where $F$ is the force, $B$ is the magnetic field strength, $I$ is the current, and $L$ is the length of the conductor. The weight of the rod, which is the force we want to balance, is given by $W = mg$, where $m$ is the mass of the rod and $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$). Setting these two forces equal to each other gives us $BIL = mg$. Solving for $I$ gives $I = \frac{mg}{BL}$.Substituting the given values: $m = 0.050 \, \text{kg}$, $g = 9.8 \, \text{m/s}^2$, $B = 0.10 \, \text{T}$, and $L = 1.0 \, \text{m}$, we get:$$I = \frac{0.050 \times 9.8}{0.10 \times 1.0} = \frac{0.49}{0.10} = 4.9 \, \text{A}$$Therefore, a minimum current of 4.9 A is needed for the magnetic force to balance the weight of the rod.

A Thin Copper Rod 1