A conducting rod of length l, mass m, and resistance R is suspended by two vertical conducting wires of lengths d each. The circuit consists of a battery ε, the horizontal fixed wires, two parallel vertical wires , and the conducting rod. All conducting wires have negligible resistance and mass. There is a conducting long cable. The cable is below the rod and at a distance x.

The current in the rod flows from right to left and the current in the cable flows from left to right. The current in the cable exerts an upward force on the rod. The rod is lifted up such that there is no tension in the vertical connecting wires.

Determine the minimum current in the cable.

###### Solution.

There are two forces acting on the rod: a downward force of gravity **mg** and an upward magnetic force **F _{B}** from the magnetic field of the current in the cable. According to Newton’s First law, for the rod at rest in the magnetic field, these forces cancel each other out and the net force on the rod is zero.

mg = F_{B} (1)

The magnetic field **B** is perpendicular to the rod (sin(90)=0), so

F_{B} =IlB (2),

where I – the current in the rod, l – the length of the rod and B – the magnitude of the magnetic field.

According to the law of Biot and Savart, the magnitude of the magnetic field at a perpendicular distance **x** from the long cable carrying a current **i** is given by

B = μ_{0}i/(2πx) (3)

We can use the Ohm’s law and get the current in the rod

I = ε/R. (4)

We substitute equations (3) and (4) into equation (2) and we get

F_{B} = εlμ_{0}i/(2πxR) (5)

Now we substitute equation (5) into equation (1)

mg = εlμ_{0}i/(2πxR)

and finally,

i = 2πxRmg/(εlμ_{0})