# EM Question on two rods and force between two and work out linear mass density.

#### riktorres

Two metallic rods of length L = 40 cm and unknown linear mass density 2 are suspended from a support using lightweight strings of length s = 8 cm, as shown in the diagram (a) below. The rods are then connected to a circuit in such a way that same current I = 20 A passes through both rods, but flows in opposite directions. When connected in such a way, the rods move away from each other and at equilibrium are separated by a distance r = 2 cm (diagram b). What is the linear mass density 1 (in kg/m) of the rods?

Attached is a picture of question.

Help would be appreciated

#### Attachments

• 191.4 KB Views: 3

#### topsquark

Forum Staff
Two metallic rods of length L = 40 cm and unknown linear mass density 2 are suspended from a support using lightweight strings of length s = 8 cm, as shown in the diagram (a) below. The rods are then connected to a circuit in such a way that same current I = 20 A passes through both rods, but flows in opposite directions. When connected in such a way, the rods move away from each other and at equilibrium are separated by a distance r = 2 cm (diagram b). What is the linear mass density 1 (in kg/m) of the rods?

Attached is a picture of question.

Help would be appreciated
The force between the two wires is $$\displaystyle F = \dfrac{ \mu _0 I_1 I_2 }{ 4 \pi r^2}$$. Since the currents are going in different directions the force is repulsive. (It would have to be otherwise the metallic rods would not separate, so this checks.) These forces will be horizontal to the floor on the diagram.

Now all we need is the mass of the rods. You can set up a quick Free Body Diagram on one of the rods. We have the weight of the rod: $$\displaystyle w = mg = \lambda Lg$$ straight down, the magnetic force on the rod F, and the tension in the wire.

Can you finish? Let us know how it goes.

-Dan

#### riktorres

Ah much appreciated for your help. It really is that simple then...I was getting confused because I had thought that the field B was a vector and there would be vector cross products involved in this questions. But obviously not - I worked it out on the attached file and am pretty sure it is correct.

Once again thank you very much Rik

#### Attachments

• 765.4 KB Views: 4