Question

Find the direction and magnitude of the force that each wire experiences in Figure \(22.58(a)\)by, using vector addition.

Short answer

The direction and magnitude of the force that each wire experiences is

Top:\({F_1} = {2.6510^{ - 4}}{\rm{ }}N/m\),\({\alpha _1} = {10.9^\circ }\)

Bottom left:\({F_2} = {3.61^{ - 4}}{\rm{ }}N/m\),\({\alpha _2} = {13.9^\circ }\)

Bottom right:\({F_3} = {3.46^{ - 4}}{\rm{ }}N/m\),\({\alpha _3} = {30^\circ }\)

Step-by-step solution

Define force

A force is a push or pull that an object experiences as a result of interacting with another object.

Step 2:Given Information

The circuit diagrams:

Step 3:Evaluating the direction and magnitude of the forces

Apply equation \(\frac{F}{l} = \frac{{{\mu _0}{I_1}{I_2}}}{{2\pi r}}\)to calculate force between wires.

After solving for every wire, we get:

- top and bottom left:\({F_1} = {\rm{ }}1\cdot{10^{ - 4}}{\rm{ }}N/m\)

- top and bottom right:\({F_2} = {\rm{ }}2\cdot{10^{ - 4}}{\rm{ }}N/m\)

- bottom left and right:\({F_3} = {\rm{ }}4\cdot{10^{ - 4}}{\rm{ }}N/m\)

Calculation using vector addition, we get:

Top:\({F_1} = {2.6510^{ - 4}}{\rm{ }}N/m\)

At angle:\({\alpha _1} = {10.9^\circ }\)to the left of vertical.

Bottom left:\({F_2} = {3.61^{ - 4}}{\rm{ }}N/m\)

At angle =\({\alpha _2} = {13.9^\circ }\)to the down of right.

Bottom right:\({F_3} = {3.46^{ - 4}}{\rm{ }}N/m\)

At angle:\({\alpha _3} = {30^\circ }\)to the down of left.

Every force is divided into its components on the two primary axes, vertical and horizontal, in order to perform vector addition. The components of every force operating on the same item of interest facing the same direction are then totalled together. They end up with two final resultant force components, one vertical and the other horizontal. Then, to determine the resulting force, let add them all up:

\({F_{total}} = \sqrt {F_{hor}^2 + {F_{vert}}} \)

And for angle:\(\alpha = {\tan ^{ - 1}}\left( {\frac{{{F_{vert}}}}{{{F_{hor}}}}} \right)\).

Therefore,

Top:\({F_1} = {2.6510^{ - 4}}{\rm{ }}N/m\),\({\alpha _1} = {10.9^\circ }\)

Bottom left:\({F_2} = {3.61^{ - 4}}{\rm{ }}N/m\),\({\alpha _2} = {13.9^\circ }\)

Bottom right:\({F_3} = {3.46^{ - 4}}{\rm{ }}N/m\),\({\alpha _3} = {30^\circ }\)