Two long, parallel wires are hung by cords of length 5.0 cm, as shown in the accompanying figure

Two long, parallel wires are hung by cords of length 5.0 cm,...

Key Concepts

Force Equilibrium

In many physical systems, objects remain at rest when the net force acting on them is zero. In problems involving suspended wires, the magnetic force acting horizontally is balanced by the horizontal component of the tension in the supporting cords, while the vertical component of tension counteracts gravitational forces.

Magnetic Force Between Currents

This concept explains that two parallel wires carrying currents exert forces on each other. The force per unit length between them is given by an expression that involves the current values and the distance between the wires, as well as the permeability of free space. Depending on the relative direction of the currents, this force can be either attractive or repulsive.

Magnetic Field of a Current-Carrying Wire

A long, straight wire generates a magnetic field around it, which forms concentric circles. The magnitude of this field at any given distance from the wire is determined by the current and is inversely proportional to the distance. This magnetic field is central in calculating the force that one wire exerts on another.

Trigonometric Decomposition of Forces

When forces act at angles, it is useful to resolve them into perpendicular components (typically horizontal and vertical). This decomposition enables the application of equilibrium conditions by relating the components of tension in the cords to the gravitational force and the magnetic force.

Transcript

00:01 Hi, in the given problem, there are two parallel current carrying conductors which are supported with the help of two chords, as shown in this figure.

00:27 These two sets of wires, these two sets of cords are supporting the two cables.

00:36 If we draw the front view of this diagram, this is one set of the cables this is one of the wire which is carrying current suppose in such a direction which is coming out perpendicular to the plane of paper and in other wire the current is in opposite direction means it is getting perpendicularly into the plane of paper this is a b and c we can assume the weight of the wire this is m -g acting vertically downward the force between these two current carrying conductors acting in acting in this manner that the wires will repel each other this is the force f this point may be assumed to be the angle these chords are making with the vertical is 6 .0 degree and here this angle will also be 6 .3 0 degree the tension in each chord is supposed to be t so it will be having two components t cost 6 .0 degree vertically upward and t sign 6 .0 degree horizontal as the wires in equilibrium are in equilibrium so we can say these forces there is no force no net force in vertically in vertical direction and in a horizontal direction no net force so we can say this t sine 6 .0 degree is equal to f the force between two parallel current carrying conductors and this t cause 6 .0 degree is equal to m g the weight of one of the wire if we divide them we'll get t sign 6 .0 degree divided by t cost 6 .0 degree is equal to f by m g so they got cancelled and we get this is 6 10 6 .0 degree now for the force between two parallel current carrying conductors this is mu not divided by 4 pi 2 i 1 i2 divided by the gap between these wires and which is bc given here divided by m g here the point to be noted is that this force is per unit length and this mass is also per unit length of the wires so this per unit length got cancelled in numerator and denominator both so finally this expression here becomes mu not by 4 pi into two i square because the magnitude of current is same in both the wires so this becomes 2 into i squared divided by bc into m g here if you look at this bc this is nothing but twice of bd and in this right angle triangle abd this bd is ab is ab sine 6 .0 degree so we can put all these values here this is mu not by 4 .0 pi into 2 i square divided by for b c this is 2 a b sign 6 .0 degree and multiplied by m g now mass of one of the wire per unit length has been given as 30 gram per meter or we can say this is 0 .03 kilogram per meter but as there are two sets of the cords which are supporting the weight of these wires so we can say mass or weight supported by each cord will become half of this as two chords are supporting each wire so effective mass of this wire supported by each cable each cord will be half of this so we will use the value m -dash as 0 .03 divided by 2 means 0 .015 kilogram per meter here so putting all these values now here for 10 6 degree this is 0 .105 is equal to for mu not upon 4 pi this is 10 dash for minus 7 2 i squared the current is missing divided by 2 for a b this is 5 .0 into 10 dash power minus 2 then sign 6 degree for mass we have just found this is 0 .015 and for g this is 9 .8 so here these values come out to be this 2 got cancelled in the numerator, this is 10 -1 minus 7 into i -square and for denominator, the overall value here comes out to be 0 .015 into 10 -bar minus.

07:04 So the value of i -square here becomes the expression for this i -square becomes i square is equal to 0 .105 multiplied by 0 .075...

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