Figure 30-36 shows five long parallel wires in the x y plane. Each wire carries a current i=3.00

Figure 30-36 shows five long parallel wires in the x y...

Key Concepts

Right-Hand Rule

The right-hand rule is a simple but essential tool for determining the directions of magnetic fields and forces in electromagnetic interactions. When applied to the magnetic field from a current, it helps in visualizing the circular field lines around a wire. Similarly, it is used to determine the direction of the magnetic force on a current-carrying wire by aligning the thumb with the current direction and the fingers with the magnetic field direction.

Superposition Principle in Magnetostatics

In magnetostatics, the net magnetic field at any point due to multiple sources is the vector sum of the fields produced by each individual source. This principle allows one to calculate the total force acting on any given wire by summing the contributions from the magnetic fields produced by the other wires, taking into account both magnitude and direction.

Magnetic Field of a Long Straight Conductor

A long straight wire carrying a current produces a magnetic field that circulates around the wire. The magnitude of this magnetic field at a distance r from the wire is given by B = (??i)/(2?r), where ?? is the permeability of free space. This relation is fundamental in determining the magnetic influence one current-carrying wire exerts on its surroundings.

Magnetic Force on a Current-Carrying Conductor

A current-carrying conductor placed in a magnetic field experiences a force given by F = iL × B, where i is the current and L is a vector in the direction of the current with magnitude equal to the length of the conductor. This concept is crucial for analyzing the interaction between the wires, as the magnetic field produced by one wire acts on the currents in the other wires, resulting in forces that can be calculated per unit length.

Transcript

00:01 In this problem, we have five parallel wires, all with the currents coming on to you, side by side wires are separated by distance d, and the goal of this problem is to get the fourth per unit length on each wire due to the other four.

00:15 So what we need first is magnetic field at each of the wire locations.

00:23 And we use the right -hand rule.

00:25 Think of grabbing the wire with your right hand, dominant direction of the current.

00:29 Fingers give you the sense of the arrows on the magnetic field lines, which are circles.

00:35 The magnetic field itself is tangent to that circle.

00:40 Now, for wire at wire one, all the other four wires, using the right -hand rule, your fingers are down at the location of one.

00:48 It means a magnetic field factor are all pointing, all four are pointing in the negative c direction.

00:56 So b at 1 is going to be b1, i mean b2 i should say, b2 plus b3 plus b4 plus b5 minus k hat and this becomes mu not i, 2 pi, 1 over d plus 1 over 2d plus 1 over 3d plus 1 over 4d.

01:30 And you can bring out the d and do a common factors in this, and don't forget the minus k hat, this becomes, this becomes mu not i, 2 pi d, 50 over 24 minus k hat.

01:52 That's what we get for the magnetic field at 1.

01:56 Now for 2, be at 2.

02:00 Now, let's not forget something.

02:04 One is now on the other side of two.

02:08 We give the right -hand rule, the magnetic field from one is straight up, positive z.

02:14 So this will become b1c -hat plus b -3 plus b -4 plus b -5 minus k -hat.

02:30 And you have here b1, so you go to b3, the same distance from wire 2.

02:38 So magnitude is going to be the same, so this should be a three here.

02:47 Let me make it clear.

02:51 That's a three.

02:54 So these two terms are gone.

02:58 So this becomes b4 plus b5 minus k hat, and this turns out to be u -0 -i, 2 ,000, pi d one half plus one third minus k hat and this works out to be u -naut i 2 pi d 56 minus k hat so that's a magnetic field at 2 b at 3 now 2 and 2 are on the left of 3 so they're going to have their contribution is positive positive z direction.

03:56 So i have b1 plus b2, k -hat, plus b4 plus b -5 minus k -hat.

04:08 But because of the position of 2 and 4 being the same around 3 and 1 and 5, we have b -1 is equal to b -5, b -2 is equal to b -4.

04:21 So everything cancels out.

04:30 B at 4.

04:33 Well, b at 4, you have three wires to the left and one to the right.

04:41 Everything's the same as what you had at 2 except the directions have switched.

04:48 So from symmetry, we know that this is just minus b at 2.

04:53 So that becomes m0 .i.

04:59 2 .5 .d.

05:01 5 ,6 k hat.

05:04 And likewise, at 5.

05:08 This is minus b at 1, and this becomes mu not i, 2 pi d, 50 over 24 khecht.