Three long, parallel conductors carry currents of I= 2.00 A. Figure P 30.11 is an end view of th

step 1 In this question, we have three long wires. In this case, all the wires, the current are pointin

Three long, parallel conductors carry currents of I= 2.00 A....

Three long, parallel conductors carry currents of $I=$ 2.00 A. Figure $\mathrm{P} 30.11$ is an end view of the conductors, with each current coming out of the page. Taking $a=$ $1.00 \mathrm{cm},$ determine the magnitude and direction of the magnetic field at points $A, B,$ and $C .$

Key Concepts

Magnetic Field Due to a Current-Carrying Conductor

A long straight conductor carrying an electric current produces a magnetic field that circles around the conductor. The magnitude of this field at a distance r from the wire is given by the formula B = (?0I)/(2?r), where ?0 is the permeability of free space and I is the current. This concept is fundamental for analyzing magnetic fields generated by currents.

Superposition Principle

When multiple currents are present, each current generates its own magnetic field. The net magnetic field at any point is the vector sum of the individual fields produced by each current. This superposition principle is essential in determining the resultant magnetic field from several sources.

Right-Hand Rule

The right-hand rule is a technique used to determine the direction of the magnetic field around a current-carrying conductor. By pointing the thumb of the right hand in the direction of the current, the curling of the fingers shows the direction of the magnetic field lines. This helps in visualizing and calculating the field orientation in different regions surrounding the conductor.

Transcript

00:01 In this question, we have three long wires.

00:04 In this case, all the wires, the current are pointing out of the page.

00:10 So we are interested to find the magnetic field at points a, b and c, three points, a, b, and c.

00:21 And then we are also given the distances, a, and a is equal to 1 cm.

00:30 Then we want to find the magnetic fuse that points a b and c okay so so that's one formula that we'll be using okay and the formula is a magnetic field due to a long wire which is mu not i or 2 pi r okay and the next thing is we need to determine the direction of magnetic field due to each wire at each point and so let's look at point a, okay, at point a.

01:09 So point a is here, you have i1, you can call this i2, and then i3, right? and then we know the distances.

01:24 So this distance will be square to cm, okay, or square to a, square to a, square to a, square to cm, okay, since a is 1 is 1, cm and this distance is 3 cm okay from the diagram okay yeah and then we need to find the individual view okay so we have i1 pointing this way and then i2 pointing this way b2 sorry b2 this is b1 okay and then b3 pointing down okay they are all determined by using right hand rule.

02:16 And then this angle is 45 degrees.

02:23 So you can find the many tube.

02:27 So v1 is equal to b2.

02:30 And then is mu not.

02:34 And the current is 2 ampere.

02:39 K, mu not times 2, by 2 pi.

02:44 And then this is square 2 times 10 to the next...