Two long insulated wires lie in the same horizontal plane. A...

Two long insulated wires lie in the same horizontal plane. A current of $20.0 \mathrm{A}$ flows toward the north in wire $A$ and a current of 10.0 A flows toward the east in wire $B .$ What are the magnitude and direction of the magnetic field at a point that is $5.00 \mathrm{cm}$ above the point where the wires cross?

Key Concepts

Magnetic Field from a Current-Carrying Wire

A long straight current-carrying wire produces a magnetic field whose magnitude at a distance r from the wire is given by B = ??I/(2?r), where ?? is the permeability of free space and I is the current. This concept is fundamental as it provides the basic quantitative relationship needed to calculate the magnetic field contribution from each wire in a configuration.

Right-Hand Rule

The right-hand rule is used to determine the direction of the magnetic field created by a current-carrying conductor. By pointing the thumb of the right hand along the direction of the current, the curled fingers indicate the circular direction of the magnetic field lines around the wire. This concept is essential to analyze the directional components of the magnetic fields from different wires.

Superposition Principle in Magnetism

The superposition principle states that the net magnetic field caused by multiple current-carrying wires is the vector sum of the individual magnetic fields produced by each wire independently. This principle is crucial when calculating the total magnetic field at a point influenced by separate field contributions.

Transcript

00:01 In this question, we have two long wires, a and b, with a current of 20 amps and 10 amps respectively.

00:09 The current of 10 amps in wire b flows towards the east.

00:14 What we want to find is the magnitude and direction of the field at a point, shown here and labelled as p, that is 5 centimetres above the point where they cross.

00:25 So what we know is the two currents in the two wires.

00:28 We know that the point p is 5 centimetres above them, so in a plane out of the paper or upwards in this diagram.

00:38 We know that due to the right -hand rule, the field at this point, due to wire a is directed east, and due to wire b is south, because you put your thumb in the direction of the current and look at the way that your fingers curve around.

00:51 So what we know is that the magnetic field due to the current in wire a can be calculated from the biosavrlo, which states that b is equal to mu -nought i, where mu -nought is a constant, i is a current, over 2 pi times r, the distance away from the wire.

01:11 So in the first wire, this is equal to mu -naut i1 over 2 pi r, which is 5 centimetres.

01:24 So we can substitute the values in.

01:27 We know mu -naut has a value of 4 pi times 10 to the minus 7.

01:34 We know the current is 20 amps in the first wire.

01:39 We know 2 pi, and the distance is 5 centimetres, so 5 times 10 to the minus 2 metres.

01:48 And thus, we can calculate that the magnitude of this field is 8 times 10 to the minus 5 tesla's, and it's acting in the east direction.

02:05 We then need to repeat this calculation for the second wire.

02:09 So we can say that b2 is equal to mu -0, so 4 pi times 10 to the minus 7, multiplied by the current in wire b, so that's 10 amps this time, over again 2 pi, multiplied by the distance from the point, so 5 times 10 to the minus 2, which gives us a value for the field of 4 times 10 to the minus 5 tesla's...

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