A long, straight wire lies along the y -axis and carries a...

A long, straight wire lies along the $y$ -axis and carries a current $I=8.00 \mathrm{A}$ in the $-y$ -direction (Fig. $\mathrm{E} 28.23$ ). In addition to the magnetic field due to the current in the wire, a uniform magnetic field $\vec{\boldsymbol{B}}_{0}$ with magnitude $1.50 \times 10^{-6} \mathrm{T}$ is in the $+x$ -direction What is the total field (magnitude and direction) at the following points in the $x z$ -plane: (a) $x=0, z=$ $1.00 \mathrm{m} ;$ (b) $x=1.00 \mathrm{m}, \quad z=0$ (c) $x=0, z=-0.25 \mathrm{m} ?$

Transcript

00:01 So in this question, we have a wire.

00:05 So let's just draw out some axes here.

00:08 So x, y, and z.

00:14 We have a wire, which is along the y axis, and it's carrying a current i in the negative y direction.

00:32 So what we're going to have is there's going to be a current from this y.

00:40 Which is going to be, it's not going to have a component in the y direction, and it's going to be mu -nought i over 2 pi r, that's going to be the magnitude of it, but let's think about its components in the x, y, and z directions.

00:59 The right -hand rule tells us that the current, that the magnetic field will be circulating in a right -handed direction around this wire, which means that its x component is negative and proportional to z, and its z component is proportional to positive x.

01:18 So the z component is proportional to positive x.

01:25 So this is going to be x over r.

01:28 So here, i'm writing this in terms of angles, basically, by dividing out by the distance you are away.

01:37 This is minus z over r.

01:40 And here r is the square root of x squared plus z squared.

01:48 So this is what the magnetic field from the y is going to look like.

01:51 And we know that it has this magnitude just from impairs law, which tells us that the integral of the magnetic field around the loop is equal to mu -nought times the current enclosed.

02:04 So you have 2 pi r times b is mu -nought i, which is what we've got here.

02:11 And then here we want a unit vector, which is proportional to minus z in the x direction, which you can see down here when z is negative, the x components is positive, and proportional to x in the z direction.

02:24 As you can see up here, it's going upwards.

02:26 This is a unit vector, because when i square these two things and add them together, i get r squared over r squared, so it's one.

02:35 But then we also have a magnetic field b0, which is 1 .5 times 10 to the much.

02:44 Minus 6 in the x direction, tesla.

02:50 So that means that the total magnetic field is the sum of these two.

02:56 So it's going to be mu -nought -i over 2 pi x squared plus z squared, because i've taken out this r, and when it hits this r, we're going to get an r -squared, and then this is going to be minus z -0 -x, and then we have to add on 1 .5 times 10 to the minus 6, 0.

03:23 So this is what we've got here.

03:26 So now we're asked to calculate it at various points.

03:31 So we have x equals 0, z equals 1.

03:38 And here we're going to have b equals mu nought i over 2 pi, and z squared is 1, x squared is 0.

03:48 And we've got minus z, so minus 1, 0, plus 1.

03:53 1 .5 times 10 to the minus 6, 0.

04:00 Okay, so now we're going to use an expression for mu -naut, which is 4 pi times 10 to the minus 7.

04:16 So we're going to use the fact that mu -naut equals 4 pi times 10 to the minus 7, and we're going to recall that i is 8 amps.

04:31 So that means that we have b is fully in the x direction.

04:37 We've got this mu -knot, which will lead us to 2 times 10 to minus 7 times 8.

04:42 So we've got 16 times 10 to the minus 7 plus 1 .5 times 10 to the minus 6, and this is in the x direction...

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