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CHAPTER 29 MAGNETIC FIELDS 819

Example 29.1 cont.

Model We are asked to find the magnetic field due to a simple current distribution, so this example is a typical problem for which the Biot–Savart law is appropriate. We must find the field contribution from a small element of current and then integrate over the current distribution from $\theta_1$ to $\theta_2$ , as shown in Figure 29.3b.

Analyse Let's start by considering a length element $d\vec{s}$ located a distance $r$ from $P$ . The direction of the magnetic field at point $P$ due to the current in this element is out of the page because $d\vec{s} \times \hat{r}$ is out of the page. In fact, because all the current elements $I d\vec{s}$ lie in the plane of the page, they all produce a magnetic field directed out of the page at point $P$ . Therefore, the direction of the magnetic field at point $P$ is out of the page and we need only find the magnitude of the field. We place the origin at $O$ and let point $P$ be along the positive $y$ axis, with $\hat{k}$ being a unit vector pointing out of the page.

From the geometry in Figure 29.3a, we can see that the angle between the vectors $d\vec{s}$ and $\hat{r}$ is $\left(\frac{\pi}{2} - \theta\right)$ radians.

Evaluate the cross product in the Biot–Savart law:

$d\vec{s} \times \hat{r} = |d\vec{s} \times \hat{r}| \hat{k} = \left[ dx \sin\left(\frac{\pi}{2} - \theta\right) \right] \hat{k} = (dx \cos \theta) \hat{k}$

Substitute into Equation 29.1:

$d\vec{B} = (dB)\hat{k} = \frac{\mu_0 I}{4\pi} \frac{dx \cos \theta}{r^2} \hat{k} \quad (1)$

From the geometry in Figure 29.3a, express $r$ in terms of $\theta$ :

$r = \frac{a}{\cos \theta} \quad (2)$

Notice that $\tan \theta = -x/a$ from the right triangle in Figure 29.3a (the negative sign is necessary because $d\vec{s}$ is located at a negative value of $x$ ) and solve for $x$ :

$x = -a \tan \theta$

Find the differential $dx$ :

$dx = -a \sec^2 \theta d\theta = -\frac{a d\theta}{\cos^2 \theta} \quad (3)$

Substitute Equations (2) and (3) into the magnitude of the field from Equation (1):

$dB = -\frac{\mu_0 I}{4\pi} \left( \frac{a d\theta}{\cos^2 \theta} \right) \left( \frac{\cos^2 \theta}{a^2} \right) \cos \theta = -\frac{\mu_0 I}{4\pi a} \cos \theta d\theta \quad (4)$

Integrate Equation (4) over all length elements on the wire, where the subtending angles range from $\theta_1$ to $\theta_2$ as defined in Figure 29.3b:

$B = -\frac{\mu_0 I}{4\pi a} \int_{\theta_1}^{\theta_2} \cos \theta d\theta = \frac{\mu_0 I}{4\pi a} (\sin \theta_1 - \sin \theta_2) \quad (29.4)$

Check the dimensions, noting that the quantity in brackets is dimensionless:

$[MQ^{-1}T^{-1}] = [MLQ^{-2}][QT^{-1}]/[L] = [MQ^{-1}T^{-1}] \text{ ☺}$

(B) Find an expression for the field at a point near a very long current-carrying wire.

Solution

We can use Equation 29.4 to find the magnetic field of any straight current-carrying wire if we know the geometry and hence the angles $\theta_1$ and $\theta_2$ . If the wire in Figure 29.3b becomes infinitely long, we see that $\theta_1 = \pi/2$ and $\theta_2 = -\pi/2$ for

Figure 29.3(a): A thin, straight wire carrying current I lies along the x-axis. Point P is located on the positive y-axis at distance a from the origin O. A small current element d→s is located at position x. The distance r from P to d→s is shown, along with the angle theta between the line segment r and the y-axis. The unit vector ŷ points from d→s towards P.

Figure 29.3(b): A thin, straight wire carrying current I lies along the x-axis. Point P is located on the positive y-axis. The angles theta_1 and theta_2 are shown, representing the angles subtended by the wire at point P. theta_1 is the angle from the positive y-axis to the left end of the wire, and theta_2 is the angle from the positive y-axis to the right end of the wire.

Figure 29.3

(Example 29.1) (a) A thin, straight wire carrying a current $I$ (b) The angles $\theta_1$ and $\theta_2$ are used for determining the net field.