Question

Consider a thin, straight wire carrying a constant current I and placed along the y-axis as shown in Figure. Determine the magnitude and direction of the magnetic field at point P on x axis due to this current if the direction of current is in y axis.

Answer 1

From the Biot–Savart law, we expect that the magnitude of the field is proportional to the current in the wire and decreases as the distance a from the wire to point P increases.

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.

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 S the current in this element is out of the page because $d \vec{s} \times \hat{r}$ is out of S 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.

Evaluate the cross product in the Biot–Savart law:

$d \vec{s} \times \hat{r} = |d \vec{s} \times \hat{r}| \hat{k} \\ = [dx sin (\frac{\pi}{2} -θ)] \hat{k} \\ = (dx cosθ) \hat{k} \\ d \vec{B} = (dB) \hat{k} = \frac{\mu_0 I}{4 \pi} \frac{dx cos θ}{r^2} \hat{k}$

From the geometry in Figure, express r in terms of u:

$r= \frac{a}{cos θ}$

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

$x= -a tan θ$

Find the differential dx:

$dx= -a sec^2 θ dθ = -\frac{a dθ}{cos^2 θ}$

Substitute equations into the magnitude of the field:

$dB=-\frac{\mu_0I}{4 \pi} (\frac{a dθ}{cos^2 θ})(\frac{cos^2 θ}{a^2}) cos θ = - \frac{\mu_0 I}{4 \pi a} cos θ dθ$

Integrate equation over all length elements on the wire, where the subtending angles range from θ₁ to θ₂ as defined in Figure:

$B= -\frac{\mu_0 I}{4 \pi a} \int^{θ_2}_{θ_1} cos θ d θ = \frac{\mu_0 I}{4 \pi a} (sin θ_1 -sin θ_2)$

We can use this result to find the magnetic field of any straight current- carrying wire if we know the geometry and hence the angles θ₁ and θ₂. Consider the special case of an infinitely long, straight wire. If the wire in b becomes infinitely long, we see that $θ_1 = \frac{\pi}{2}\; and\; θ_2=-\frac{\pi}{2}$ for length elements ranging between positions $x = -\infty \;and \;x =+ \infty$ . Because $(sin θ_1 -sin θ_2) = [sin \frac{\pi}{2} -sin (-\frac{\pi}{2})]=2$

$B= \frac{\mu_0 I}{2 \pi a}$

The magnitude of the magnetic field is proportional to the current and decreases with increasing distance from the wire, as expected.