Q.Derive an expression for the magnetic induction at a point due to an infinitely long straight conductor carrying current. Write the expression for the magnetic induction when the conductor is placed in a medium of permeability '$\mu$'.

Applying the Biot-Savart law to every current element of an infinitely long straight wire and integrating over its full length gives $B=\mu_0 I/2\pi a$ at perpendicular distance $a$, becoming $B=\mu I/2\pi a$ in a medium of permeability $\mu$.

Setup

Consider a long straight conductor $XY$ carrying a steady current $I$. Let $P$ be a point at perpendicular distance $a$ from the wire, with $O$ the foot of the perpendicular from $P$ onto the wire. Consider a small current element $I\,dl$ of the wire located at distance $l$ from $O$. Let $r$ be the distance from this element to $P$, and let $\theta$ be the angle between the direction of the current element $d\vec{l}$ and the vector $\vec{r}$ joining the element to $P$. It is convenient to use instead the angle $\phi$ between $OP$ (the perpendicular) and the line joining the element to $P$, so that $\theta = 90^\circ-\phi$ and hence $\sin\theta=\cos\phi$.

Biot-Savart law for the element

$dB = \frac{\mu_0}{4\pi}\cdot\frac{I\,dl\sin\theta}{r^2} = \frac{\mu_0}{4\pi}\cdot\frac{I\,dl\cos\phi}{r^2}$

directed perpendicular to the plane containing the wire and $P$ (into or out of the page, by the right-hand rule); all elements of the straight wire give a field at $P$ in the same direction, so the magnitudes simply add.

Geometry substitution

From the right triangle formed by $O$, the element, and $P$: $l = a\tan\phi \Rightarrow dl = a\sec^2\phi\,d\phi$, and $r = a\sec\phi \Rightarrow r^2 = a^2\sec^2\phi$.

Substituting:

$dB = \frac{\mu_0}{4\pi}\cdot\frac{I(a\sec^2\phi\,d\phi)\cos\phi}{a^2\sec^2\phi} = \frac{\mu_0 I}{4\pi a}\cos\phi\,d\phi$

Integration

Let $\phi_1$ and $\phi_2$ be the angles subtended at $P$ by the two ends $X$ and $Y$ of the wire, measured from the perpendicular $OP$ (taken as positive on either side). Integrating over the length of the wire:

$B = \int dB = \frac{\mu_0 I}{4\pi a}\int_{-\phi_1}^{\phi_2}\cos\phi\,d\phi = \frac{\mu_0 I}{4\pi a}\left(\sin\phi_1+\sin\phi_2\right)$

This is the general result for a straight conductor of finite length, giving the field at a point in terms of the angles subtended by its ends.

Infinitely long conductor

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