Q.(a) Derive an expression for electric field intensity due to an electric dipole at a point on its axial line. OR

(a) The axial field of a dipole is obtained by vector-adding the fields of its two point charges along the axis, giving $E_{axial}=\dfrac{2kpr}{(r^2-a^2)^2}$, which reduces to $2kp/r^3$ far from the dipole. (b) The field of an infinite straight current-carrying wire is obtained by integrating the Biot–Savart law over its length, giving $B=\mu_0 I/2\pi a$. Both alternatives are answered in full below.

(a) Electric field on the axial line of a dipole

Consider an electric dipole consisting of charge $-q$ at point $A$ and charge $+q$ at point $B$, separated by a distance $2a$, with centre $O$. The dipole moment is $\vec p = q(2a)$, directed from $-q$ to $+q$. Let $P$ be a point on the axial line (the line through $A$, $O$, $B$ extended), at a distance $r$ from $O$, on the side of the $+q$ charge, with $r>a$.

The distance from $B$ ($+q$) to $P$ is $(r-a)$, and the distance from $A$ ($-q$) to $P$ is $(r+a)$.

The field at $P$ due to $+q$ (pointing away from $B$, i.e. away from the dipole, along the axis) has magnitude

$E_1 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{(r-a)^2}$

The field at $P$ due to $-q$ (pointing towards $A$, i.e. towards the dipole, opposite to $E_1$) has magnitude

$E_2 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{(r+a)^2}$

Since $E_1$ and $E_2$ point in opposite directions along the axis, and $E_1 > E_2$ (because $r - a < r + a$), the resultant field is along $E_1$'s direction, with magnitude

$E_{axial} = E_1 - E_2 = \dfrac{q}{4\pi\varepsilon_0}\left[\dfrac{1}{(r-a)^2} - \dfrac{1}{(r+a)^2}\right] = \dfrac{q}{4\pi\varepsilon_0}\cdot\dfrac{(r+a)^2-(r-a)^2}{(r^2-a^2)^2}$

Using $(r+a)^2-(r-a)^2 = 4ar$,

$E_{axial} = \dfrac{q}{4\pi\varepsilon_0}\cdot\dfrac{4ar}{(r^2-a^2)^2} = \dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{2(q\cdot 2a)r}{(r^2-a^2)^2} = \dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{2pr}{(r^2-a^2)^2}$

since $p = q(2a)$. The field points in the direction of $\vec p$ (from $-q$ to $+q$).

For a point far from the dipole ($r \gg a$), $a^2$ can be neglected compared with $r^2$, giving the familiar short-dipole approximation:

$E_{axial} \approx \dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{2p}{r^3}$

(b) Magnetic induction due to an infinitely long straight current-carrying conductor

Consider a long straight conductor carrying current $I$, and a point $P$ at perpendicular distance $a$ from the wire. Let $O$ be the foot of the perpendicular from $P$ to the wire. Consider a small current element $I\,d\vec l$ on the wire at a distance $l$ from $O$, and let $\vec r$ be the vector from this element to $P$, with $r=\sqrt{a^2+l^2}$.

By the Biot–Savart law, the magnetic field at $P$ due to this element is

$dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\theta}{r^2}$ …