Question

Consider a straight piece of length $x$ of a wire carrying current $i$. Let $P$ be a point on the perpendicular bisector of the piece, situated at a distance $d$ from its middle point. Show that for $d \gg \,x$, the magnetic field at $P$ varies as $\dfrac{1}{{{d^2}}}$ whereas for $d \ll \,x$, it varies as $\dfrac{1}{d}$.

Answer 1

Hint: here, the wire will be in the form of conductor carrying current $i$ which is of the length $x$ . Also, we will draw a perpendicular bisector on the wire from point $P$. Now, to find the magnetic field on the wire we will use Biot-Savart’s law in case of current carrying wire.

Formula used:
The formula of the magnetic field is given by
$B = \dfrac{{{\mu _0}i}}{{2\pi d}}\sin \theta $
Here, ${\mu _0}$ is the permeability in free space, $i$ is the current in the wire and $d$ is the distance of the point from the wire, where electric field is to be calculated.

Complete step by step answer:
Consider a straight piece of wire of length $x$ and carrying current $i$. Here, we will consider a point $P$ which will be on the perpendicular bisector of the wire. This point will be situated at a distance $d$ from the middle point of the wire.
Therefore, the current $ = i$
And the distance of the point $P$ from the wire, $ = d$
Now, the magnetic field induced on the wire is shown below
$B = \dfrac{{{\mu _0}i}}{{2\pi d}}\sin \theta $
Now, for the perpendicular bisector of the wire, the magnetic field will be as shown below
$B = \dfrac{{{\mu _0}i}}{{2\pi d}}2\sin \theta $
Now, we will use the identity as shown below
$B = \dfrac{{{\mu _0}i}}{{2\pi d}}\dfrac{{2 \times \dfrac{x}{2}}}{{\sqrt {{x^2} + 4{d^2}} }}$
$ \Rightarrow \,B = \dfrac{{{\mu _0}i}}{{4\pi d}}\dfrac{{2x}}{{\sqrt {{x^2} + 4{d^2}} }}$
Now, as given in the question, we have to calculate in the cases, when $d \gg \,x$ and $d \ll \,x$
Now, for $d \gg x$ , the $x$ can be neglected relative to $d$ , therefore, the magnetic field induced in the wire will become
$B = \dfrac{{{\mu _0}i}}{{4\pi d}}\dfrac{{2x}}{{\sqrt {4{d^2}} }}$
$ \Rightarrow \,B = \dfrac{{{\mu _0}i}}{{4\pi d}}\dfrac{{2x}}{{2d}} = \dfrac{{{\mu _0}i}}{{4\pi d}}\dfrac{x}{d}$
$ \Rightarrow \,B \propto \dfrac{1}{{{d^2}}}$
Which is the required condition.
Now, for $d \ll x$ , the $d$ can be neglected relative to $x$, therefore, the magnetic field induced in the wire will become
$B = \dfrac{{{\mu _0}i}}{{4\pi d}}\dfrac{{2x}}{{\sqrt {{x^2}} }}$
$ \Rightarrow \,B = \dfrac{{{\mu _0}i}}{{2\pi d}}$
$ \therefore \,B \propto \dfrac{1}{d}$
This is the required equation.

Note:Here, we have taken the length of the wire as $\dfrac{x}{2}$ because the perpendicular bisector will divide the wire in two parts. Also, the magnetic field on the perpendicular bisector is given by
$B = \dfrac{{{\mu _0}i}}{{2\pi d}}\left( {\sin {\theta _1} + \sin {\theta _2}} \right)$
But, in the above question, both the angles will be the same. That is why, we have taken the magnetic field as $B = \dfrac{{{\mu _0}i}}{{2\pi d}}2\sin \theta $.