Suppose a straight current carrying conductor AB, carrying an upward current `I`, is placed on the paper. P is a point at a perpendicular distance R from conductor, shown in Fig.

According to Biot-Savart's law, the field at P due to element `dl` is

`therefore` = `mu_0/(4pi)\ \(Idlsintheta)/r^2 = (mu_0I)/(4pi) ((dlsintheta)/r)1/r`

Now from the figure, `dlsintheta = rdphi` and `cosphi = R/r`

`therefore``dB = (mu_0I)/(4pi)\ \cosphi/(R) dphi`

Hence magnetic field due to the whole conductor

`B = int_(-theta_2)^(theta_1) dB = int_(-theta_2)^(theta_1) (mu_0I)/(4piR) cosphidphi = (mu_0I)/(4pi R)(sintheta_1 + sintheta_2)`

`therefore B = (mu_0I)/(4piR)(sintheta_1 + sintheta_2)`

(a) At an axial point and
(b)At the center of the loop

(a) `text(At an axial point :)`

Shown in (Fig.2) Consider a circular loop of radius a carrying a current `i`. We have to find the magnetic field at a point P on the axis of the loop at a distance d from its center O . In the above figure the loop is perpendicular to the plane of the figure while its axis is in the plane of the figure. The current comes out of the plane at M and goes into it at N. Consider the current element `idvecl` of the wire at M. The vector joining the element to the point P is `vecr = vec(MP)` . The magnetic field at P due to this current element is

`dvecB = mu_0/(4pi)\ \(dvecl xx vecr)/r^3`

As `dvecl` is perpendicular to the plane of the figure, `dvecl xx vecr` must be in the plane. The figure shows the direction of `dvecl` according to the rules of vector product. The magnitude of the field is

dB = `mu_0/(4pi) (idl)/r^2=(mu_0i)/(4pi)\ \ (dl)/(a^2 + d^2)`

The component along the axis is

`dBcos theta = (mu_0ia)/(4pi)\ \ (dl)/(a^2 + d^2)^(3/2)` .............................(1)

Now consider the diametrically opposite current element at N. The field due to this element will have the same magnitude dB and its direction will be along the dotted arrow shown in the figure. The two fields due to the elements at M and N have a resultant along the axis of the loop. Dividing the loop in such pairs of diametrically opposite elements, we conclude that the resultant magnetic field at P must be along the axis. The resultant field at P can, therefore, be obtained by integrating the right hand side of (1), i.e.,

`B = int (mu_0ia)/(4pi)\ \ 1/((a^2 + d^2)^(3/2)) dl`

= `(mu_0ia)/(4pi) \ \1/((a^2 + d^2)^(3/2)) xx 2pia`

`B = mu_0/2 \ \ (ia^2)/((a^2 + d^2)^(3/2))`

The right-hand thumb rule can be used to find the direction of the field.

(b) `text(At the center of the loop :)`

The magnetic field at the center of the loop is simply obtained by putting `d = 0` in above eqn., thus obtaining

`B = (mu_0i)/(2a)`

`text(Concentric Circular Loops)` ( N = 1)

(i) Coplanar and concentric : It means both coils are in same plane with common centre

(a) Current in same direction
(b) Current in opposite direction shown in (Fig.3)

`B_1 = mu_0/(4pi) 2pi i(1/r_1 + 1/r_2)`
`B_2 = mu_0/(4pi) 2pi i [1/r_1 - 1/r_2]`

Note: (i) `B_1/B_2 = ((r_1 + r_2)/(r_2 - r_1))`

(ii) Non-coplanar and concentric : Plane of both coils are perpendicular to each other

Magnetic field at common centre shown in (Fig.4)

`B = sqrt(B_1^2 + B_2^2) = mu_0/(2r)sqrt(i_1^2 + i_2^2)`

Arc making an angle `theta` at the centre : To measure the magnetic field at the centre of a circular arc of radius R carrying current `I` making an angle `theta` at the centre, we have to take a very small element at angle `alpha` making `dalpha` at the centre, shown in (Fig.5)

Thus,
`dB = (mu_0I)/(4pi) . (Rdalpha sin90^0)/R^2 = (mu_0I)/(4piR).dalpha`

`B = int_0^theta (mu_0I)/(4piR) dalpha = (mu_0Id theta)/(4piR)`

Or `B= mu_0/(4pi). (I theta)/R` (shown in Fig.6)

Magnetic field at the centre of circular coil is denoted by

`B_0 = (mu_0/(4pi) . (2pi i)/R)`

A solenoid is a long cylindrical helix which is obtained by winding closely a large number of turns of insulated copper wire over a tube of card board or china clay. When electric current is passed through it. a magnetic field is produced around and within the solenoid. If `n` be the number of turns per unit length or the solenoid, then number of turns in width `dx = ndx`. If `i` current is passing through it then magnitude of magnetic induction at P, shown in(Fig.7)

`db = (mu_0(ndx)iR^2)/(2(R^2 + x^2)^3)` along the axis, shown in (Fig.8)

Net magnetic induction `B = intdB = int_(-l_2)^(l_1) (mu_0niR^2dx)/(2(R^2 + x^2)^(3/2))` , shown in (Fig.9)

`B=(mu_0ni)/2[l_1/(sqrt(R^2 + l_1^2)) + l_2/(sqrt(R^2 + l_2^2))]`

= `(mu_0ni)/2[cos theta_1 + cos theta_2]`

or `B = (mu_0ni)/2[cos alpha - cos beta]` , shown in (Fig. 10)

Which is directed along the axis.

`text(Case)` 1: For an ideal solenoid (i.e., solenoid of infinite length) R<< l

=> `B_(i n) = mu_0ni` (i.e. same everywhere) `B_(out) = 0`

`text(Case)` 2: Semi-infinitely long solenoid (at one end)

`B = (mu_0ni)/2[cos 0 + cos (pi/2)] = (mu_0ni)/2`