Here charge q is moving with velocity v.

Magnetic field at point P is,

$$\overrightarrow B = \left( {{{{\mu _0}} \over {4\pi }}} \right)q\left( {{{\overrightarrow v \times \overrightarrow r } \over {{r^3}}}} \right)$$

Here $$\mu_0$$ = Permeability of free space.

and $${{{\mu _0}} \over {4\pi }} = {10^{ - 7}}$$

$$\overrightarrow r $$ = position vector of point p with respect to charge

[Note :-

1. Direction of $$\overrightarrow r $$ is always from object towards point.

2. q will be present in the equation with sign.

3. Direction of magnetic field ($$\overrightarrow B $$) is towards the direction of $$\overrightarrow v $$ $$\times$$ $$\overrightarrow r $$ which is inside the plane when charge is positive. If charge is negative then direction of $$\overrightarrow B $$ is opposite of $$\overrightarrow v $$ $$\times$$ $$\overrightarrow r $$.]

Magnitude of $$\overrightarrow B $$ is,

$$B = \left( {{{{\mu _0}} \over {4\pi }}} \right)\left( {{{qvr\sin \theta } \over {{r^3}}}} \right)$$

$$ = \left( {{{{\mu _0}} \over {4\pi }}} \right)\left( {{{qv\sin \theta } \over {{r^2}}}} \right)$$

here $$\theta$$ = Angle between $$\overrightarrow v $$ and $$\overrightarrow r $$.

Magnetic field at point P due to current carrying element $$\overrightarrow {dl} $$,

$$\overrightarrow B = \left( {{{{\mu _0}} \over {4\pi }}} \right)i\left( {{{\overrightarrow {dl} \times \overrightarrow r } \over {{r^3}}}} \right)$$

It's magnitude is,

$$B = \left( {{{{\mu _0}} \over {4\pi }}} \right)\left( {{{idl\sin \theta } \over {{r^2}}}} \right)$$

Note :

(1) Direction of $$\overrightarrow {dl} $$ is same as direction of current.

(2) $$\theta$$ = Angle between $$\overrightarrow {dl} $$ and $$\overrightarrow {r} $$

$$B = {{{\mu _0}i} \over {4\pi r}}\left( {\sin \alpha + \sin \beta } \right)$$

Direction of B :

Note :-

On the axis of the wire magnetic field = 0.

Case 1 :

Infinite long wire (Very Long wire) :

$$\alpha ,\beta \to 90^\circ $$

$$\therefore$$ $$B = {{{\mu _0}i} \over {4\pi r}}(\sin 90^\circ + \sin 90^\circ )$$

$$ = {{{\mu _0}i} \over {4\pi r}}(1 + 1)$$

$$ = {{{\mu _0}i} \over {2\pi r}}$$

Graph :

Case 2 :

One end is at infinite and other end is in front.

Here $$\alpha=90^\circ$$ and $$\beta=0^\circ$$

$$B = {{{\mu _0}i} \over {4\pi r}}(\sin 90^\circ + \sin 0^\circ )$$

$$ = {{{\mu _0}i} \over {4\pi r}}$$

Case 3 :

When drawn perpendicular from P don't lie on the wire

Here, magnetic field at point P is,

$$B = {{{\mu _0}i} \over {4\pi r}}(\sin\alpha-\sin\beta)$$

Circular coil means N number of turns.

Circular loop means 1 turn.

Magnetic field at centre O,

$$B = {{{\mu _0}Ni} \over {2r}}$$

Here,

N = Number of turns (It can be integer or fraction)

i = Amount of current flowing through coil's one turn

R = Radius of circular coil

Direction of B found by right hand thumb rule. Curl your right hand's finger in the direction of current then thumb gives direction of magnetic field on axis as well as entire plane.

Circular loop means 1 turn. So, N = 1.

Magnetic field at centre O,

$$B = {{{\mu _0}i} \over {2r}}$$

Here,

i = Amount of current flowing through loop

R = Radius of circular loop

Direction of B found by right hand thumb rule. Curl your right hand's finger in the direction of current then thumb gives direction of magnetic field on axis as well as entire plane.

Half loop means $${1 \over 2}$$ turn. So, N = $${1 \over 2}$$.

Magnetic field at centre O,

$$\therefore$$ B at centre $$ = {{{\mu _0} \times {1 \over 2} \times i} \over {2R}} = {{{\mu _0}i} \over {4R}}$$

Here,

Here, $$N=\frac{1}{2}$$

i = Amount of current flowing through loop

R = Radius of half loop

Direction of B found by right hand thumb rule. Curl your right hand's finger in the direction of current then thumb gives direction of magnetic field on axis as well as entire plane.

Quarter loop means $${1 \over 4}$$ turn. So, N = $${1 \over 4 }$$.

Magnetic field at O,

$$\therefore$$ B at centre $$ = {{{\mu _0} \times {1 \over 4} \times i} \over {2R}} = {{{\mu _0}i} \over {8R}}$$

Here,

Here, $$N=\frac{1}{4}$$

i = Amount of current flowing through loop

R = Radius of quarter loop

Direction of B found by right hand thumb rule. Curl your right hand's finger in the direction of current then thumb gives direction of magnetic field on axis as well as entire plane.

We know, magnetic field due to full circular loop is, $$B = {{{\mu _0}i} \over {2R}}$$

$$\therefore$$ Due to 2$$\pi$$ radiun angle of circular loop $$B = {{{\mu _0}i} \over {2R}}$$

$$\Rightarrow$$ Due to 1 radiun angle of circular loop $$B = \left( {{{{\mu _0}i} \over {2R}}} \right) \times {1 \over {2\pi }}$$

$$\Rightarrow$$ Due to $$\theta$$ radiun angle of circular loop $$B = \left( {{{{\mu _0}i} \over {2R}}} \right)\left( {{\theta \over {2\pi }}} \right)$$

$$\therefore$$ Here B at O $$ = \left( {{{{\mu _0}i} \over {2R}}} \right)\left( {{\theta \over {2\pi }}} \right)$$

And $$\theta$$ is in radius.

Here coil is with N turns.

Magnetic field on the axis is

$$B = {{{\mu _0}Ni{R^2}} \over {2{{({R^2} + {x^2})}^{3/2}}}}$$

Graph :-

It look like paper tape.

Magnetic field at centre O :-

Take a element of dx width at a distance x.

Here width of coil = b $$-$$ a

Total turns = N

$$\therefore$$ Turns per unit width = $$\frac{N}{b-a}$$

$$\therefore$$ Turns in dx width (dN) = $$\left(\frac{N}{b-a}\right)$$dx

At centre O, magnetic field due to dx width coil is

$$dB = {{{\mu _0}dNi} \over {2x}}$$

$$ = {{{\mu _0}Ni} \over {2(b - a)}}dx$$

$$\therefore$$ Total magnetic field due to entire spiral coil of width (b $$-$$ a) is

$$B = \int\limits_a^b {dB} $$

$$ = \int\limits_a^b {{{{\mu _0}Ni} \over {2(b - a)x}}dx} $$

$$ = {{{\mu _0}Ni} \over {2(b - a)}}\int\limits_a^b {{{dx} \over x}} $$

$$ = {{{\mu _0}Ni} \over {2(b - a)}}\left[ {\ln x} \right]_a^b$$

$$ = {{{\mu _0}Ni} \over {2(b - a)}}\ln \left( {{b \over a}} \right)$$