Given,
Radius, r =  0.51 Å = 0.51 x 10^–10 m 
velocity, v = 2 x 10⁵ m/s

(i) Equivalent current due to orital motion of electron is given by, 

             

             $\mathrm{I} = \frac{1.6 \times {10}^{-19} \times 2 \times {10}^5}{2 \times 3.14 \times 0.51 \times {10}^{-10}}$  

             $$\begin{gathered} \mathrm{I} = \frac{3.2 \times {10}^{-4}}{3.2028} \\ = 0.99 \times {10}^{-4} \\ = {10}^{-4} \mathrm{A} \end{gathered}$$ 

(ii) Magnetic field produced at the centre of the electron is,
                     $\boxed{\mathrm{B} = \mathrm{\mu nI}}$ 

                      $\mathrm{B} = 4\mathrm{\pi }\times {10}^{-7}\times 1\times {10}^{-4}$ 

                      $$\begin{gathered} \mathrm{B} = 4\times 3.14\times {10}^{-11} \\ =12.56 \times {10}^{-11} \\ = 1.256 \times {10}^{-10}\mathrm{T}. \end{gathered}$$ 

(iii) Magnetic moment assosciated with the electron

           $$\begin{gathered} \boxed{\mathrm{M} = \mathrm{IA}} \\ = \mathrm{I} \left({\mathrm{\pi r}}^2\right) \\ = {10}^{-4}\times 3.14\times (0.51 \times {10}^{-10}{)}^2 \\ = 3.14 \times 0.2601 \times {10}^{-4} \times {10}^{-20} \end{gathered}$$ 

           $$\begin{gathered} \mathrm{M} = 0.816 \times {10}^{-24} \\ = 8.16 \times {10}^{-25} {\mathrm{Am}}^2 \end{gathered}$$

Biot-savart law states that, the magnetic field induction dB at a point due to current element is 

                $dB = \frac{{\mu }_o}{4\pi } \frac{I dl sin\theta }{r^2}$ 
where,
I.dl is the current element,
r is the distance of point from the current elemnt and,
$\mathrm{\theta } \mathrm{is} \mathrm{the} \mathrm{angle} \mathrm{between} \stackrel{\rightharpoonup }{\mathrm{dl} }\mathrm{and} \stackrel{\rightharpoonup }{\mathrm{r}}.$

Expression for magnetic field at the centre of a circular coil :

Consider a circular coil of radius 'r' with center O. Let, I be the current flowing in the circular coil. Assume, that the current coil is made of a large number of current elements each of length 'dl' .
current element = I.dl 
Now, as per Biot Savart law we have,  

               $dB = \frac{{\mu }_o}{4\pi } \frac{I dl sin\theta }{r^2}$
Since, the angle between $\stackrel{\rightharpoonup }{\mathrm{dl}} and \stackrel{\rightharpoonup }{r}$ is 90^0.
Therefore,
                     $dB = \frac{{\mu }_o}{4\pi } \frac{I dl}{r^2}$ 
Thus, 

B= $\int dB = \int \frac{{\mu }_o}{4\pi } \frac{I dl }{r^2} = \frac{{\mu }_o}{4\pi } \frac{I}{r^2}\int dl$
But, 
$$\begin{gathered} \int dl = total length of circular coil \\ = circumference of current loop = 2\pi r \end{gathered}$$ 

Therefore,
                  $\mathrm{B} = \frac{{\mathrm{\mu }}_{\mathrm{O}}}{4\mathrm{\pi }} \frac{2\mathrm{\pi nI}}{\mathrm{r} }$ 
where,
n is the number of turns in the coil.

The magnetic field lines.
The field lines of (a) a bar magent, (b) a current carrying finite solenoid and (c) electric dipole. At large distances, the field lines are very similar. The curves labelled (i) and (ii) are closed Gaussian surfaces.
There is a basic difference between magnetic and electric field lines. In case of the electric field of an electric dipole, the electric lines of force originate from positive charge and end at the negative charge.

In case of a bar magnet, the magnetic field lines are closed loops, i.e., magnetic field lines do not start or end anywhere.

Let I = current through the loop PQRS
a, b = sides of the rectangular loop
A = ab = area of the loop
N = number of turns in the loop

According to Fleming's left-hand rule, the side PQ experiences a normal inward force, F₁ = laB and side SR experiences a normal outward force, F₂ = IaB. These two equal and opposite forces form a couple which exerts a torque given by
τ = Force x Perpendicular distance
= IaB x b sin θ
= IB (ab) sin θ
= IB A sin θ    [ ∵ A = ab]
As the coil has N turns, so
τ = NBA sin θ
Conversion of galvanometer into ammeter: A galvanometer can be converted into an ammeter by connecting a low resistance S in parallel with it. This low resistance is called shunt.

Let I_g be the current with which the galvanometer gives full scale deflection. As galvanometer and shunt are connected in parallel, so
P.D. across the galvanometer = P.D. across the shunt
or                     
or                        
Hence by connecting a shunt of resistance R_s across the galvanometer, we get an ammeter of desired range.