Derive an expression for the torque acting on a loop of N turns, area A, carrying current I, when held in a uniform magnetic field.
With the help of circuit, show how a moving coil galvanometer can be converted into an ammeter of given range. Write the necessary mathematical formula.  

Let, I be the current through the loop PQRS.
Given, a and b are the sides of the rectangular loop.
Area of the loop, A = ab 
Number of turns in the loop = N

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
                  ${\mathrm{I}}_{\mathrm{g}}{\mathrm{R}}_{\mathrm{g}} = (\mathrm{I}-{\mathrm{I}}_{\mathrm{g}}){\mathrm{R}}_{\mathrm{s}}$ 

                    ${\mathrm{R}}_{\mathrm{s}} = \frac{\mathrm{I}}{\mathrm{I}-{\mathrm{I}}_{\mathrm{g}}}\times {\mathrm{R}}_{\mathrm{g}}$
Hence by connecting a shunt of resistance R_s across the galvanometer, we get an ammeter of desired range.