• # question_answer (i) What do you mean by current sensitivity of a moving coil galvanometer. On what factor does it depend? (ii) State two reasons why a galvanometer cannot be used as such to measure high order current in a given circuit? (iii) The current sensitivity of a moving coil galvanometer is 5 div/mA and voltage sensitivity is 20 div/volt. Find the resistance of the galvanometer. Or (i) Write any three characteristics, a ferromagnetic substance should possess, if it is to be used to make a permanent magnet. Give one example of such a material. (ii) The susceptibility of a magnetic material is - 0.085. Identify the magnetic type of the material. (iii) A Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core of relative permeability 800. What is the magnet field (B) in the core for a magnetising current of 1.2 A?

(i) Current sensitivity of a moving coil galvanometer Current sensitivity of a moving coil galvanometer is defined as the deflection of the coil per unit current flowing through it. Current sensitivity,          $({{I}_{s}})=\frac{\theta }{I}=\frac{NAB}{C}$
(iii) Given,          ${{I}_{s}}=5\,div/mA=5\times {{10}^{3}}div/A,$ ${{V}_{s}}=20div/V$ We know that resistance,  $R=\frac{{{I}_{s}}}{{{V}_{s}}}=\frac{5\times {{10}^{3}}}{20}$ [by Ohm's law] $\Rightarrow$               $R=250\,\Omega$ Or  (i) A ferromagnetic substance should possess the following three characteristics to be used to make a permanent magnet. (a) High permeability (b) High retentivity (c) High coercivity For example, Steel and cobalt.
(ii) As susceptibility of the substance is negative. Therefore, it is a diamagnetic substance. (iii) A rowland ring is a circular ring of a magnetic material over which is wound a torroidal solenoid. The magnitude of the magnetic field in the core is given by $\because$       $B=\frac{\mu \,nl}{2\pi r}$ and       n = 3500 is the number of turns Now relative permeability,            ${{\mu }_{r}}=\frac{\mu }{{{\mu }_{0}}}$ $\Rightarrow$   $\mu ={{\mu }_{0}}{{\mu }_{r}}=4\pi \times {{10}^{-7}}\times 800$ $\therefore$ $B=4\pi \times {{10}^{-7}}\times 800\times \frac{3500\times 1.2}{2\times 3.14\times 15\times {{10}^{-2}}}$                                     [given r = 15 cm] $\Rightarrow$   B = 4.48 T