• # question_answer A galvanometer having 30 division scale and $100\,\Omega$ resistance is connected in series to the battery of emf 3 V through a resistance of $200\,\Omega ,$ shows full scale deflection. Find the figure of merit of the galvanometer in microampere. Or A circular segment of radius 10 cm, subtends an angle of $60{}^\circ$ at its centre. A current of 9 A is flowing through it. Find the magnitude and direction of the magnetic field produced at the centre.

Here, n = 30, $G=100\,\Omega ,$ E = 3 V, $R=200\,\Omega ,$ k = ? Total resistance   $=G+R=100+200=300\,\Omega$             ${{I}_{g}}=\frac{E}{G+R}=\frac{3}{300}=\frac{1}{100}A$             $k=\frac{{{I}_{g}}}{n}=\frac{1/100}{30}$ or         $k=(1/3)\times {{10}^{-3}}A/division$ or         $k=(1/3)\times {{10}^{-3}}\times {{10}^{6}}\mu A/division$ or         $k=333.3\mu A/division$ Or Given,   r = 10 cm = 0.1 m             $\theta =60{}^\circ =\pi /3rad$             I = 9 A, B = ? Here, CD is the circular segment with centre O, in the plane of paper. Let l = length of segment CD. It subtends an angle $60{}^\circ =(\pi /3)$at 0, i.e. $\alpha =\pi /3\,rad.$ Now magnetic field induction at O due to current through segment CD will be $B=\frac{{{\mu }_{0}}l}{4\pi r}\alpha =\frac{{{\mu }_{0}}l}{4\pi r}.\frac{\pi }{3}$ or  $={{10}^{-7}}\times \frac{9}{0.1}\times \frac{3.14}{3}=9.42\times {{10}^{-6}}T$ The direction of magnetic field induction, according to right hand rule will be downwards perpendicular to the plane of paper.