A) decrease by \[{{30}^{o}}\]
B) decrease by \[{{15}^{o}}\]
C) increase by \[{{15}^{o}}\]
D) increase by \[{{30}^{o}}\]
Correct Answer: B
Solution :
When current is passed through tangent galvanometer, a magnetic field is created given by \[B=\frac{{{\mu }_{0}}n\,I}{2\,r}\] where I is current, n is number of turns and r is the radius. When it is set in plane, then from tangent law, \[B={{B}_{H}}\tan \theta \] \[\Rightarrow \] \[\frac{{{\mu }_{0}}\,n\,I}{2\,r}={{B}_{H}}\,\tan \theta \] \[\Rightarrow \] \[I=\left( \frac{2\,r\,{{B}_{H}}}{{{\mu }_{0}}\,n} \right)\tan \theta \] \[\Rightarrow \] \[I=K\tan \theta \]. where K is reduction factor. Given, \[{{I}_{2}}=\frac{{{I}_{1}}}{\sqrt{3}},{{\theta }_{1}}={{45}^{o}}\] \[\therefore \] \[\frac{{{I}_{1}}}{{{I}_{1}}/\sqrt{3}}=\frac{\tan {{45}^{o}}}{\tan {{\theta }_{2}}}\] \[\Rightarrow \] \[\tan {{\theta }_{2}}=\frac{1}{\sqrt{3}}\] \[\Rightarrow \] \[{{\theta }_{2}}={{\tan }^{-1}}\left( \frac{1}{\sqrt{3}} \right)={{30}^{o}}\] Hence, deflection decreases by\[{{45}^{o}}-{{30}^{o}}={{15}^{o}}\].
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