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A current of 0.5 A produces a deflection of 60° in a tangent galvanometer. The current that produces a deflection of 30° in the same galvanometer is

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a

$\frac{0 .5}{2} A$

b

$\frac{0 .5}{3} A$

c

$\frac{0.5}{\sqrt{2}} A$

d

$\frac{0 .5}{\sqrt{3}} A$

answer is B.

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Detailed Solution

$\begin{array}{l} B = B_{H} \tan θ = \frac{μ_{0} i}{2 r} \Rightarrow i \propto \tan θ \\ \frac{i_{1}}{i_{2}} = \frac{\tan θ_{1}}{\tan θ_{2}} \Rightarrow \frac{0 .5}{i_{2}} = \frac{\tan 60^{\circ}}{\tan 30^{\circ}} \\ \Rightarrow i_{2} = \frac{0 .5 \times \frac{1}{\sqrt{3}}}{\sqrt{3}} \Rightarrow i_{2} = \frac{0 .5}{3} A \end{array}$

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