The deflection produced in a tangent galvanometer, whose coil has a resistance of \(9 \Omega\) is…
The deflection produced in a tangent galvanometer, whose coil has a resistance of \(9 \Omega\) is \(30^{\circ}\). The potential difference across the coil is \(4.5 \mathrm{~V}\). If the number of turns in the coil is 10 , the radius of the coil is (Given, \(B_{\mathrm{H}}=3.14 \times 10^{-5} \mathrm{~T}\) )
\(2 \sqrt{3} \times 10^{-2} \mathrm{~m}\)
\(10 \sqrt{3} \times 10^{-2} \mathrm{~m}\)
\(6 \times 10^{-2} \mathrm{~m}\)
\(3.5 \times 10^{-2} \mathrm{~m}\)
Solution
For tangent galvanometer,
Deflection, \(\theta=30^{\circ}\)
Resistance of coil, \(R=9 \Omega\)
Potential difference across coil,
\(V=4.5 \mathrm{~V}\)
Number of turns in the coil
\(\begin{aligned}
N & =10 \\
B_H & =3.14 \times 10^{-5} \mathrm{~T}
\end{aligned}\)
Current flowing through the coil,
\(I=\frac{V}{R}=\frac{4.5}{9}=0.5 \mathrm{~A}\)
We know that, current flowing through the galvanometer is given as
\(\begin{aligned}
I & =\frac{2 r B_H \tan \theta}{\mu_0 N} \\
\Rightarrow \quad r & =\frac{\mu_0 N I}{2 B_H \tan \theta} \\
& =\frac{4 \pi \times 10^{-7} \times 10 \times 0.5}{2 \times 3.14 \times 10^{-5} \times \tan 30^{\circ}} \\
& =10 \sqrt{3} \times 10^{-2} \mathrm{~m}
\end{aligned}\)