A tangent galvanometer has a coil with 50 tur | vedclass

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(B) For a tangent galvanometer,the current $I$ is given by the formula: $I = \frac{2rB_H}{\mu_0 n} 	an 	heta$,where $r$ is the radius,$B_H$ is the h

Vedclass · Accepted Answer

$48.2$

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(B) For a tangent galvanometer,the current $I$ is given by the formula: $I = \frac{2rB_H}{\mu_0 n} 	an 	heta$,where $r$ is the radius,$B_H$ is the horizontal component of the earth's magnetic field,$n$ is the number of turns,and $	heta$ is the deflection.Rearranging for $	an 	heta$: $	an 	heta = \frac{I \mu_0 n}{2r B_H}$.Given: $I = 0.1$ $A$,$n = 50$,$r = 4$ $cm = 0.04$ $m$,$B_H = 7 	imes 10^{-5}$ $T$,and $\mu_0 = 4\pi 	imes 10^{-7}$ $T \cdot m/A$.Substituting the values: $	an 	heta = \frac{0.1 	imes 4\pi 	imes 10^{-7} 	imes 50}{2 	imes 0.04 	imes 7 	imes 10^{-5}}$.$	an 	heta = \frac{0.1 	imes 4 	imes 3.14159 	imes 10^{-7} 	imes 50}{0.08 	imes 7 	imes 10^{-5}} = \frac{6.283 	imes 10^{-6}}{0.56 	imes 10^{-5}} = \frac{0.6283}{0.56} \approx 1.122$.Therefore,$	heta = 	an^{-1}(1.122) \approx 48.2^o$.

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