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(C) The formula for a tangent galvanometer is $I = K 	an 	heta$,where $K = \frac{2r B_H}{\mu_0 n}$.Given: $I = 10 \, mA = 10 	imes 10^{-3} \, A$,$\

Vedclass · Accepted Answer

$570$

Vedclass · Answer

(C) The formula for a tangent galvanometer is $I = K 	an 	heta$,where $K = \frac{2r B_H}{\mu_0 n}$.Given: $I = 10 \, mA = 10 	imes 10^{-3} \, A$,$	heta = 45^{\circ}$,$B_H = 3.6 	imes 10^{-5} \, T$,$r = 10 \, cm = 0.1 \, m$.Since $	an 45^{\circ} = 1$,we have $I = K$,so $n = \frac{2r B_H}{\mu_0 I}$.Substituting the values: $n = \frac{2 	imes 0.1 	imes 3.6 	imes 10^{-5}}{4\pi 	imes 10^{-7} 	imes 10 	imes 10^{-3}}$.$n = \frac{0.72 	imes 10^{-5}}{4 	imes 3.14159 	imes 10^{-10}} = \frac{0.72 	imes 10^5}{12.566} \approx 5729.7 / 10 = 573$ (approx $570$ based on standard approximation $\pi \approx 3.14$ and $4\pi \approx 12.56$).Using $\pi = 3.14$,$n = \frac{0.72 	imes 10^{-5}}{12.56 	imes 10^{-10}} = \frac{72000}{125.6} \approx 573$. Given the options,$570$ is the correct choice.

$A$ tangent galvanometer shows a deflection of $45^{\circ}$ when a $10 \, mA$ current passes through it. If the horizontal component of the Earth's magnetic field is $3.6 \times 10^{-5} \, T$ and the radius of the coil is $10 \, cm$,find the number of turns in the coil.

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