In a moving coil galvanometer a coil of area π cm2 and 10 windings is used. Magnetic field strength applied on the coil is 1 tesla and torsional stiffness of the torsional spring is 6×10−6N.m/rad. For marking the 90o space is equally divided into 10 parts as shown. If the least count of this galvanometer in mA is k, then find 10k.

(NBA)i=Cθ i=CθnBA=(6×10−6)10×1×π×10−4(π2)So, current corresponding 1 part=3010=3mA

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In a moving coil galvanometer a coil of area $π c m^{2}$ and 10 windings is used. Magnetic field strength applied on the coil is 1 tesla and torsional stiffness of the torsional spring is $6 \times 10^{- 6} N . m / r a d$ . For marking the $90^{o}$ space is equally divided into 10 parts as shown. If the least count of this galvanometer in mA is k, then find 10k.

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Detailed Solution

$(N B A) i = C θ i = \frac{C θ}{n B A} = \frac{(6 \times 10^{- 6})}{10 \times 1 \times π \times 10^{- 4}} (\frac{π}{2})$

So, current corresponding 1 part

$= \frac{30}{10} = 3 m A$

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