Question

moving coil galvanometer (MCG) has 10 turns each of length
12 cm and breadth 8 cm. The coil of M.C. G. carries a current
of 125uA and is kept perpendicular to the uniform magnetic
field of induction 10 T. The twist constant of phosphor bronze
fibre is 12x10 Nm/degree. Calculate the deflection
produced​

Answer 1

deflection produced​ is  10°

Explanation:

given data

turn = 10

length  = 12 cm

breadth = 8 cm

current  = 125 uA

uniform magnetic  field of induction = $10^{-2}$  T

twist constant = 12 x $10^{-9}$ Nm/degree

solution

Current flowing through coil of MCG

current I = $(\frac{c}{nBA} )\theta$    .................................1

so here $\theta$  will be

$\theta$ = $(\frac{nBA}{c} )l$      

put here value we get

$\theta$ = $\frac{10\times 10^{-2}\times 12 \times 10^{-2}\times 8 \times 10^{-2}\times 125 \times 10^{-6}}{12\times 10^{-9}}$    

$\theta$ = 10°

so deflection produced​ is  10°

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current produce a deflection

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