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A coil of a moving Coil galvanometer twi...

A coil of a moving Coil galvanometer twists through `90^@` when a current of one microampere is passed through it. If the area of the coil is `10^-4 m^2` and it has `100` turns, calculate the magnetic field of the magnet of the galvanometer. Given, k = `10^-8`N -m /degree.

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The correct Answer is:
90
`i=(k/(NBA))phi`
`:. B=(kphi)/(NiA)=((10^-8)(90))/(100xx10^-6xx10^-4)`
`=90T`
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