The coil of a moving coil galvanometer has 5 turns and each turn has an effective area of `2 xx 10^(-2) m^(2)` . It is suspended in a magnetic field whose strength is `4 xx 10^(-2)` Wb `m^(-2)` . If the torsional constant K of the suspension fibre is ` 4xx 10^(9)` N m `deg^(-1)` .
(a) Find its current sensitivity in degree per micro - ampere
(b) Calculate the voltage sensitivity of the galvanometer for it to havefull scale deflection of 50 divisions for 25 mV.
(c ) Compute the resistance of the galvanometer .

Explanation

The number of turns of coil is 5 turns
The area of each coil is ` 2xx 10^(-2) m^(2)`
Strength of the magnetic field is 4 `xx 10^(-2) ` Wh `m^(-2)`
Torsional constant is 4 `xx 10^(-9) ` N m `deg^(-10`
`I_(x) = ("N A B")/( K) = (5 xx 2 xx 10^(-2) xx 4 xx 10^(-2))/( 4 xx 10^(-9)) = 10^(@)` divisions per ampere
`I mu`A = 1 microampere = `10^(-6)` ampere
Therefore,
` I_(x) = 10^(6) ("div")/(A) = 1 ("div")/(10^(-6) A) = 1 ("div")/(mu A)`
`I_(x) ` = 1div `(mu A)^(-1)`
(b) Voltage sensitivity
`V_(x) = (theta)/(V) = (50"div")/(25 m V) = 2xx 10^(3) "div" V^(-1)`
(c) The resistance of the galvanometer is
`R_(g) = (I_(s))/(V_(s)) = (10^(6) ("div")/(A))/( 2 xx 10^(3) ("div")/(A)) = 0.5 xx 10^(3) (V)/(A) = 0.5 k Omega`