A tangent galvanometer shown a deflection of `45^@` when `10 mA` of current is passed through it. If the horizontal component of the earth's magnetic field is `B_H = 3.6 xx 10^(-5) T` and radius of the coil is `10cm`, find the number of turns in the coil.

Explanation

Given B_(H) = 3.6 xx 10^(-6)T `
`theta = 45^(@) , I = 10 mA`
`= 10^(-3)A`
`n = ? , r = 10 cm = 0.1 m`
We know
`B_(H) tan theta = (mu_(0) in)/(2r )`
`n = B_(H) tan theta xx 2 t)/(mu_(0)i)`
`= (3.6 xx 10^(-6) xx 2 xx 1 xx 10^(-1))/(4 pi xx 10^(-7) xx 10^(-2))`
` = 0.57332 xx 10^(8) = 573 turns`