A magnetic needle is arranged at the centre of a current carrying coil having 50 turns with radius of coil `20 cm` arranged along magnetic meridian. When a current of `0.5mA` is allowed to pass through the coil the deflection is observed to be `30^(@)`. Find the horizontal component of earth's magnetic field

Explanation

`B = B_H tan theta (mu_0 "ni" )/(2r tan theta) = B_H`
`B_H = (4pi xx 10^(-7) xx 50 xx 5 xx 10^(-4) xx sqrt3)/(2(10^(-1) )(1) )`
` = 5 sqrt3 pi xx 10^(-8) T = 26.35 xx 10^(-8) T = 2.635 xx 10^(-7) T`