A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of `45^(@)` with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east.
(a) Determine the horizontal component of the earth's magnetic field at the location.
(b) The current in the coil is reversed, and the coil is rotated `90^(@)` about its vertical axis by an the needle. Take the magnetic declination at the places to be zero.

Explanation

(a) Here `n = 30, r = 12 cm = 12 xx 10^(-2) cm`
i = 0.35 amp, H = ?
As is clear from fig. the needle can point west to east only when `H = B sin 45^(@)`.
Where B = Magnetic field strength due to current in coil `= (mu_(0))/(4pi)(2pi n i)/(r)`
`H = (mu_(0))/(4pi).(2pi n i)/(r) sin 45^(@)`
`= 10^(-7) xx (2pi xx 30 xx 0.35)/(12 xx 10^(-2)) xx (1)/(sqrt(2))`
`= 2 xx (22)/(7) xx (30 xx35)/(12 xx sqrt(2)) xx 10^(-7)`
`H = 3.9 xx 10^(-5)` tesla
(b) When current in coil is reversed and coil is turned through `90^(@)` anticlock wise, the direction of needle will reverse (i.e., it will point from East to West). This follow from the figure.