A compass needle free to turn in a horizontal plane is placed at the centre of a 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 `0.35amp.`, the needle points west to east.
(a) Determine the horizontal component of earth's magnetic field at the location.
(b) The current in the coil is reversed and the coil is rotated about its vertical axis by an angle of `90^@` in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.

Explanation

Number of turns in the circular coil, N=30
Radius of the circular coil, r=12cm=0.12m
Current in the coil, I=0.35A
Angle of dip, `delta=45^(@)`
(a) The magneetic field due to current I, at a distance r, is given as:
`B=(mu_(@) 2piNI)/(4pir)`
where
`mu_(0)=` permeability of free space=`4nxx10^(-7)TmA^(-1)`
`thereforeB=(4pixx10^(-7)xx2pixx30xx0.35)/(4pixx0.12)`
`=5.49xx10^(-5)T`
The compass needle points from west to East. Hence, the horizontal compoennt of earth's magnetic field is given as:
`B_(H)=B sin delta`
`=5.49xx10^(-5)sin 45^(@)=3.88xx10^(-5)T=0.388G`
(b) When the current in the coil is reversed and the coil is rotated about its vertical axis by an angle of `90^(@)`. the needle will reverse its original direction. in this case, the needle will point from E ast to west.