5.20(a). 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 \degree with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east.  Determine the horizontal component of the earth’s magnetic field at the location.

Answers (1)


Number of turns in the coil, n = 30

Radius of coil, r = 12cm = 0.12m

Current in the coil, I = 0.35A

The angle of dip, \delta = 45o

We know, Magnetic fields due to current carrying coils, B = \mu_{0}nI/2r

B = 4\pi \times10^{-7}\times30\times0.35/2\times0.12

= 5.49 \times 10^{-5}T

Now, Horizontal component of the earth’s magnetic field = Bsin\delta

= 5.49 \times 10^{-5}T× sin45^o

=3.88\times10^{-5}T (Hint: Take sin45as 0.7)

=0.388 G