Question

A circular coil of 10 turns and diameter 12cm carries a current I. The coil is placed with its plane in the magnetic meridian of the earth. A small, magnetic needle placed at the center of the coil makes 30 oscillations per minute about a vertical axis. When the current is cut off, it makes 15 revolutions per minute. If the horizontal component of the earth’s magnetic flux density is 2.0 x 10-5 T, calculate the magnitude of the current I. (Assume that the square of frequency of oscillations is proportional to the magnetic flux density)

Answer 1

Let $I$ be the current in coil

Then flux density at the centre of coil due to current will be

$$\frac{\mu_oNI}{2R}=\frac{4\pi\times10^{-7}\times 10I}{2\times 6\times 10^{-2}}$$

Now applying two condition when current is on and current is off

And

Square of oscillatiDon frequency directly proportional to flux density

Then

$$\frac{(30)^2}{(15)^2}=\frac{resultant\ field}{earth\ field}=\frac{\sqrt{(1.0472\times 10^{-4}I)^2+(2\times 10^{-5})^2}}{2\times 10^{-5}}$$

$$60\times 10^{-10}=1.0966\times 10^{-8}I^2$$

$$Or$$

$$0.5471=I^2$$

Or

$$I=0.74\ A$$