Question

Question 5.20: 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 45º 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 about its vertical axis by an angle of 90º in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.

Class 12 - Physics - Magnetism And Matter Magnetism And Matter Page-203

Answer 1

Here,n = 30, r = 12cm = 12 × 10^-2 m
i = 0.35A , horizontal component of earth's magnetic field, H = ?
As it is clear from figure shown the needke can point west to east only when
H = Bsin45°
where B = magnetic field strength due to current in coil = $\frac{\mu_0}{4\pi}\frac{2\pi ni}{r}$

$\therefore\:H=\frac{\mu_0}{4\pi}\frac{2\pi ni}{r}sin45^{\circ}\\\\=10^{-7}\times\frac{2\pi\times30\times0.35}{12\times10^{-2}}.\frac{1}{\sqrt{2}}\\\\=2\times\frac{22}{7}\times\frac{30\times35}{12\times\sqrt{2}}\times10^{-7}=3.9\times10^{-5}T$

(b) when current in coil is reversed and oil is turned through 90° anticlockwise, the direction of needle will reverse (e.g., it will point from east to west ).this follows from figure shown.