Physics - Magnetism Question with Solution | TestHub

Given, number of turns in the coil n = 30

Current in the coil I = 0.35 A

Radius of circular coil = 12 cm = 0.12 m



Let N - S be the line of magnetic of 45^o with the magnetic meridian. C₁ and C₂ be the plane of coil. The needle points west to east. The magnetic field produced due to the coil is B. The direction of B is along the axis of coil i.e., it makes an angle of 45^o with east. The needle points west to east only if the direction of magnetic field B is at 45^o from east.

Use the formula, magnetic field produced by a current carrying coil

$\text{B}=\frac{{\mu }_0}{4\pi }·\frac{2\pi \text{I}}{\text{r}}=\frac{10\times 2\times 3\text{.}14\times 0\text{.}35\times 30}{0\text{.}12}$

= 1.83 x 10^-6x 30 = 5.49 x 10^-5T

The horizontal component of magnetic field

H = B sin 45^o= 5.49 x 10^-5$\times \frac{1}{\sqrt{2}}$

= 3.9 x 10^-5T.