Question

A coil of radius 10 cm and resistance 40 Ω has 1000 turns. It is placed with its plane vertical and its axis parallel to the magnetic meridian. The coil is connected to a galvanometer and is rotated about the vertical diameter through an angle of 180°. Find the charge which flows through the galvanometer if the horizontal component of the earth's magnetic field is B_H = 3.0 × 10⁻⁵ T.

Answer 1

Given:
Radius of the coil, r = 10 cm = 0.1 m
Resistance of the coil, R = 40 Ω
Number of turns in the coil, N = 1000
Angle of rotation, θ = 180°
Horizontal component of Earth's magnetic field, B_H = 3 × 10⁻⁵ T
Magnetic flux, ϕ = NBA cos 180°
⇒ ϕ = −NBA
= −1000 × 3 × 10⁻⁵ × π × 1 × 1 × 10⁻²
= 3π × 10⁻⁴ Wb
dϕ = 2NBA = 6π × 10⁻⁴ Wb
$e=\frac{d\mathrm{\varphi }}{dt}=\frac{6\pi \times {10}^{-4}}{dt}$ V
Thus, the current flowing in the coil and the total charge are:
$$\begin{gathered} i=\frac{e}{R}=\frac{6\mathrm{\pi }\times {10}^{-4}}{40dt}=\frac{4.71\times {10}^{-5}}{dt} \\ Q=\frac{4.71\times {10}^{-5}\times dt}{dt} \\ =4.71\times {10}^{-5}\mathrm{C} \end{gathered}$$