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.

#### Solution

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

\[e = \frac{d\phi}{dt} = \frac{6\pi \times {10}^{- 4}}{dt}\]

Thus, the current flowing in the coil and the total charge are:-

\[i = \frac{e}{R} = \frac{6\pi \times {10}^{- 4}}{40dt} = \frac{4 . 71 \times {10}^{- 5}}{dt}\]

\[Q = \frac{4 . 71 \times {10}^{- 5} \times dt}{dt}\]

\[ = 4 . 71 \times {10}^{- 5} C\]