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 BH = 3.0 × 10−5 T.

Answer 1

Charge is Product of Current and Time

Explanation:

Given:

Radius of the coil, r = 10 cm = 0.1 m

Resistance of the coil, R = $40 \Omega$

Number of turns in the coil, N = 1000

Angle of rotation, $\theta$ = 180

Horizontal component of Earth's magnetic field, $B_H = 3 \times 10^-^5 T$

Magnetic flux, $\phi$=  NBA cos 180°

$\phi$ = - NBA

      = $1000 \times 3\times 10^5\times\pi\times 1\times 1\times 10^-^2$

      = $3\pi \times 10^-^4$ Wb

$d\phi = 2NBA = 6\pi \times 10^-^4$ Wb

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

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

$i = \frac {e}{R} = \frac {6\pi \times 10^-^4}{40dt} = \frac {4.71 \times 10^-^5}{dt}$

$Q = \frac {4.71 \times 10^-^5 \times dt}{dt}$

   = $4.71 \times 10^-^5 C$