    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.

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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, BH = 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 $e=\frac{d\mathrm{\varphi }}{dt}=\frac{6\pi ×{10}^{-4}}{dt}$ V Thus, the current flowing in the coil and the total charge are: $i=\frac{e}{R}=\frac{6\mathrm{\pi }×{10}^{-4}}{40dt}=\frac{4.71×{10}^{-5}}{dt}\phantom{\rule{0ex}{0ex}}Q=\frac{4.71×{10}^{-5}×dt}{dt}\phantom{\rule{0ex}{0ex}}=4.71×{10}^{-5}\mathrm{C}$  Suggest Corrections  0      Related Videos   Torque on a Magnetic Dipole
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