NEET-XII-Physics
38: Electromagnetic Induction
- #23A 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.Ans : 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 \times {10}^{-4}}{dt}`` V
Thus, the current flowing in the coil and the total charge are:
`` 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\,}``
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