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.

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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, 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:

$i=\frac{e}{R}=\frac{6\mathrm{\pi}\times {10}^{-4}}{40dt}=\frac{4.71\times {10}^{-5}}{dt}\phantom{\rule{0ex}{0ex}}Q=\frac{4.71\times {10}^{-5}\times dt}{dt}\phantom{\rule{0ex}{0ex}}=4.71\times {10}^{-5}\mathrm{C}$

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

Magnetic flux, ϕ = NBA cos 180°

⇒ ϕ = −NBA

= −1000 × 3 × 10

= 3π × 10

dϕ = 2NBA = 6π × 10

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

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

$i=\frac{e}{R}=\frac{6\mathrm{\pi}\times {10}^{-4}}{40dt}=\frac{4.71\times {10}^{-5}}{dt}\phantom{\rule{0ex}{0ex}}Q=\frac{4.71\times {10}^{-5}\times dt}{dt}\phantom{\rule{0ex}{0ex}}=4.71\times {10}^{-5}\mathrm{C}$

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