Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
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# 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. - Physics

Sum

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

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

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

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

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

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

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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 16 Electromagnetic Induction
Q 23 | Page 307
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