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.
Given:
Radius of coil r=10cm
Resistance of the coil R=40Ω
No. of turns in the coil N=1000
Horizontal component of earth’s magnetic field=
Angle of rotation θ =180°
Magnetic flux through the circular coil ϕ can be given by formula
…. (i)
Where B=magnetic field intensity
A=area of cross section
N=no. of turns in the coil
θ =angle between area vector and magnetic field
Initially, angle between area vector and magnetic field is 0°
Therefore, initial flux through the coil is
When it is rotated by 180° flux passing through the coil is given by
Now,
by faraday’s law of electromagnetic induction
Where
ϵ =emf produced
ϕ =flux of magnetic field
therefore, emf produced in the coil is given by
………(ii)
Current passing through the loop (i) of resistance R is
using eq.(ii)
Charge flowing through the galvanometer (Q) in time dot is given by formula
Putting the values of N, B, A and R we get,
Therefore charge which flows through the galvanometer is 