Question :
A closed coil consists of 500 turns on a rectangular frame of an area of 4.0cm2 and has a resistance of 500Ω . The coil is kept with its plane perpendicular to a uniform magnetic field of 0.2wb/m2 , the amount of charge flowing through the coil if it is turned over. (Rotate through 1800 )
Solution :
To understand the Question based on the Closed Coil
Given parameters are:
Number of turns n=500
area A = 4cm2 = 4×10−4m2
resistance R = 500Ω
magnetic field B = 0.2wb/m2
We have to calculate the amount of charge flowing through the coil.
Magnetic flux is given by the following formula.
ϕ = nBAcosθ
Where n is a number of turns, B is a magnetic field, A is the area, and θ is the angle between the area vector and the magnetic field.
So, let us calculate the change in magnetic flux when it is turned over (rotated by 180 -degrees).
△ϕ=nBAcos0−(nBAcos180)
Let us simplify the expression.
△ϕ = nBA + nBA = 2nBA = 2 × 500 × 0.2 × 4 × 10−4
△ϕ = 8×10−2
The charge flowing through the area is given by the following formula.
△Q = △ϕ/resistance = 8×10−2/500 =1.6×10−4C =0.16×10−3C
