Question

In a coil of resistance 10 Ω, the induced current developed by changing magnetic flux through it, is shown in figure as a function of time. The magnitude of change in flux through the coil in weber is

• A. 8
• B. 2
• C. 6
• D. 4

Hint:

We use Faraday's equation of the emf induced in a coil subjected to the changing magnetic flux and Ohm’s law to solve this question.

Answer 1

The Correct Option is B

Area under i-t graph = q
q = ½ x 4 x 0.1 = 0.2 C
As \(q=\frac{\Delta\phi}{R}\)
Δϕ = qR 
= (0.2 C)(10 Ω)
So, Δϕ = 2 weber
Hence, option B) 2 is the correct answer.

Answer 2

I = \(|{1 \over R} {d \phi \over dt} |\)

∣dϕ∣ = ∣IRdt∣

dϕ = Area of the triangle x R

= 2 Wb

In a coil of resistance 10 Ω, the magnitude of change in flux through the coil in weber is 2 Wb.

Answer 3

We use Faraday's equation of the emf induced in a coil subjected to the changing magnetic flux and Ohm’s law to solve this question.
e = -dφ/dt
Taking the EMF magnitude, we get
e = dφ/dt
We know that Ohm's law states that V = IR
As the voltage across the coil is equal to the emf induced - 
v = e
e = IR
So, 
 dφ/dt = IR
dφ = IRdt
Integrating the sides, we get
∫dφ = ∫IRdt
As the coil's radius is constant - 
Δφ = R ∫Idt
The integration of a dependent variable with respect to the independent variable is the area under the curve of the dependent variable. So the integrand value of the relation will be equal to the area of the triangle in the figure in question. 
A = ½ × b × h
According to the figure, b = 0.1 s and h = 4 A
So, ∫Idt = ½ x 0.1 x 4
= 0.2
Δφ = 0.2R
As the radius of the coil is given as 10Ω
We get, Δφ = 2 Wb
Thus, the magnitude of change in flux through the coil in Weber is 2.