A coil of wire with 50 turns lies in the plane of the page and has an initial area of 0.210 m2. The coil is now stretched to have no area in 0.100 s. What is the magnitude and direction of the average value of the induced emf if the uniform magnetic field points into the page and has a strength of 1.50 T?

Explanation:

Given that,

A coil has 50turns

N=50

Initial area of plane is A1=0.21m²

The coil was stretch for t=0.1seconds to have a no area(A2)

Now the loop is said to be stretched to have no area. This means that, A2=0

Magnetic field B= 1.5T

EMF?

EMF is given as

ϵ=−d(ϕB)/dt

We expand the magnetic flux here as: since ϕB=NBA

ϵ=−d/dt[NBA]

The change in area is re-expressed as:

ϵ=−NB(A2−A1) / Δt

Where N = 50, ∆t = 0.1 second

A2 = 0 A1 = 0.21m², B =1.5T

Then,

ϵ= —50 × 1.5 ( 0—0.21) / 0.1

ϵ= —75 × — 0.21 /0.1

Note —×— = +

ϵ = 15.75/0.1

ϵ = 157.5 Volts

b. Direction

Using the right hand rule, The direction of the emf is clockwise and point our thumb into the page.

Given Information:

Number of Turns = N = 50

Area of plane = A₁ = 0.210 m²

Area of plane = A₂ = 0 m²

time = Δt = 0.100 sec

Magnetic field = B = 1.50 T

Required Information:

Induced emf = ξ = ?

Induced emf = 157.5

Explanation:

From the Faraday's law the induced emf in the coil is given by  

ξ = -dΦ/dt  

Where Φ is the magnetic flux  and is given by

Φ = NAB

Where B is the magnetic field A is the area and N is the number of turns

ξ = -d(NAB)/dt

ξ = -NBd(A)/dt

Taking derivative yields

ξ = -NB(A₂ - A₁)/Δt

Where A₂ is area when coil was stretched to have no area that is A₂ = 0 and A₁ is the initial area of the coil.

ξ = -50*1.50*(0 - 0.210)/0.100

ξ = 157.5 V

Therefore, the magnitude to the induced emf is 157.5 V and the direction can be found using the right hand rule. The induced emf direction will be clockwise.