ToughSTEM
Sign Up
Log In
ToughSTEM
A question answer community on a mission
to share Solutions for all STEM major Problems.
Cant find a problem on ToughSTEM?
0
A piece of copper wire is formed into a two circular loop of radius 12 cm. A magnetic field is oriented parallel to the normal to the loop, and it increases from 0 to 0.60 T in a time of 0.45 s. The wire has a resistance per unit length of 0.15 ohms (Please show your work!)

a) What is the value of EMF induced in the copper coil?

b)What is the average electrical energy dissipated in the resistance of the wire during this period of time?
Edit
Added Tue, 04 Aug '15
Community
1
Comment
Add a Comment
Solutions
0
This is an application of Faraday's Law.

EMF = N d ( B A ) / dt, where B is the magnetic field which can be changing with time and A is the area of the loop that maybe changing with time too. N is the number of loop/coils

In this problem N= 1, A is constant and B is changing - from 0 to 0.60 T in 0.45 s. So the above simplifies to:

EMF = N d ( B A ) / dt = A * dB/dt

We will assume it is a linear increase from 0 T to 0.60 T do the have change :

(dB/dt)_avg = (Bf - Bi)/ ?t = (0.60 T - 0T) / 0.45s = 0.60/0.45 T/s = 4/3 T/s
A = pi * r^2 = pi * (12 cm)^2 = pi * (.12 m)^ = .0144 pi m^2

EMF = pi * 0.0144 m^2 * 4/3 T/s

If R is the resistance in the wire,

R = rho * L where rho is the resistance per unit length (ohm/m)
The length is the circumference of the circular loop,
L = pi * diameter = pi * 2 * radius
R = 2 * pi * radius * rho

If a voltage, EMF, is applied across a resistance of R, the power dissipated in the resistance is:

I = V/R
P = V^2 / R
P = EMF^2 / R = (A * dB/dt)^2 / R = ( pi * radius^2) (dB/dt)^2 / (2 * pi * radius * rho)
P = ( pi * radius^2)^2 (dB/dt)^2 / (2 * pi * radius * rho)
P = pi/2 * radius^3 * (dB/dt)^2 / rho
P = pi/2 * (0.12m)^3 (4/3 T/s)^2 / (3.3x10^-2 ohm/m)
P = pi/2 *(0.12)^3 * (4/3)^2 / (3.3) * 10^2 m^4 * T^2/s^2 / ohm

Let's simplify the units:
T = V s/ m^2
T^2 = V^2 s^2 /m^4
m^4/s^2 T = V^2

P = pi/2 *(0.12)^3 * (4/3)^2 / (3.3* 10^-2) V^2 / ohm
P = pi/2 *(0.12)^3 * (4/3)^2 / (3.3) * 10^2 W
P = 0.146 W
Rounding,
P = 0.15 W
Edit
Added Tue, 04 Aug '15
Community
1
Comment
Add a Comment
Add Your Solution!
Close

Click here to
Choose An Image
or
Get image from URL
GO
Close
Back
Add Image
Close
What URL would you like to link?
GO
α
β
γ
δ
ϵ
ε
η
ϑ
λ
μ
π
ρ
σ
τ
φ
ψ
ω
Γ
Δ
Θ
Λ
Π
Σ
Φ
Ω
Copied to Clipboard

Add Your Solution
Sign Up
to interact with the community. (That's part of how we ensure only good content gets on ToughSTEM)
OR
OR
ToughSTEM is completely free, and its staying that way. Students pay way too much already.
Almost done!
Please check your email to finish creating your account!
Welcome to the Club!
Choose a new Display Name
Only letters, numbers, spaces, dashes, and underscores, are allowed. Can not be blank.
Great! You're all set, .
A question answer community on a mission
to share Solutions for all STEM major Problems.
Why
The Purpose
How
The Community
Give Feedback
Tell us suggestions, ideas, and any bugs you find. Help make ToughSTEM even better.