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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?

Added Tue, 04 Aug '15

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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

Added Tue, 04 Aug '15

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