A piece of copper wire is formed into a single circular loop of radius 9.1 cm. A magnetic field is oriented parallel to the normal to the loop, and it increases from 0 to 0.90 T in a time of 0.66 s. The wire has a resistance per unit length of 2.9 x 10-2 / m. What is the average electrical energy dissipated in the resistance of the wire

Given Information:

Radius of circular loop = r = 9.1 cm = 0.091 m

Change in time = Δt = 0.66 seconds

Change in magnetic field = ΔB = 0.90 T

Resistance of wire per unit length = R = 2.9x10⁻² Ω/m

Number of turns = N = 1

Required Information:

Electrical energy dissipated = E = ?

Answer:

Electrical energy dissipated = 50.09x10⁻³ Joules

Step-by-step explanation:

We know that energy is given by

E = Pt

Where power is given by

P = ξ²/R

Where ξ is the induced EMF in the wire and is given by

ξ = - NΔΦ/Δt

Where ΔΦ is the change in flux and is given by

ΔΦ = ΔBAcosφ

Where φ is the angle between magnetic field and circular loop

A = πr² and R = 2.9x10⁻²*2πr

Substituting the above relations into the energy equation and simplifying yields,

E = [-Nπr²cosφ (ΔB/Δt) ²]*t/R

E = [-1*π (0.091) ²*cos (0) (0.90/0.66) ²*0.66]/2.9x10⁻²*2π * (0.091)

E = 0.050094 Joules

E = 50.09x10⁻³ Joules

Therefore, the average electrical energy dissipated in the circular loop of the wire is 50.09x10⁻³ Joules.