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.
