Question

If we were give a resistance of 0.75x10^-3 Ω, but not a graph of thermal energy changing over time (no E value given), what would the thermal energy in the loop be?

In Fig. 30-53a, a circular loop of wire is concentric with a solenoid and lies in a plane perpendicular to the solenoid's central axis. The loop has radius 6.00 cm. The solenoid has radius 2.00 cm, consists of 8000 turns/min, and has a currentol varying with time tas given in Fig. 30-53 b, where the vertical axis scale is set by = 1.00 A and the horizontal axis scale is set by t, 2.0 s Figure 30-53cshows, as a function of time, the energy Eth that is transferred to thermal energy of the loop: the vertical axis scale is set by E,-100.0 nJ. What is the loop's resistance? Discussion The magnitude of the emf is determined by the rate at which the magnetic flux within it changes with time. Magnetic flux is dependent on both current and energy. The graphs represent the change in current as well as the change in energy with respect to time, thereby implying a change in magnetic flux. This points to the use of Faraday's law to find the magnitude of the emf (in volts), which is then used to find the resistance of the loop. Equations and Knowns radius of loop = η = 6.00 cm = .06 m radius of solenoid = r, = 2.00 cm = .02 m nurnber of turns = n = 8000 turns. current of solenoid = i = 1.00 A time = t = 2.0 s thermal energy transferred to loop = E = 100.0 n] = 80.0×10-9 J minute magnetic flux = Φ B = BA magnitude of magnetic field = B-Honi area of solenoid = A = πr2 magnitude of induced emf (Faraday's law) = ε dt resistance = R =-=-(power ! s in J /s) ぼ) Solution First, find the magnitude of the induced emf, measured in volts (4mx10-7ZY8000AM EN1.00A)(n)(02m ). 6.3×10-6 v urns 2.0 s Now that we have the magnitude of the emf, we can find the resistance V2 dE (6.3×10-6 V)2 80.0×10-9 0.99×10-3 The loop's resistance is 0.99×10-3 Ω.

Answer 1

Joule's law: $H=I^{2}Rt/J$ where; H= Amount of heat/Thermal energy; I= current; t=time; J= Mechanocal equivalent of heat. In S.I. system J=1; and for C.G.S. system J=4.2J/Cal;

Thus $H=I^{2}Rt/J=[1\times (0.99\times 10^{-3})\times 2] Joule=1.98\times 10^{-3}J$ $\approx 4.71\times 10^{-4}Cal$