Question

1a m2 and langth 3.00 m has a current of 2.40 A uniformly distributed across that area A coppar wire of cross-sectional area 2.20 in 35.0 min? nd across that area. (a) what is the magnit ส What ts the magnitude of the electric field along the wire?(b) How much electrical energy s transferred to the nergy is tranaferred to a energy Units (a) Number (b) Number Units

Answer 1

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Given data, Cross sectional area (A) = 2.20 times 10^-6 m^2 Length of wire (L) = 3m Current through the wire (I) = 2.40 A Magnitude of Electric field along the core: We know that, Electric field as E = V/L E = IR/L (Because = V = IR) E = (I times rho L/A)/L (Because R = rho L/A) rho = resistivity of copper = 1.7 times 10^-8 Ohm m E = [2.40 times 1.7 times 10^-8 times 3/2.20 times 10^-6]/3 E = 0.01854545 V/m electrical energy to thermal energy in t = 35.0 min we know that Energy (E) = P times t (P = power P = VI) E = V times I times t = (IR) I times t (Because V = IR) E = I^2 R times t = > E = (2.40)^2 times [1.7 times 10^-8 times 3/2.2 times 10^-6] times t E = 0.1335272 times 35 times 60 (Because 1 min = 60 sec) E = 280.40727 J