Question

2:49 ) (1 point) THE PROBLEM Standard electrical wires used in houses have a diameter of d 2.053 mm and are approved for currents up to 20.0 A. If the current in a particular wire is I-18 A, calculate the magnetic field strength (a) inside the wire, -0.147 mm from the centre. (b) at the surface of the wire. (c) outside the wire, at distance L-0.373 mm from the surface of the wire PAPER SOLUTION INTERPRET Identify all of the true statements A. This is a job for Gauss's law for magnetism. B. We should integrate over the complete surface area of a circle centred at the wires centre C. We should integrate around a circle centred at the wire's centre. D. This is a job for Ampere's law. E. None of the above. DEVELOP In your answers below, type u0 to represent o. (a) Derive an algebraic expression for the magnetic field strength at the inside location. (b) Derive an algebraic expression for the magnetic field strength at the surface.

Answer 1

"d = 2.053 mm rightarrow R = d/2 = 1.0265 mm I_0 = 18 A Because Amperes law is given by contourintegral_c B d l = mu_0 I INTERPRET - We should integrate around a circle centered at the wires center Because B is constant, so expression becomes B contourintegral_c d l = mu_0 5 This is job of Amperes law as we know that, ""it states that for any closed loop path, the sum of length elements times magnetic field in the direction of the length element is equal to permeability times the electric current in the leep DEVELEP I_in For r = 0.147 mm is I = I_0/pi R^2 pi r^2 = 18/(1.0265)^2 (0.147)^2 = 0.3691384 A Therefore B_in contourintegral d u = mu_0 I B_m (2 pi r) = mu_0 I rightarrow B_in = mu_o (0.3691384)/(2 pi x 0.147 times 10^-3) rightarrow B_in = 399.66 mu_0 Test"