Question

Questions 16-17 The figure shows the cross-section of a very long coaxial cable. The radius of the inner conductor part of the cable is a, the outer radius of the insulator is b2a and the outer radius of the conductor part is c, 2a <e<3a. The current density in both conductive regions is constant The internal current, I, flows into the plane of the page and the current in the outer part, 1- 2I, flows out of the page plane. (take 3 16. What is the magnitude of the magnetic field, in terms of o, I and a, generated by these currents at the point r μο! 12a (b) o 24a 18a (e) ol 2α 17, what is the magnitude of the magnetic field, in terms of μο 1 and α, generated by these currents at the point r 3a? μο! 12a (e) Ho e (b) 2Ta (d) 18a Questions 18-20 A rectangular conductive loop is placed near a long, straight wire at a distance a carrying an alternating current of 1(t)-4xìn(wot). Here, wo--.T-0.02 s ls the period of the alternating current. Both the wire and the loop are in the same plane. The slde I lengths of the loop are Li-α and L2 :-2a, and the resistance R 20 Ω is connected between the two open ends of the loop (take ln2-0.6 and π-3). 18, what is the magnetic flux, in terms of flo. L and a, passing through the loop at t-T/4? What is the voltage, in terms of flo, lo and a, between the ends of the resistor at t-T/2? 19. 20. What is the current, in terms of o, I and a, passing through reslstance at t T/2?

Please solve all of them clearly and separately thank you

Answer 1

Consider an Amperian loop of radius r = a/2 Using Ampere�s law contourintegral B^vector middot ds^vector = mu_0 I_enc Due to symmetry contourintegral B^vector middot ds^vector = B(2 pi r) = B(2 pi a/2) I_enc = current in the loop, the loop encloses part of inner conductor only = I/pi a^2 middot pi(a/2)^2 = I/4 therefore B(2 pi a/2) = mu_0 I/4 B = mu_0 I/4 pi a = mu_0 I/12 a (pi = 3) option a This time we choose a loop of radius r = 3a Using Ampere�s law contourintegral B^vector middot ds^vector = mu_0 I_enc, I_enc = I_2 + (-I_1) The loop encloses both current I_enc = 2I - I = I contourintegral B^vector middot ds^vector = B(2 pi r) = B(2 pi(3a)) so B(6 pi a) = mu_0 I B = mu_0 I/6 pi a = mu_0 I/18 a option e