Question

1 2 3 4 5 6 A solid cylindrical conducting shell of inner radius a = 5.4 cm and outer radius b - 7.8 cm has its axis aligned with the z. axis as shown. It carries a uniformly distributed current 12 = 5.6 A in the positive z-direction. An inifinte conducting wire is located along the z-axis and carries a currently = 2.4 A in the negative z-direction. R(707d, 7070) SC6d, 6d) 1) What is By(P), the y-component of the magnetic field at point P, located a distance d - 24 cm from the origin along the x-axis as shown? T Submit You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. 2) What is where the integral is taken along the dotted path shown in the figure above: first from point P to point Rat (x,y) - (0.7070, 0.707d), and then to point S at (x,y) - (0.6d, 0.6d). T-m Submit You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. 3) What is B (T), the y-component of the magnetic field at point T, located at (x,y) = (-6.2 cm,0), as shown? T Submit You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question.

Answer 1

1) Applying the Ampere's law: ſ Bdl = Holenc B(2nd) = M.(12 - 11) then: B - Moll₂-4) 2nd Using the giving data: (4110-7)(5.6 -2.4) B 21(0.24) B = 2.67 x 10-6T 2) Let's separate the integral into two parts: R B - d = 3 - dĩ + B dĩ The second integral is zero, because the magnetic field is perpendicular to the vector dl TẾ - di = TB di = BL = Bed In this case: Then: 8 di = (2.6 X 107°)( ) (0.24) TB di = 4.90 x 10-7 Tm 3) Applying the Ampere's law. R_Mo(11 – Izenclosed) 2nr Where: 12(r2 – a) Izenclosed b2 - a2 Using the giving data: (5.6) (6.22 – 5.42) Izenclosed 7.82 – 5.42 12 enclosed = 1.64 A Then: (4110-7)(2.4 - 1.64) 21(0.062) B = 2.45 x 10-6T