This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: ELE401 - Field Theory: Tutorial Problems Week 8 1. An infinitely long line of current Io [A] passes through the point A(6, 0, 5) in the direction, -3ax + 4ay . - Determine the expression for the magnetic intensity vector H at point P (11, 10, 0). z
P(0, 0, z) z=c z' z=b dz'
K = K a 0 y x
Figure 1: A circle of line current 2. Starting with the expression for a circle of line current located on the circle (see Figure 1) x2 + y 2 = a2 at z = z as sensed at a point P (0, 0, z) as: - H = Ia2 2[a2 + (z - z )2 ] 2
3 az [A/m] (1) (a) Convert this into a suitable expression for an incremental ring of - - surface current density K Having a width of dz . ( K = K a ) - (b) Integrate to solve for the H -field due to a solenoid of radius a and extending from z = b to z = c. 1 - (c) Show that the H -field due to an infinitely long solenoid (i.e. c and b -) becomes, - H = K az [A/m] - 3. If J =
Io a2 az (2) - [A/m2 ] in the conductor (see Figure 2), but J = 0 outside. - - (a) Solve for H1 inside the conductor, and H2 outside the conductor, using Ampere's Circuital Law and symmetry. - - (b) Solve B1 and B2 inside and outside the conductor respectively. (c) Solve for the flux crossing the surface,
a 2 2a; = 0 z L; - - B dS .
s = 0 (3) - - (d) If A1 and A2 are given for the two regions, inside and outside the conductor respectively: - A1 = - A2 = -Io o 2 az [W b/m] 4a2 Io o a 1 ln - az [W b/m]. 2 2 (4) (5) - - - Confirm that B = A in both regions. (e) Confirm that the flux calculation in (3c) can also be done using: = - - A d l . (6) z J Non-magnetic o everywhere Conductor a
Figure 2: A conductor 2 ...
View Full Document
- Spring '08
- Magnetic Field, conductor, circuital law, infinitely long line, magnetic intensity vector