Tutorial 3 - Solutions Electromagnetic Waves

Transcription

1 Tutorial 3 - Solutions Electromagnetic Waves You can find formulas you require for vector calculus at the end of this tutorial. 1. Find the Divergence and Curl of the following functions - (a) k r ˆr f = k r 3 f = 0 (b) r R ˆθ f = 0 f = Rẑ (c) y x +y ˆx + x +y f = 0 f = 0 (d) y x +y ˆx + x +y 4xy f = (x + y ) y x f = (x + y ) You could try plotting these functions to get an intuitive feel on how fields with zero curl or zero divergence look like. Here are the plots for part (c) and (d). The others are fairly trivial to plot 1

2 Figure 1: y x +y ˆx + x +y Figure : y x +y ˆx + x +y. Find the gradient of the following functions - (a) xe (x +y ) : ((1 x )ˆx xyŷ)e (x +y ) (b) (cos x + cos y) : (( + cos(x) + cos(y))sin(x))ˆx + (( + cos(x) + cos(y))sin(y))ŷ 3. For a harmonic plane wave propagating in a source free medium, show that Maxwell s equation reduce to the following - E = ωµ H H = ωɛ E k E = 0 k H = 0

3 Maxwell s equations E = µ H t B = ɛ E t E = 0 H = 0 Any electromagnetic wave in free space will be of the form E = E 0 e j(ωt k r )ˆη and H = H 0 e j(ωt k r ) ˆζ k r = kx x + k y y + k z z Then all differentials will transform to - Similarly, E t jωe E E = x + E y + E z = j(k x + k y + k z ) E = jk E E = j E Substituting the above results in Maxwell s equations, we will get the following results - j E = jωµ H j B = jωɛ E j k E = 0 j k H = 0 Which will give us the following results - E = ωµ H H = ωɛ E k E = 0 k H = 0 3

4 4. Assume that there is a region with cylindrical symmetry in which the conductivity is given by σ = 1.5e 150ρ ks. An electric field E = 30ẑ V in present in the region. m m Find the following - (a) Find J. J = σ E = e 150ρ ẑ (b) Find the total current crossing the surface ρ < ρ 0 and z = 0. I = J ds = π ρ e 150ρ ρdρdφ = (π)( ) 1 e 150ρ 0 (150ρ 0 + 1) 500 = 1.6(1 e 150ρ 0 (150ρ 0 + 1)) A (c) Calculate H. We know that the magnetic field is circular in nature. Thus, by using Ampere s circuit law we have, H dl = Ienclosed H(πρ) = 1.6(1 e 150ρ (150ρ + 1)) H = (1 e 150ρ (150ρ + 1)) ρ Am 1 5. Two infinitely long parallel filaments each carry 50 A in the ẑ direction. If the filaments lie in the plane y = 0 at x = 0 and x = 5 mm. Find the vector force per meter length of the filament passing through the origin. Do you know that the SI system uses a similar definition to standardize 1 Ampere? What changes can you make to the given system to define 1 A? B = µ 0 I πr ; F = I( dl B ) Substituting the values and solving the equation, we will get - F dl = BI = µ 0I πr = 0.1ˆx N m 4

5 1 Ampere is defined by the SI system as the constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular crosssection, and placed 1 m apart in vacuum, would produce between these conductors a force equal to 10 7 newtons per metre of length. 6. For the electric field E = E 0 e kx cos( 10 8 t y)ẑ in free space, find the value(s) of k for which the field satisfies both of Maxwells curl equations. Satisfying the Maxwell s curl equation is the same as satisfying the wave equation. E µ 0 ɛ 0 E t = 0 (E 0 e kx cos( 10 8 t y)) = µ 0 ɛ 0 (E 0 e kx cos( 10 8 t y)) t k = k = ± 3 m 1 7. Using Gauss law, find the electric field inside a sphere which carries a charge density proportional to the distance from the origin, ρ = kr. Plot E vs r. Assume k to be some positive constant. E ds = q enclosed Since it is spherically symmetric, the equation can be transformed to the following, ρr sin θdrdθdφ 4πr V E = ɛ kr r sin θdrdθdφ 4πr V E = ɛ 4πr E = kr4 4π 4ɛ E = kr 4ɛ ˆr ɛ r<radius of sphere 5

6 Figure 3 8. Derive Gauss law from Coulomb s law. What property of Coulomb s law is exploited to get such a simple equation of Gauss law. Using Coulomb s law, we have - 1 q E (r) = 4πɛ r ˆr E ds = 1 4πɛ q r ˆr ds If we take the charge as charge enclosed within a volume, we can take it out of the integral to get the integral in the form of a solid angle. E q enclosed 1 ds = 4πɛ r ˆr ds We know this integral goes to 4π, for a point inside a closed surface. Thus, we have - E ds = q enclosed ɛ 9. Find the magnetic field of a very long solenoid, consisting of n closely wound turns per unit length on a cylinder of radius R and carrying a steady current of I. 6

7 Figure 4 It can easily be shown that the magnetic field of this long solenoid will only be along the axis of the cylinder. We will take two closed loops as shown above and apply Ampere s law. For loop 1 - B dl = µ0 I enclosed = 0 B(a) = B(b) = B( ) But we know that the magnetic field of a solenoid is 0 at large distances. Thus, B(r) = 0 for r > R For loop - B dl = µ0 I enclosed = µ 0 nli B(R ) B(R + ) = µ 0 nli B(r) = µ 0 nli for r < R Thus, B = { µ0 nliẑ r <R 0 r >R 7