Radiation Zone Approximation We had seen that the expression for the vector potential for a localized cuent distribution is given by AA (xx, tt) = μμ 4ππ ee iiiiii dd xx eeiiii xx xx xx xx JJ (xx ) In near field approximation, this led to expression that was familiar to us in magnetostatics excepting for the time dependent multiplicative factor ee iiiiii, leading to our describing the situation as quasistationary. We will not have much to say about the intermediate zone as this case is very complicated. However, the far field approximation or the radiation zone for which the distance of the point of observation from the source is much greater than both the typical source dimension and the wavelength of radiation is of great interest to us and will be described in the following. Let us write xx = nn. We expand xx xx and 1 xx xx as follows: xx xx = ( 2 + xx 2 2nn xx nn xx = 1 2 nn xx Radiation Lecture4: Electromagnetic Theory Professor D. K. Ghosh, Physics Department, I.I.T., Bombay 1 2 + xx 2 2 Thus for situations where xx. In a similar way, we have, keeping terms of the order of 1/ in Binomial expansion 1 xx xx = 1 1 2 nn 1 xx + xx 2 2 2 = 1 nn xx + 2 + 1 nn 2 xx xx 2 2 Using the above, we can write, retaining terms upto 1 2, ee iiii xx xx xx xx = eeiiiiii 1 iiiinn xx 1 nn xx + 2 1
= eeiiiiii 1 + 1 iiii nn xx Electric Dipole Approximation The lowest order approximation is to ignore the second term within bracket of the above. This gives The integral above is rewritten by observing, AA (xx, tt) = μμ eeiiiiii ee iiiiii 4ππ dd xx JJ (xx ) xx ii JJ = ( xx ii ) J + xx ii JJ = ( xx ii ) J x i ρ t = JJ ii + iiiixx ii ρρ where we have used xx ii = nn ii and the fact that the time dependence is given byee iiiiii. We replace each component of JJ (xx ) in the expression for vector potential by xxii JJ iiiiiixx ii. The integration over the first term is done by first converting it to a surface integral using the divergence theorem and letting the surface go toinfinity. Since the source dimension is confined to a limited region of space, this integral is zero and we are then left with, AA (xx, tt) = μμ eeiiiiii ee iiiiii 4ππ dd xx iiiixx ρρ The integral xx ρρdd xx is the electric dipole moment of the source. Hence, we can write, where we have used ωω = cccc and μμ εε = 1 cc 2. AA (xx, tt) = μμ 4ππ = ii kk ee iiiiii pp 4ππεε cc ee iiiiii eeiiiiii We can now obtain the expressions for the magnetic field and the electric field from the vector potential. Using the fact that pp is a constant vector, we can write, iiiipp HH = 1 ii kk AA eeiiiiii = μμ 4ππμμ εε cc pp = iicccc eeiiiiii 4ππ pp = iiiiii iiiiiiiiiiii eeiiiiii 4ππ 2 pp 2
ee iiiiii = cckk2 ii 1 + pp 4ππ kkkk Thus the magnetic field direction is transverse to the direction of. We will see, however, that the electric field has a longitudinal component as well. which gives εε EE = iiiiεε EE = HH EE = iiμμ cc kk HH = iiμμ cc cckk 2 ii eeiiiiii 1 + kk 4ππ kkkk 2 ( pp ) = iiii 4ππεε eeiiiiii 2 + iieeiiiiii ( pp ) + eeiiiiii kkkk 2 + iieeiiiiii ( pp ) kkkk We have, eeiiiiii 2 + iieeiiiiii = iiii kkkk 2 eeiiiiii eeiiiiii ii kk 4 eeiiiiii ( pp ) = ( pp ) ( )pp + (pp ) ( )pp The first and the last terms of above are zero because pp is a constant vector. We are left with ( pp ) = pp + pp = 2pp Substituting these, the electric field becomes EE = iiii iiii 4ππεε 2 eeiiiiii eeiiiiii ii kk 4 eeiiiiii ( pp ) 2 eeiiiiii 2 + iieeiiiiii pp kkkk We reorganize the terms in the following fashion. Keep the first term within the first parenthesis in tact. For the second and the third terms, expand the vector triple product EE = ( pp ) = ( pp ) = [( pp ) pp ] iiii iiii 4ππεε 2 eeiiiiii ( pp ) 2 eeiiiiii + ii kk eeiiiiii [( pp ) pp ] 2 eeiiiiii 2 = ee iiiiii 4ππεε kk2 ( pp ) + iiii ii iiii ii 1 + pp 1 + ( pp ) kkkk kkkk = ee iiiiii 4ππεε kk2 ( pp ) + iiii ii 1 + (pp ( pp ) ) kkkk + iieeiiiiii pp kkkk
Note that, while the magnetic field was transverse, the electric field has a longitudinal component as well. Both these fields go as 1 for large distances. However, for smaller distances, while the magnetic field vary as 1 1 2, the electric field goes as (the last term ). In the radiation zone, we are only interested in large distance behaviour. In this regime, the field expressions are, and the coesponding electric field is given by so that EE, HH and are mutually perpendicular. HH = cckk2 ee iiiiii ( pp ) 4ππ EE = ee iiiiii 4ππεε kk2 ( pp ) = ccμμ HH How does the emitted power, i.e., the average value of Poynting vector vary? SS = 1 2 Re EE HH = 1 2 ccμμ Re HH HH = 1 2 ccμμ HH 2 = cc kk 4 μμ 1 2ππ 2 ( pp )2 2 = ωω4 μμ 1 2ππ 2 cc 2 pp2 sin 2 θθ The amount of power flowing per unit solid anglein the direction of (θθ, φφ) is given by dddd ddω = 2 SS = ωω4 μμ 2ππ 2 cc pp2 sin 2 θθ and the total radiated power is obtained by integrating over (θθ, φφ). Since there is no azimuthal dependence, the integration over φφ gives 2ππ and we get, ωω4 μμ PP = 2ππ 2 cc pp2. 2ππ sin θθ dddd = ωω4 μμ 16ππππ pp2 4 = ωω4 μμ 12ππππ pp2 ππ 4
The following figure shows the power pattern in dipole approximation. It can be seen that the two lobes are symmetric about the dipole. The distance from the dipole to the lobe at any point gives the relative power in that direction with respect to the maximum. The total radiated power varies as the fourth power of the frequency. pp dddd θ Magnetic Dipole and Electric Quadrupole Approximation We had started our discussion on radiation zone with the following expression for the vector potential, AA (xx, tt) = μμ 4ππ ee iiiiii dd xx eeiiii xx xx xx xx JJ (xx ) Retaining terms up to order 1 2, we had approximated ee iiii xx xx xx xx = eeiiiiii 1 + 1 iiii nn xx In the dipole approximation, we only retained the first term within the square bracket and ignored the remaining. Our next approximation is concerned with the second term. We will not write down the dipole term so that what we are discussing below is the coection to the dipole radiation. We consider AA (xx, tt) = μμ eeiiiiii ee iiiiii 4ππ 1 iiii dd xx nn JJ (xx xx ) where we have taken out factors which are not concerned with the integration variable outside the integration. 5
In order to perform the integration, let us write the dot product in an expanded form. Restricting only to the integration (the outside constants will be inserted later) JJ (xx ) xx jj xx jj dd xx Using the argument given before, as the ettent of the source is limited, by extending the surface integral to infinite distances, we have, from the divergence theorem, xx ii xx jj JJ dd xx = If we write this by expanding the divergence term, we have, xx ii xx jj JJ = xx ii xx jj JJ + JJ xx ii xx jj = xx ii xx jj JJ + xx ii JJ jj + xx jj JJ ii = xx ii xx jj iiiiii + xx ii JJ jj + xx jj JJ ii where, in the last step, we have used the equation of continuity. Inserting this into i-th component of the integral, JJ ii (xx ) xx jj xx jj dd xx = xx jj xx ii xx jj JJ iiiiii xx ii xx jj xx ii JJ jj dd xx As mentioned before, the integral over the divergence is zero and we are left with JJ ii (xx ) xx jj xx jj dd xx = xx jj iiiiii xx ii xx jj + xx ii JJ jj dd xx We replace the right hand side with half the term and add half of the left hand side to it JJ ii (xx ) xx jj xx jj dd xx = 1 2 xx jj iiiiρρ xx ii xx jj + (xx ii JJ jj xx jj JJ ii ) dd xx The term (xx ii JJ jj xx jj JJ ii ) is easily seen to be component of a cross product and can be written as (xx ii JJ jj xx jj JJ ii ) = εε iiiiii xx JJ kk where εε iiiiii is the Levi-Civita symbol which is + 1 if ii, jj, kk is an even permutation of 1,2 and, is 1 if it is an odd permutation and is zero if any pair of symols are equal. Using εε iiiiii xx jj xx JJ kk = xx xx JJ ii 6
We also have, xx jj xx jj = xx xx Thus JJ ii (xx ) xx jj xx jj dd xx = 1 xx xx JJ dd xx ii xx xx xxii ρρ ωωωω xx 2 ii 2 Adding three components JJ (xx ) xx jj xx jj dd xx = xx mm ii xx xx xx ρρ ωωωω xx 2 where the magnetic moment vector mm is given by mm = IIAA = 1 2 IIIIll = 1 2 JJ dd The term xx mm is the magnetic dipole term and the remaining term is the electrical quadrupole term. Substituting these in the expression for the vector potential for the magnetic dipole term becomes, AA mm (xx, tt) = μμ eeiiiiii ee iiiiii 4ππ 1 iiii nn mm The magnetic field is now readily calculated. Suppressing the time variation, HH mm (xx ) = 1 μμ AA mm (xx ) = 1 4ππ eeiiiiii 1 iiii 2 mm = 1 4ππ eeiiiiii 1 iiii 1 2 ( mm ) 4ππ eeiiiiii 1 iiii 2 ( mm ) ee iiiiii 1 iiii eeiiiiii 2 = iiii + kk2 ee iiiiii 2 eeiiiiii nn 4 ee iiiiii 1 iiii eeiiiiii 2 ( mm ) = iiii 2 + kk2 ee iiiiii = iiii eeiiiiii 2 + kk2 ee iiiiii eeiiiiii nn (nn mm ) eeiiiiii [nn (nn mm ) mm ] 7
and ( mm ) = ( mm ) ( )mm + (mm ) ( )mm = mm + mm = 2mm Combining these, HH mm (xx ) = 2 4ππ eeiiiiii 1 iiii 1 2 mm 4ππ iiii eeiiiiii 2 + kk2 ee iiiiii eeiiiiii [nn (nn mm ) mm ] = 1 kk 2 ee iiiiii nn (nn mm ) + 1 4ππ 4ππ eeiiiiii 1 iiii 2 [nn (nn mm ) mm ] where we have restored the structure nn (nn mm ) for the central term of the second expression retaining the expanded form for the remaining. In the radiation zone, the first term, which varies as 1 at large distances dominate and we have, HH mm (xx ) = 1 kk 2 ee iiiiii nn (nn mm ) 4ππ The coesponding electric field can be determined from Faraday s law, EE = BB = iiiibb = iiii AA so that EE = iiiiaa = iiiiii 4ππ = kkkkkk 4ππ = μμ cckk 2 4ππ ee iiiiii ee iiiiii nn mm ee iiiiii nn mm 1 iiii nn mm where we have retained only the leading term in radiation zone approximation. The power radiated can be calculated in the same manner as we did for the electric dipole case, PP = μμ cckk 4 12ππ mm 2 [The above formulae were for two specific cases of radiating systems, viz. oscillating electric dipole and oscillating magnetic dipole. The power radiated from an arbitrary cuent and charge distribution is given by the formula PP ee = μμ 6ππππ pp 2 PP mm = μμ 6ππππ mm 2 8
where the double dots indicate double differentiation with time. For the specific cases mentioned above, these boil down to respective formulae on taking the time average. ] Radiation Lecture4: Electromagnetic Theory Professor D. K. Ghosh, Physics Department, I.I.T., Bombay Tutorial Assignment 1. A spherically symmetric charge distribution oscillates in a radial direction such that it remains spherically symmetric at all time. Show that no radiation is emitted. 2. An electric dipole of strength 1 7 C-m oscillates at an angular frequency of 1 8 rad/s. The dipole is placed along the z-axis. (a) Find the strength of the electric field measured by an observer located at (5,,) km. (b) What is the phase difference between the oscillation frequency of the dipole and that seen by this observer? (c) What is the average power measured at this location? (d) What would be the average power measured if the observer was located at (,,5) km instead of the position defined in (a)? (e) What is the total rate at which the dipole radiates power?. A wire of length 1 m is bent in a circular loop to form a magnetic dipole antenna. The cuent in the dipole varies as II = II sin ωωωω with ωω = 1 8 rad/s and II = 2 A. Determine the electric and the magnetic field as seen by an observer at a large distance from the dipole at an angle θθ to the axis of the loop. Calculate the total power radiated. Solutions totutorial Assignment 1. Spherical symmetry implies that both the electric and the magnetic field remains radial at all times. Using Gauss s law, this implies, EE ( ) = QQ iiiiiiiiiiii and BB ( ) =. This implies the 4ππ 2 εε Poyntingvector SS =, hence there is no radiation. 2. Since the distance from the source is large, it is reasonable to assume that the observer is in the radiation zone. In this region we use cckk EE = ccμμ HH 2 ee iiiiii = ccμμ [( pp ) ] 4ππ cc 2 kk 2 ee iiiiii = μμ 4ππ pp [(xx zz ) xx ] = μμ cc 2 kk 2 ee iiiiii 4ππ pp zz = μμ 4ππππ ωω2 pp ee iiiiii zz 9
where we have used the fact that the observer is located along the x-axis. (a) The field is thus directed opposite to the direction of the dipole moment vector, consi8stent with the fact that the dipole moment vector is along the direction from the negative to the positive charge. The strength of the electric field obtained by substituting given values is.18 V/m. (b) As discussed in the last lecture, the wave generated at time t aives at tt at the observer s cc location. Thus the phase difference between the wave at the source and that at the location of the observer is ωωωω = cc 1 radians. (c) Since the observer is in the equatorial plane θθ = 9. The power measured at this location is ωω 4 μμ 1 2ππ 2 cc 2 pp2 sin 2 θθ = 4. 1 5 Watts. (d) Along the axis of dipole θθ = so that no power is radiated in this direction. (e) The total power radiated is given by ωω 4 μμ 12ππππ pp2 = 9 1 W.. Take the dipole along the x-y plane with its axis being along the z-axis. Consider the position vector of the of observer to be in the x-z plane, nn = cos θθzz + sin θθxx The magnetic moment vector is given by mm = II AAcos ωωωω zz, where A is the area of the loop. The complex magnetic field is given by HH = 1 kk 2 4ππ eeii(kkkk ωωωω ) nn (nn mm ) = 1 kk 2 4ππ eeii(kkkk ωωωω ) II AA sin θθ(sin θθzz cos θθxx ) The electric field is given by EE = μμ cckk 2 4ππ eeii(kkkk ωωωω ) nn mm = μμ cckk 2 4ππ eeii(kkkk ωωωω ) II AA sin θθyy The average Poynting vector is given by SS = 1 2 Re EE HH = = kk4 μμ 2ππ 2 2 II 2 AA 2 sin 2 θθ nn ωω 4 μμ = 2ππ 2 cc 2 II 2 AA 2 sin 2 θθ nn The power flowing out per unit solid angle is given by dddd ddω = 2 nn SS = ωω4 μμ 2ππ 2 cc II 2 AA 2 sin 2 θθ The total power radiated is obtained by integrating over the solid angle and is given by 1
PP = ωω4 μμ 2ππ 2 cc II 2 AA 2 sin 2 θθ2ππ sin θθθθθθ ππ = ωω4 μμ 2ππ 2 cc II 2 AA 2 2ππ 4 = ωω4 μμ 12ππcc II 2 AA 2 The area of the loop is obtained from its circumference of 1m to be 1 4ππ. Substituting II = 2 and other constants, we get the total power radiated to be.25 W. Radiation Lecture4: Electromagnetic Theory Professor D. K. Ghosh, Physics Department, I.I.T., Bombay Self Assessment Questions 1. Two point charges, q each, are attached to the ends of a rigid rod of length ll. The rod is rotated with an angular speed ωω about an axis through the centre of the rod and perpendicular to it. Calculate the electric and the magnetic dipole moment of the system taking the centre of the rod as the origin. What is the total power radiated by this system? 2. An electric dipole of strength pp initially lies along the x-axis centred at origin. The dipole rotates in the x-y plane about the z-axis at an angular frequency ωω. Calculate the radiation field and the radiated power as seen by an observer in the x-z plane at an angle θθ to the z-axis.. Neutron stars are a class of very dense stars consisting primarily of neutrons. The neutron star rotates with an angular frequency ωω about an axis which makes an angle αα with the direction of the magnetic moment. Assume the radius of the star to be 1 km and the angular frequency of rotation to be 1 rad/s about the z-axis which makes angle of αα = 45 with the magnetic moment vector of the neutron star. Take the strength of the magnetic field at the magnetic equator to be 1 8 T. Calculate the total power radiated. 11
Solutions to Self Assessment Questions 1. The electric dipole moment about the centre is zero as the position vector of the two (equal) charges are always directed oppositely. Magnetic dipole moment can be calculated by observing that charge 2q is rotating with a time period TT = 1, which is equivalent to a 2ππππ cuent4ππππππ. The area of the cuent loop is ππll 2 zz. Thus the magnetic moment is given by mm = 4ππ 2 ll 2 qqqqzz. Both the electric and the magnetic dipoles are constant in time and hence there is no radiation emitted by the system. 2. The dipole vector can be written as pp = pp [cos ωωωω xx + sin ωωωω yy ] = pp Re[(xx + iiyy )ee iiiiii ] Since the observer is in x-z plane, y-direction is perpendicular to the direction of observation. We can write = cos θθzz + sin θθxx. Note that (xx + iiyy ) = cos θθyy + ii yy Using these, we can write the complex magnetic field as HH = pp cckk 2 4ππππ eeii(kkkk ωωωω ) (xx + iiyy ) The coesponding electric field is given by EE = ccμμ HH cc 2 kk 2 μμ = pp 4ππππ cc 2 kk 2 μμ = pp 4ππππ The average Poynting vector is given by = pp cckk 2 4ππππ eeii(kkkk ωωωω ) [cos θθyy + ii yy ] eeii(kkkk ωωωω ) [cos θθyy + ii yy ] eeii(kkkk ωωωω ) [cos θθ(yy ) + iiyy ] SS = 1 2 Re EE HH 2 = pp cc kk 4 μμ 2ππ 2 2 (1 + cos2 θθ) The power flowing out per unit solid angle is given by dddd ddω = 2 SS = cc kk 4 μμ 2ππ 2 pp 2 (1 + cos 2 θθ) The total power radiated is given by integrating the above over the solid angle. Since there is no azimuthal dependence, we have, PP = cc kk 4 μμ 2ππ 2 pp 2 (1 + cos 2 θθ) 2ππππππ = cc kk 4 μμ 6ππ pp 2 ππ = ωω4 μμ 6ππππ pp 2 12
. The magnetic field of a dipole at is given by the expression BB = μμ [mm cos θθ mm ] 4ππ whereθθ is the angle between and mm. At the equator, θθ = 9 so that the strength of the magnetic field is BB = μμ 4ππ mm Substituting the given values, the magnitude of the magnetic field is mm = 1 27 Am 2. As the star rotates about an axis which does not coincide with the magnetic moment vector, the direction of the magnetic moment vector precesses about the z-axis. This accounts for the radiation. It was pointed out that the magnetic dipole radiation power is proportional to the square of the second derivative of the magnetic moment. We can write ω mm The magnetic moment vector as mm = mm sin αα(cos ωωωωxx + sin ωωωω yy ) + mm cos αα zz α It is seen that mm 2 = mm 2 ωω 4 sin 2 αα. The power radiated is given by PP mm = μμ 6ππππ mm 2 Substituting numerical values, one gets the power radiated to be 1.2 1 29 W. 1