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9 Apr 2009: When the normal to the coil makes an angle θ = ωt to B as shown, so that n = cos θksin θi,then the flux of B through the coil is. ... B dS = N B nA = N B cos θA.

20 Jan 2009: Repeat the calculation in spherical geometry, i.e. for concentric spheres of radiir = a, b, b > a. ... For (iii), from a decent diagram of spherical polars, find k = er cos θ eθ sin θ. ].

20 Jan 2009: MATHEMATICAL TRIPOS PART IB. ELECTROMAGNETISM: Examples 3. 1. It is given, as a consequence of the Maxwell equation E = B/t, that. ... b. ]. sin(kz ωt), Bz =nπA. bcos. [ nπy. b. ]. cos(kz ωt),. where n is an integer and A is constant.

18 Feb 2009: E = (E0, 0, 0) cos(kz ωt), B = (0,1. cE0, 0) cos(kz ωt), (9). ... for E0 real,. (Ephys)x = 2E0 sin kz sin ωt, and (Bphys)y = (2/c)E0 cos kz cos ωt.

24 Aug 2009: he, = sin 0 cos 8 ( A 2 - sin2 0)B sin 0 cos $, he, = - sin2 8 ( A 2 - sin2 0 ) t cos 0 cos $, he, = (A2 - sin2 8)j ... Gal. ( 6 b ). ( 6 c ). ( 7 ). and the orbit equation (1) reduces to. 4 = [ C O S 0 (A2 - sin2 6) cos 91 with 0 constant.

8 Jul 2009: B′ cos KX instead of A and B, with solution. N (0) ce(. ... A′ B′. 2sin 2T. B′. (2)2cos 2T. )cos KX. (5.10). Note that the oscillating part has to be retained in the present analysis up to the secondorder.

13 Jul 2009: d&e 2( 1 - a2)8 2ar ------= dr ylr 4 -QB & (J&B - 1). ... Substituting our solution for the streamfunction, we find a stress. fr,, = -p Kr-2’“-2)(3 - n>n cos%ze,.

30 Nov 2009: f (2k1)(1) f (2k1)(1)],. 1. 1f (x) sin(bx) dx cos b. ... where. F (a, b) = cos(a b) exp. (. a2. 4z2. ) [.

1 Dec 2009: 01. ]. µ cos ωt, (5.1). thereby, using our notation,. h(x) =. [0 1k b. ]. x, f(x) =. [01. ]. In this case, since the system is already linear, the matrixU is ... Putting everything together, we have. ψ1(t) = p1,0 µf(p0,0) cos ωt.

31 Jul 2009: E. J. Hinch and L. G. Leal. f = cos [( (8h/r2) - 1 - 3n2)) In C constant]. ... 2 2 b ) C3r2 sin r cos r (Cz cos2 r - 1). (

23 Jul 2009: with associated stress fields. u,., = - 2pEe2{Fp2 (#Gp4 5 H p a ) cos 40)) a,., = -2pEe2(. = ... R ( 0 , z ; e ) = R,(z) [ 1 e f ( z ) cos 20 O(e2)].

14 Jul 2009: u = (A cos t, 0) on y = 1,. u = 0 on y = cos x. ... U0 cos ωt. R = 200 mm. h. (a). (b) (c).

5 Mar 2009: ũi =. {0 on y = 1,16A cos x sin t on y = 0. ... Lagrangian drift near a wavy boundary 17. (a) (b). U0 cos ωt particles h.

10 Jul 2009: The O(1) solution is. Φ0(η, T0, T1) = B(T0, T1) cos η, B(0, 0) = 1,. ... H̃(θ, T ) h̃(θ, T ) 1 γ cos θ 32 γ2 cos 2θ.

14 Jul 2009: Figure 4(b) shows that the numerical error is first order in the time-step. ... a ) The computed probability distributions P ( x ) , with the dashed curve being 0.9937 exp(- cos 2nx/2n). (

23 Jul 2009: At leading order we find. a,, = a;, - - pf sin 8/r2, a,, = goz - pf cos B/r2. ... grr N 2,uf sin 8 / r -,ufy'( 1 3 cos 28)/r2, and.

21 Jul 2009: d A a d d b (AF) = - q ’ - < ’ t c o ’ ( t q ) ) - 2q2-(c0t cos(tcp)). ... follows to leading order for high frequencies. - [cos (pw3 sin2 qw?]] (n 1 b) d2r.