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27 Feb 2020: 2 S)zByBz. Bx. Bz. K cos(Kz). sin(Kz+) 1. z+. ,. vy = v2az2. ... BxBz. K cos(Kz). sin(Kz+). The meridional (x and z) components of B are called the poloidal magnetic field, while the azimuthal(y) component is called the toroidal magnetic field.

4 Oct 2013: TA sin θATA cos θA. µgδxT. (3). Now. TB sin θBTB cos θB. ... T(t) = cos jmnct and sin jmnct. Putting all this together i.e.

18 Oct 2021: The wavefunction (2.12) can be viewed as a sum of sine and cos functions and describes a complex-valued wave of wavelength. ... 2. x. The most general solution has the form. x. (. t. ). =. A. cos. (. ω. (. t. -. t.

12 Jun 2024: x(, ) = ( cos , sin , ) (4.18). with a   b and 0  < 2. ... equation tells us that. µ0I =. Z. C. B dx =. Z 2. 0. d b() =) B(x) =µ0I. 2( sin , cos , 0).

9 Oct 2023: F8. (a) i. (b) 1. (c) 12(1 i). (d) e(cos 1 isin 1). ... F13. (a)sin 6θ sin 5θ sinθ. 2(1 cosθ). (b). cos 6θ cos 5θ cosθ 12(1 cosθ).

11 Feb 2014: α =. C. A dx. α =. C. A dx =. B dS. Aϕ =g. 4πr. 1 cos θsin θ. B =g. ... 1 cos θsin θ. B =g. 4πr2̂r. References. [1] Gibbons and Rychenkova, 9608085.

5 Dec 2008: So general solution becomes. f = A(cos 2φ 1) B(sin 2φ 2φ). ... The boundary condition at φ = α gives. A(cos 2α 1) B(sin 2α 2α) = 12.

8 Nov 2019:   cos φ sin φ 0sin φ cos φ 00 0 1. ... ψ̇ φ̇ cos θ. . (b) [optional extra] Calculate the matrix B = R>Ṙ and verify that it is antisymmetric.

12 Nov 2023: b) 2(cos θ θ sin θ)/ cos2 θ. (c) t(1 2 ln t). ... a tan1 y. a). H14. (b) J = 1b(1 eax cos bx) a.

28 Nov 2008: B. B3 = B cos θ. B2 = B sin θ cos ψ. ... H = 2. 2m2 12 B σ cos θ. 12B. 2 sin2 θ 12.

3 Oct 2023: a)d. dx(sin x) (b). d. dθ(cos θ). (c)d. dt(tan t) (d). ... x2 4)dx (b) 20. (xa)2dx. (c) π0eiθdθ (d). π0. cos(x)dx. (e)π/4π/4.

8 Jul 2021: ψ. x. (x(1)s ,y(1)s ). = b cos(x t ), (18). ... ψ. x. (x(2)s ,y(2)s ). = b cos(x t ), (20).

14 Oct 2014: obtain:. Φ(ϕ) = A sin(λϕ) B cos(λϕ),. with positve constants A, B. ... Aλrλ Bλr. λ) sin(λϕ) (Cλrλ Dλr. λ) cos(λϕ). E0 F0 ln(r).

18 Nov 2019: q1 = Q1 cos λ P2 sin λ, q2 = Q2 cos λ P1 sin λ,. ... p1 = Q2 sin λ P1 cos λ, p2 = Q1 sin λ P2 cos λ.

26 Jun 2024: or,. (x) = B cos kx |x| < a (2.34). where B is again a normalisation constant and k is, like , related to the energy, now. ... automatically patched at x = a as well. We have. (x) cts at x = a ) B cos ka = Aea.

30 Oct 2008: k=0. (1)k[2ω(n m)]2k2 [f. (2k1)(b) cos 2ω(n m)b. f(2k1)(a) cos 2ω(n m)a]}. ... m. ). {. k=0. (1)k[ω(2n 2m 1)]2k1. [f(2k)(b) cos ω(2n 2m 1)b f(2k)(a) cos ω(2n 2m 1)a].

23 Jan 2004: B(t) = v(t). [. ω1 sin ωt cos ωt. ]. [ cos ωt ω1 sin ωt ].(2.3). ... ω3,. b(2)2 (ω) =. 4(1 cos ω). ω2+. 8 sin ω.

23 Dec 2020: vr = rν cos φ[c(ν 1)Pν1 b(ν 1) cos θ Pν. ], ... 2.12a). vθ = rν cos φ[cP′ν1 2a csc θ Pν b.

16 Oct 2023: λk(A) =λk(C). λk(B)=. 12µsin2 12πkh1 2µsin2 12πkh. |λk(A)| 1, k = 1.M. ... iγ where γ = 2 cos(πh) for = 1,.

8 Oct 2015: x = A cos ωt B sin ωt a cos(ωt ϵ) Peiωt Qeiωt, (2). ... The solution isx = A cos ωt B sin ωt. Initially, x = a and ẋ = 0, so.