Solution 2 i ph o 36
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1. The document provides solutions to problems from the 36th International Physics Olympiad regarding absolute measurements of electrical quantities. It includes derivations of formulas for induced electromotive force, magnetic flux, power, magnetic field, resistance, electromotive force, and force between coils. 2. Formulas are derived for the magnetic flux through a coil, induced electromotive force, instantaneous power, total magnetic field, resistance, induced electromotive force on disks, relationship between resistance and current, force between parallel wires, and force and equilibrium position of a coil-magnet system. 3. Solutions involve derivations of relevant formulas, application of concepts like Kirchhoff's laws, and analytical solutions to
Solution 2 i ph o 36
- 1. 36th International Physics Olympiad. Salamanca (España) 2005 Solution Th 2 Page 1 of 4 R.S.E.F. Th 2 ABSOLUTE MEASUREMENTS OF ELECTRICAL QUANTITIES SOLUTION 1. After some time t, the normal to the coil plane makes an angle ω t with the magnetic field iBB rr 00 = . Then, the magnetic flux through the coil is SBN rr ⋅= 0φ where the vector surface S r is given by ( )jtitaS rrr ωωπ sincos2 += Therefore tBaN ωπφ cos0 2 = The induced electromotive force is dt dφ ε −= ⇒ tBaN ωωπε sin0 2 = The instantaneous power is =P ε 2 /R , therefore ( ) R BaN P 2 2 0 2 ωπ = where we used 2 1 sin 1 sin 0 22 =>=< ∫ T dtt T t ωω 2. The total field at the center the coil at the instant t is it BBB rrr += 0 where iB r is the magnetic field due to the induced current ( )jtitBB ii rrr ωω sincos += with a IN Bi 2 0µ = and I = ε / R Therefore t R BaN Bi ω ωπµ sin 2 0 2 0= The mean values of its components are R BaN t R BaN B tt R BaN B iy ix 4 sin 2 0cossin 2 0 2 020 2 0 0 2 0 ωπµ ω ωπµ ωω ωπµ == == And the mean value of the total magnetic field is j R BaN iBBt rrr 4 0 2 0 0 ωπµ += The needle orients along the mean field, therefore R aN 4 tan 2 0 ωπµ θ =
- 2. 36th International Physics Olympiad. Salamanca (España) 2005 Solution Th 2 Page 2 of 4 R.S.E.F. Finally, the resistance of the coil measured by this procedure, in terms of θ , is θ ωπµ tan4 2 0 aN R = 3. The force on a unit positive charge in a disk is radial and its modulus is BrBvBv ω==× rr where B is the magnetic field at the center of the coil a I NB 2 0µ = Then, the electromotive force (e.m.f.) induced on each disk by the magnetic field B is ∫ === b DD bBdrrB 0 2 ' 2 1 ωωεε Finally, the induced e.m.f. between 1 and 4 is ε = εD + ε D' a Ib N 2 2 0 ωµ ε = 4. When the reading of G vanishes, 0=GI and Kirchoff laws give an immediate answer. Then we have RI=ε ⇒ a b NR 2 2 0 ωµ = 5. The force per unit length f between two indefinite parallel straight wires separated by a distance h is. h II f 210 2π µ = for III == 21 and length aπ2 , the force F induced on C2 by the neighbor coils C1 is 20 I h a F µ = 6. In equilibrium dFxgm 4= Then 204 I h da xgm µ = (1) so that 21 04 / da xhgm I ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = µ
- 3. 36th International Physics Olympiad. Salamanca (España) 2005 Solution Th 2 Page 3 of 4 R.S.E.F. 7. The balance comes back towards the equilibrium position for a little angular deviation δϕ if the gravity torques with respect to the fulcrum O are greater than the magnetic torques. δϕ δδ µδϕδϕ cos 11 2cossin 2 0 d zhzh IaxgmlMg ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + + − >+ Therefore, using the suggested approximation δϕ δµ δϕδϕ cos1 4 cossin 2 22 0 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ +>+ h z h Ida xgmlMg Taking into account the equilibrium condition (1), one obtains δϕ δ δϕ cos h z xgmsinlgM 2 2 > Finally, for d zδ ϕδϕδ =≈ sintan dxm hlM z 2 <δ ⇒ dxm hlM z 2 max =δ O l x δϕ δϕ δz h + δz h - δz h + δz h - δz Mg mg d G
- 4. 36th International Physics Olympiad. Salamanca (España) 2005 Solution Th 2 Page 4 of 4 R.S.E.F. Th 2 ANSWER SHEET Question Basic formulas and ideas used Analytical results Marking guideline 1 R P dt d SBN 2 0 ε ε Φ Φ = −= ⋅= rr ( ) R BaN P tBaN 2 sin 2 0 2 0 2 ωπ ωωπε = = 0.5 1.0 2 x y i i B B I a N B BBB = = += θ µ tan 2 0 0 rrr θ ωπµ tan4 2 0 aN R = 2.0 3 rv BvE ω= ×= rrr a I NB 2 0µ = ∫= b rdE 0 rr ε a Ib N 2 2 0 ωµ ε = 2.0 4 IR=ε a b NR 2 2 0 ωµ = 0,5 5 h II f ′ = π µ 2 0 20 I h a F µ = 1.0 6 dFxgm 4= 21 04 / da xhgm I ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = µ 1.0 7 maggrav ΓΓ > dxm hlM z 2 max =δ 2.0