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..[r]
(1)36th International Physics Olympiad Salamanca (España) 2005
Solution Th Page of R.S.E.F
Th ABSOLUTE MEASUREMENTS OF ELECTRICAL QUANTITIES SOLUTION
1 After some time t, the normal to the coil plane makes an angle ωt with the magnetic field Br0 =B0ir Then, the magnetic flux through the coil is
S B
N r ⋅r
= 0
φ
where the vector surface Sr is given by Sr=πa2(cosωtir+sinωt rj)
Therefore φ=Nπa2B0cosωt
The induced electromotive force is
dt dφ
ε =− ⇒ ε =Nπa2B0ωsinωt
The instantaneous power is P=ε2 /R , therefore
( )
R B a N P
2
2 ω
π = where we used
2 sin
1 sin
0
2 >= =
< ∫T tdt
T
t ω
ω
2 The total field at the center the coil at the instant t is
i
t B B
Br = r0+ r
where Bi
r
is the magnetic field due to the induced current Bri =Bi(cosωtir+sinωtrj)
with
a I N Bi
2
0
µ
= and I = ε / R
Therefore t
R B a N
Bi µ π ωsinω
2
0
=
The mean values of its components are
R B a N t
R B a N B
t t R
B a N B
iy ix
4 sin
2
0 cos sin
0 2
0
0
ω π µ ω ω π µ
ω ω ω π µ
= =
= =
And the mean value of the total magnetic field is
j
R B a N i B
Brt r r
4
0 0
ω π µ + =
The needle orients along the mean field, therefore
R a N
4 tan
2 π ω
(2)36th International Physics Olympiad Salamanca (España) 2005
Solution Th Page of R.S.E.F
Finally, the resistance of the coil measured by this procedure, in terms of θ, is
µ πθω
tan
2 0N a
R=
3 The force on a unit positive charge in a disk is radial and its modulus is vr×Br =vB=ωrB
where B is the magnetic field at the center of the coil
a I N B
2
0
µ =
Then, the electromotive force (e.m.f.) induced on each disk by the magnetic field B is
∫ =
=
= D b
D B rdr B b
0
2
' 2
1 ω ω
ε ε
Finally, the induced e.m.f between and is ε = εD+ εD'
a I b N
2
2
0 ω
µ ε=
4 When the reading of G vanishes, IG =0 and Kirchoff laws give an immediate answer Then we have
ε =IR ⇒
a b N R
2
2
0 ω
µ =
5 The force per unit lengthfbetween two indefinite parallel straight wires separated by a distance h is
h I I
f
2π µ =
for I1 =I2 =I and length 2πa, the force Finduced on C2 by the neighbor coils C1 is
I2
h a F =µ
6 In equilibrium
mgx=4Fd Then
I2
h d a x
g
m = µ (1)
so that
2
4
/
d a
x h g m
I ⎟⎟
⎠ ⎞ ⎜⎜
⎝ ⎛ =
(3)36th International Physics Olympiad Salamanca (España) 2005
Solution Th Page of 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 δ
δ µ
δϕ
δϕ cos 1 cos
sin d
z h z h I a x
g m l
Mg ⎟⎟
⎠ ⎞ ⎜⎜
⎝ ⎛
+ + − >
+
Therefore, using the suggested approximation
δϕ δ
µ δϕ
δϕ cos cos
sin
2 2
0
⎟ ⎟ ⎠ ⎞ ⎜
⎜ ⎝ ⎛
+ >
+
h z h
I d a x
g m l
Mg
Taking into account the equilibrium condition (1), one obtains δϕ
δ
δϕ cos
h z x g m sin l g M
2
>
Finally, for
d z
δ ϕ δ ϕ
δ ≈sin =
tan
d x m
h l M z
2
<
δ ⇒
d x m
h l M z
2 max =
δ
O
l x
δϕ
δϕ δz
h + δz h - δz
h + δz h - δz Mg
mg
d
(4)36th International Physics Olympiad Salamanca (España) 2005
Solution Th Page of R.S.E.F
Th ANSWER SHEET
Question Basic formulas and ideas used Analytical results Marking
guideline
1
R P
dt d
S B N
2
ε
ε Φ
Φ
= − =
⋅
= r r
( )
R B a N P
t B
a N
2 sin
2 2
ω π
ω ω π ε
=
= 0.5
1.0
2
x y i
i
B B I a
N B
B B B
= =
+ =
θ µ
tan
0
r r r
θ ω π µ
tan
2
0N a
R= 2.0
3
r v
B v E
ω =
× =r r r
a I N B
2
µ =
∫
= bEdr
0
r r
ε
a I b N
2
2 ω
µ
ε= 2.0
4 ε=RI
a b N R
2
2 ω
µ
= 0,5
5
h I I
f = ′
π µ
2
0
I h
a
F =µ 1.0
6 mgx=4Fd
2
4
/
d a
x h g m
I ⎟⎟
⎠ ⎞ ⎜⎜
⎝ ⎛ =
µ 1.0
7 Γgrav >Γmag
d x m
h l M z
2 max =
δ 2.0