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