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]
Trang 1Th 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 Br B ir
0
0 = Then, the magnetic flux through the coil is
S B
⋅
φ
where the vector surface Sr
is given by Sr a ( ir t rj)
ω ω
π 2 cos +sin
=
Therefore φ=Nπa2B0cosωt
The induced electromotive force is
dt
dφ
The instantaneous power is P=ε2 /R , therefore
R
B a N P
2
2 0
π
where we used
2
1 sin
1 sin
0 2
dt t T
ω
2 The total field at the center the coil at the instant t is
i
+
= 0
where Bri
is the magnetic field due to the induced current Bri B i( ir trj)
ω
=
with
a I N
B i
2
0 µ
= and I = ε / R
R B a N
2
0 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 2
0
2 0 2
0
2 0
0
2 0
ω π µ ω ω π µ
ω ω ω π µ
=
=
=
=
And the mean value of the total magnetic field is
R B a N i B
4
0 2 0 0
ω π µ
+
=
The needle orients along the mean field, therefore
R a N
4 tan
2
µ
θ =
Trang 2Finally, the resistance of the coil measured by this procedure, in terms of θ , is
ω π µ
tan 4
2
0N a
3 The force on a unit positive charge in a disk is radial and its modulus is
vr×Br =v B=ωr B
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
=
0
2
ε ε
Finally, the induced e.m.f between 1 and 4 is ε = εD+ εD'
a
I b N
2
2
0 ω µ
4 When the reading of G vanishes, I G =0 and Kirchoff laws give an immediate answer Then we have
a
b N R
2
2
µ
=
5 The force per unit lengthf between two indefinite parallel straight wires separated by a distance h is
h
I I
2π
µ
=
for I1 =I2 =I and length 2πa , the force Finduced on C2 by the neighbor coils C1 is
h
a
F µ
=
6 In equilibrium
m g x=4F d
Then
h
d x
g
so that
2 1
0
4
/
d
x h m
⎠
⎞
⎜⎜
⎝
⎛
= µ
Trang 37 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
z h z h I a x
g m l
Mg + > ⎜⎜⎛ − + + ⎟⎟⎞
Therefore, using the suggested approximation
δϕ δ
µ δϕ
sin
2
2 2
0
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ +
>
+
h
z h
I d x
g m l
Mg
Taking into account the equilibrium condition (1), one obtains
δϕ δ
h
z x g m sin l g M
2
2
Finally, for
d z
δ ϕ δ ϕ
tan
d m
h l M z
2
<
d 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
G
Trang 4Th 2 ANSWER SHEET
Question Basic formulas and ideas used Analytical results Marking
guideline
1
R P dt d
S B N
2
0
ε
Φ
=
−
=
⋅
R
B a N P
t B
a N
2
sin 2 0 2 0 2
ω π
ω ω π
ε
=
1.0
2
x y i
i
B B
I a
N B
B B B
=
=
+
=
θ
µ
tan
2
0 0
r r r
θ
ω π µ
tan 4
2
0N a
3
r v
B v E
ω
=
×
=r r r
a
I N B
2 0
µ
=
∫
= b E d r
0
r r
ε
a I b N
2
2
µ
4 ε=R I
a
b N R
2
2
µ
5
h
I
=
π
µ
2
I h
a
2 1
0
4
/
d
x g m
d m h l M z
2 max =