State and derive ohms law in form of mass, number density and relaxation time
Answers
Answer:
Explanation:
Drift Velocity: The average velocity with which the free electron drift under the influence of an electric field.
V
d
=−
m
e
E
τ
Or, V∝I
(ii) As
V
d
=−
m
e
E
τ....(i)
Current flowing through the conductor ---
I=nAV
d
e....(ii)
from equation (i) and (ii).
I=nA(
m
eE
τ)e
⟹I=
m
nAe
2
Eτ
....(iii)
If V is potential difference applied across the two ends of the conductor, then
E=
l
V
Putting this E in equation (iii),
I=
ml
nAe
2
Vτ
⟹
I
V
=
ne
2
τ
m
A
l
According to ohm's law,
I
V
=R (resistance of the conductor)
R=
ne
2
τ
m
A
l
....(iv)
But, R=ρ
A
l
.....(v)
Comparing (iv) and (v)
ρ=
ne
2
τ
m
Resistivity of a conductor depends on the following factors:
(1) It is inversely proportional to the number of free electrons per unit volume (n) of the conductor.
(2) It is inversely proportional to the average relaxation time (τ) of the free electrons in the conductor.
(iii) Being an alloy, its resistivity is very high and temperature coefficient of resistance is very low.
(i) Principle of potentiometer: The potential difference across any two points of current carrying wire, having uniform cross-sectional area and material, of the potentiometer is directly proportional to the length between the two points.
That is, V∝l
Proof:
V=IR=I(
A
ρl
)
i.e., V=(
A
Iρ
)l
For uniform current and cross-sectional area, we have
A
ρl
=constant= potential drop per unit length
i.e., V∝l
(ii) Given: potentiometer circuit AB is a uniform wire of length 1m and resistance 10Ω
To find the potential gradient along the wire and balance length AO(=1).
Solution: According to the given criteria,
E=2V,R=15Ω,R
AB
=10Ω
Potential difference across the wire, =
25
2
×10=0.8V/m.
Therefore, potential gradient =
1
0.8
=0.8V/m
Potential difference across AO =
1.5
1.5
×0.3=0.3V
Therefore, Length, AO =
0.8
0.3
×100=37.5cm
which is the reqired balance length of the wire
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