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1 | 2490-2493 | 1 Charges +Q and –Q put at the ends
of a metallic cylinder The electrons will drift
because of the electric field created to
neutralise the charges The current thus
will stop after a while unless the charges +Q
and –Q are continuously replenished FIGURE 3 |
1 | 2491-2494 | The electrons will drift
because of the electric field created to
neutralise the charges The current thus
will stop after a while unless the charges +Q
and –Q are continuously replenished FIGURE 3 2
Illustrating the
relation R = rl/A for
a rectangular slab
of length l and area
of cross-section A |
1 | 2492-2495 | The current thus
will stop after a while unless the charges +Q
and –Q are continuously replenished FIGURE 3 2
Illustrating the
relation R = rl/A for
a rectangular slab
of length l and area
of cross-section A Rationalised 2023-24
Physics
84
identical to the first and the same current I flows through
both |
1 | 2493-2496 | FIGURE 3 2
Illustrating the
relation R = rl/A for
a rectangular slab
of length l and area
of cross-section A Rationalised 2023-24
Physics
84
identical to the first and the same current I flows through
both The potential difference across the ends of the
combination is clearly sum of the potential difference
across the two individual slabs and hence equals 2V |
1 | 2494-2497 | 2
Illustrating the
relation R = rl/A for
a rectangular slab
of length l and area
of cross-section A Rationalised 2023-24
Physics
84
identical to the first and the same current I flows through
both The potential difference across the ends of the
combination is clearly sum of the potential difference
across the two individual slabs and hence equals 2V The
current through the combination is I and the resistance
of the combination RC is [from Eq |
1 | 2495-2498 | Rationalised 2023-24
Physics
84
identical to the first and the same current I flows through
both The potential difference across the ends of the
combination is clearly sum of the potential difference
across the two individual slabs and hence equals 2V The
current through the combination is I and the resistance
of the combination RC is [from Eq (3 |
1 | 2496-2499 | The potential difference across the ends of the
combination is clearly sum of the potential difference
across the two individual slabs and hence equals 2V The
current through the combination is I and the resistance
of the combination RC is [from Eq (3 3)],
2
2
C
V
R
R
=I
=
(3 |
1 | 2497-2500 | The
current through the combination is I and the resistance
of the combination RC is [from Eq (3 3)],
2
2
C
V
R
R
=I
=
(3 4)
since V/I = R, the resistance of either of the slabs |
1 | 2498-2501 | (3 3)],
2
2
C
V
R
R
=I
=
(3 4)
since V/I = R, the resistance of either of the slabs Thus,
doubling the length of a conductor doubles the
resistance |
1 | 2499-2502 | 3)],
2
2
C
V
R
R
=I
=
(3 4)
since V/I = R, the resistance of either of the slabs Thus,
doubling the length of a conductor doubles the
resistance In general, then resistance is proportional to
length,
R
∝l
(3 |
1 | 2500-2503 | 4)
since V/I = R, the resistance of either of the slabs Thus,
doubling the length of a conductor doubles the
resistance In general, then resistance is proportional to
length,
R
∝l
(3 5)
Next, imagine dividing the slab into two by cutting it
lengthwise so that the slab can be considered as a
combination of two identical slabs of length l, but each
having a cross sectional area of A/2 [Fig |
1 | 2501-2504 | Thus,
doubling the length of a conductor doubles the
resistance In general, then resistance is proportional to
length,
R
∝l
(3 5)
Next, imagine dividing the slab into two by cutting it
lengthwise so that the slab can be considered as a
combination of two identical slabs of length l, but each
having a cross sectional area of A/2 [Fig 3 |
1 | 2502-2505 | In general, then resistance is proportional to
length,
R
∝l
(3 5)
Next, imagine dividing the slab into two by cutting it
lengthwise so that the slab can be considered as a
combination of two identical slabs of length l, but each
having a cross sectional area of A/2 [Fig 3 2(c)] |
1 | 2503-2506 | 5)
Next, imagine dividing the slab into two by cutting it
lengthwise so that the slab can be considered as a
combination of two identical slabs of length l, but each
having a cross sectional area of A/2 [Fig 3 2(c)] For a given voltage V across the slab, if I is the current
through the entire slab, then clearly the current flowing
through each of the two half-slabs is I/2 |
1 | 2504-2507 | 3 2(c)] For a given voltage V across the slab, if I is the current
through the entire slab, then clearly the current flowing
through each of the two half-slabs is I/2 Since the
potential difference across the ends of the half-slabs is V,
i |
1 | 2505-2508 | 2(c)] For a given voltage V across the slab, if I is the current
through the entire slab, then clearly the current flowing
through each of the two half-slabs is I/2 Since the
potential difference across the ends of the half-slabs is V,
i e |
1 | 2506-2509 | For a given voltage V across the slab, if I is the current
through the entire slab, then clearly the current flowing
through each of the two half-slabs is I/2 Since the
potential difference across the ends of the half-slabs is V,
i e , the same as across the full slab, the resistance of each
of the half-slabs R1 is
1
2
2 |
1 | 2507-2510 | Since the
potential difference across the ends of the half-slabs is V,
i e , the same as across the full slab, the resistance of each
of the half-slabs R1 is
1
2
2 ( /2)
V
V
R
R
I
I
=
=
=
(3 |
1 | 2508-2511 | e , the same as across the full slab, the resistance of each
of the half-slabs R1 is
1
2
2 ( /2)
V
V
R
R
I
I
=
=
=
(3 6)
Thus, halving the area of the cross-section of a conductor doubles
the resistance |
1 | 2509-2512 | , the same as across the full slab, the resistance of each
of the half-slabs R1 is
1
2
2 ( /2)
V
V
R
R
I
I
=
=
=
(3 6)
Thus, halving the area of the cross-section of a conductor doubles
the resistance In general, then the resistance R is inversely proportional
to the cross-sectional area,
1
R
∝A
(3 |
1 | 2510-2513 | ( /2)
V
V
R
R
I
I
=
=
=
(3 6)
Thus, halving the area of the cross-section of a conductor doubles
the resistance In general, then the resistance R is inversely proportional
to the cross-sectional area,
1
R
∝A
(3 7)
Combining Eqs |
1 | 2511-2514 | 6)
Thus, halving the area of the cross-section of a conductor doubles
the resistance In general, then the resistance R is inversely proportional
to the cross-sectional area,
1
R
∝A
(3 7)
Combining Eqs (3 |
1 | 2512-2515 | In general, then the resistance R is inversely proportional
to the cross-sectional area,
1
R
∝A
(3 7)
Combining Eqs (3 5) and (3 |
1 | 2513-2516 | 7)
Combining Eqs (3 5) and (3 7), we have
l
R
∝A
(3 |
1 | 2514-2517 | (3 5) and (3 7), we have
l
R
∝A
(3 8)
and hence for a given conductor
l
R
=ρA
(3 |
1 | 2515-2518 | 5) and (3 7), we have
l
R
∝A
(3 8)
and hence for a given conductor
l
R
=ρA
(3 9)
where the constant of proportionality r depends on the material of the
conductor but not on its dimensions |
1 | 2516-2519 | 7), we have
l
R
∝A
(3 8)
and hence for a given conductor
l
R
=ρA
(3 9)
where the constant of proportionality r depends on the material of the
conductor but not on its dimensions r is called resistivity |
1 | 2517-2520 | 8)
and hence for a given conductor
l
R
=ρA
(3 9)
where the constant of proportionality r depends on the material of the
conductor but not on its dimensions r is called resistivity Using the last equation, Ohm’s law reads
I l
V
I
R
ρA
=
×
=
(3 |
1 | 2518-2521 | 9)
where the constant of proportionality r depends on the material of the
conductor but not on its dimensions r is called resistivity Using the last equation, Ohm’s law reads
I l
V
I
R
ρA
=
×
=
(3 10)
Current per unit area (taken normal to the current), I/A, is called
current density and is denoted by j |
1 | 2519-2522 | r is called resistivity Using the last equation, Ohm’s law reads
I l
V
I
R
ρA
=
×
=
(3 10)
Current per unit area (taken normal to the current), I/A, is called
current density and is denoted by j The SI units of the current density
are A/m2 |
1 | 2520-2523 | Using the last equation, Ohm’s law reads
I l
V
I
R
ρA
=
×
=
(3 10)
Current per unit area (taken normal to the current), I/A, is called
current density and is denoted by j The SI units of the current density
are A/m2 Further, if E is the magnitude of uniform electric field in the
conductor whose length is l, then the potential difference V across its
ends is El |
1 | 2521-2524 | 10)
Current per unit area (taken normal to the current), I/A, is called
current density and is denoted by j The SI units of the current density
are A/m2 Further, if E is the magnitude of uniform electric field in the
conductor whose length is l, then the potential difference V across its
ends is El Using these, the last equation reads
GEORG SIMON OHM (1787–1854)
Georg Simon Ohm (1787–
1854) German physicist,
professor at Munich |
1 | 2522-2525 | The SI units of the current density
are A/m2 Further, if E is the magnitude of uniform electric field in the
conductor whose length is l, then the potential difference V across its
ends is El Using these, the last equation reads
GEORG SIMON OHM (1787–1854)
Georg Simon Ohm (1787–
1854) German physicist,
professor at Munich Ohm
was led to his law by an
analogy
between
the
conduction of heat: the
electric field is analogous to
the temperature gradient,
and the electric current is
analogous to the heat flow |
1 | 2523-2526 | Further, if E is the magnitude of uniform electric field in the
conductor whose length is l, then the potential difference V across its
ends is El Using these, the last equation reads
GEORG SIMON OHM (1787–1854)
Georg Simon Ohm (1787–
1854) German physicist,
professor at Munich Ohm
was led to his law by an
analogy
between
the
conduction of heat: the
electric field is analogous to
the temperature gradient,
and the electric current is
analogous to the heat flow Rationalised 2023-24
Current
Electricity
85
E l = j r l
or, E = j r
(3 |
1 | 2524-2527 | Using these, the last equation reads
GEORG SIMON OHM (1787–1854)
Georg Simon Ohm (1787–
1854) German physicist,
professor at Munich Ohm
was led to his law by an
analogy
between
the
conduction of heat: the
electric field is analogous to
the temperature gradient,
and the electric current is
analogous to the heat flow Rationalised 2023-24
Current
Electricity
85
E l = j r l
or, E = j r
(3 11)
The above relation for magnitudes E and j can indeed be cast in a
vector form |
1 | 2525-2528 | Ohm
was led to his law by an
analogy
between
the
conduction of heat: the
electric field is analogous to
the temperature gradient,
and the electric current is
analogous to the heat flow Rationalised 2023-24
Current
Electricity
85
E l = j r l
or, E = j r
(3 11)
The above relation for magnitudes E and j can indeed be cast in a
vector form The current density, (which we have defined as the current
through unit area normal to the current) is also directed along E, and is
also a vector j (ººººº j E/E) |
1 | 2526-2529 | Rationalised 2023-24
Current
Electricity
85
E l = j r l
or, E = j r
(3 11)
The above relation for magnitudes E and j can indeed be cast in a
vector form The current density, (which we have defined as the current
through unit area normal to the current) is also directed along E, and is
also a vector j (ººººº j E/E) Thus, the last equation can be written as,
E = jr
(3 |
1 | 2527-2530 | 11)
The above relation for magnitudes E and j can indeed be cast in a
vector form The current density, (which we have defined as the current
through unit area normal to the current) is also directed along E, and is
also a vector j (ººººº j E/E) Thus, the last equation can be written as,
E = jr
(3 12)
or, j = s E
(3 |
1 | 2528-2531 | The current density, (which we have defined as the current
through unit area normal to the current) is also directed along E, and is
also a vector j (ººººº j E/E) Thus, the last equation can be written as,
E = jr
(3 12)
or, j = s E
(3 13)
where s º1/r is called the conductivity |
1 | 2529-2532 | Thus, the last equation can be written as,
E = jr
(3 12)
or, j = s E
(3 13)
where s º1/r is called the conductivity Ohm’s law is often stated in an
equivalent form, Eq |
1 | 2530-2533 | 12)
or, j = s E
(3 13)
where s º1/r is called the conductivity Ohm’s law is often stated in an
equivalent form, Eq (3 |
1 | 2531-2534 | 13)
where s º1/r is called the conductivity Ohm’s law is often stated in an
equivalent form, Eq (3 13) in addition to Eq |
1 | 2532-2535 | Ohm’s law is often stated in an
equivalent form, Eq (3 13) in addition to Eq (3 |
1 | 2533-2536 | (3 13) in addition to Eq (3 3) |
1 | 2534-2537 | 13) in addition to Eq (3 3) In the next section, we
will try to understand the origin of the Ohm’s law as arising from the
characteristics of the drift of electrons |
1 | 2535-2538 | (3 3) In the next section, we
will try to understand the origin of the Ohm’s law as arising from the
characteristics of the drift of electrons 3 |
1 | 2536-2539 | 3) In the next section, we
will try to understand the origin of the Ohm’s law as arising from the
characteristics of the drift of electrons 3 5 DRIFT OF ELECTRONS AND THE ORIGIN
OF RESISTIVITY
As remarked before, an electron will suffer collisions with the heavy fixed
ions, but after collision, it will emerge with the same speed but in random
directions |
1 | 2537-2540 | In the next section, we
will try to understand the origin of the Ohm’s law as arising from the
characteristics of the drift of electrons 3 5 DRIFT OF ELECTRONS AND THE ORIGIN
OF RESISTIVITY
As remarked before, an electron will suffer collisions with the heavy fixed
ions, but after collision, it will emerge with the same speed but in random
directions If we consider all the electrons, their average velocity will be
zero since their directions are random |
1 | 2538-2541 | 3 5 DRIFT OF ELECTRONS AND THE ORIGIN
OF RESISTIVITY
As remarked before, an electron will suffer collisions with the heavy fixed
ions, but after collision, it will emerge with the same speed but in random
directions If we consider all the electrons, their average velocity will be
zero since their directions are random Thus, if there are N electrons and
the velocity of the ith electron (i = 1, 2, 3, |
1 | 2539-2542 | 5 DRIFT OF ELECTRONS AND THE ORIGIN
OF RESISTIVITY
As remarked before, an electron will suffer collisions with the heavy fixed
ions, but after collision, it will emerge with the same speed but in random
directions If we consider all the electrons, their average velocity will be
zero since their directions are random Thus, if there are N electrons and
the velocity of the ith electron (i = 1, 2, 3, N ) at a given time is vi, then
1
0
1
N
i
i
v =
=∑
N
(3 |
1 | 2540-2543 | If we consider all the electrons, their average velocity will be
zero since their directions are random Thus, if there are N electrons and
the velocity of the ith electron (i = 1, 2, 3, N ) at a given time is vi, then
1
0
1
N
i
i
v =
=∑
N
(3 14)
Consider now the situation when an electric field is
present |
1 | 2541-2544 | Thus, if there are N electrons and
the velocity of the ith electron (i = 1, 2, 3, N ) at a given time is vi, then
1
0
1
N
i
i
v =
=∑
N
(3 14)
Consider now the situation when an electric field is
present Electrons will be accelerated due to this
field by
a= – E
e
m
(3 |
1 | 2542-2545 | N ) at a given time is vi, then
1
0
1
N
i
i
v =
=∑
N
(3 14)
Consider now the situation when an electric field is
present Electrons will be accelerated due to this
field by
a= – E
e
m
(3 15)
where –e is the charge and m is the mass of an electron |
1 | 2543-2546 | 14)
Consider now the situation when an electric field is
present Electrons will be accelerated due to this
field by
a= – E
e
m
(3 15)
where –e is the charge and m is the mass of an electron Consider again the ith electron at a given time t |
1 | 2544-2547 | Electrons will be accelerated due to this
field by
a= – E
e
m
(3 15)
where –e is the charge and m is the mass of an electron Consider again the ith electron at a given time t This
electron would have had its last collision some time
before t, and let ti be the time elapsed after its last
collision |
1 | 2545-2548 | 15)
where –e is the charge and m is the mass of an electron Consider again the ith electron at a given time t This
electron would have had its last collision some time
before t, and let ti be the time elapsed after its last
collision If vi was its velocity immediately after the last
collision, then its velocity Vi at time t is
−
=
+
E
V
v
i
i
i
e
t
m
(3 |
1 | 2546-2549 | Consider again the ith electron at a given time t This
electron would have had its last collision some time
before t, and let ti be the time elapsed after its last
collision If vi was its velocity immediately after the last
collision, then its velocity Vi at time t is
−
=
+
E
V
v
i
i
i
e
t
m
(3 16)
since starting with its last collision it was accelerated
(Fig |
1 | 2547-2550 | This
electron would have had its last collision some time
before t, and let ti be the time elapsed after its last
collision If vi was its velocity immediately after the last
collision, then its velocity Vi at time t is
−
=
+
E
V
v
i
i
i
e
t
m
(3 16)
since starting with its last collision it was accelerated
(Fig 3 |
1 | 2548-2551 | If vi was its velocity immediately after the last
collision, then its velocity Vi at time t is
−
=
+
E
V
v
i
i
i
e
t
m
(3 16)
since starting with its last collision it was accelerated
(Fig 3 3) with an acceleration given by Eq |
1 | 2549-2552 | 16)
since starting with its last collision it was accelerated
(Fig 3 3) with an acceleration given by Eq (3 |
1 | 2550-2553 | 3 3) with an acceleration given by Eq (3 15) for a
time interval ti |
1 | 2551-2554 | 3) with an acceleration given by Eq (3 15) for a
time interval ti The average velocity of the electrons at
time t is the average of all the Vi’s |
1 | 2552-2555 | (3 15) for a
time interval ti The average velocity of the electrons at
time t is the average of all the Vi’s The average of vi’s is
zero [Eq |
1 | 2553-2556 | 15) for a
time interval ti The average velocity of the electrons at
time t is the average of all the Vi’s The average of vi’s is
zero [Eq (3 |
1 | 2554-2557 | The average velocity of the electrons at
time t is the average of all the Vi’s The average of vi’s is
zero [Eq (3 14)] since immediately after any collision,
the direction of the velocity of an electron is completely
random |
1 | 2555-2558 | The average of vi’s is
zero [Eq (3 14)] since immediately after any collision,
the direction of the velocity of an electron is completely
random The collisions of the electrons do not occur at
regular intervals but at random times |
1 | 2556-2559 | (3 14)] since immediately after any collision,
the direction of the velocity of an electron is completely
random The collisions of the electrons do not occur at
regular intervals but at random times Let us denote by
t, the average time between successive collisions |
1 | 2557-2560 | 14)] since immediately after any collision,
the direction of the velocity of an electron is completely
random The collisions of the electrons do not occur at
regular intervals but at random times Let us denote by
t, the average time between successive collisions Then
at a given time, some of the electrons would have spent
FIGURE 3 |
1 | 2558-2561 | The collisions of the electrons do not occur at
regular intervals but at random times Let us denote by
t, the average time between successive collisions Then
at a given time, some of the electrons would have spent
FIGURE 3 3 A schematic picture of
an electron moving from a point A to
another point B through repeated
collisions, and straight line travel
between collisions (full lines) |
1 | 2559-2562 | Let us denote by
t, the average time between successive collisions Then
at a given time, some of the electrons would have spent
FIGURE 3 3 A schematic picture of
an electron moving from a point A to
another point B through repeated
collisions, and straight line travel
between collisions (full lines) If an
electric field is applied as shown, the
electron ends up at point B¢ (dotted
lines) |
1 | 2560-2563 | Then
at a given time, some of the electrons would have spent
FIGURE 3 3 A schematic picture of
an electron moving from a point A to
another point B through repeated
collisions, and straight line travel
between collisions (full lines) If an
electric field is applied as shown, the
electron ends up at point B¢ (dotted
lines) A slight drift in a direction
opposite the electric field is visible |
1 | 2561-2564 | 3 A schematic picture of
an electron moving from a point A to
another point B through repeated
collisions, and straight line travel
between collisions (full lines) If an
electric field is applied as shown, the
electron ends up at point B¢ (dotted
lines) A slight drift in a direction
opposite the electric field is visible Rationalised 2023-24
Physics
86
time more than t and some less than t |
1 | 2562-2565 | If an
electric field is applied as shown, the
electron ends up at point B¢ (dotted
lines) A slight drift in a direction
opposite the electric field is visible Rationalised 2023-24
Physics
86
time more than t and some less than t In other words, the time ti in
Eq |
1 | 2563-2566 | A slight drift in a direction
opposite the electric field is visible Rationalised 2023-24
Physics
86
time more than t and some less than t In other words, the time ti in
Eq (3 |
1 | 2564-2567 | Rationalised 2023-24
Physics
86
time more than t and some less than t In other words, the time ti in
Eq (3 16) will be less than t for some and more than t for others as we go
through the values of i = 1, 2 |
1 | 2565-2568 | In other words, the time ti in
Eq (3 16) will be less than t for some and more than t for others as we go
through the values of i = 1, 2 N |
1 | 2566-2569 | (3 16) will be less than t for some and more than t for others as we go
through the values of i = 1, 2 N The average value of ti then is t
(known as relaxation time) |
1 | 2567-2570 | 16) will be less than t for some and more than t for others as we go
through the values of i = 1, 2 N The average value of ti then is t
(known as relaxation time) Thus, averaging Eq |
1 | 2568-2571 | N The average value of ti then is t
(known as relaxation time) Thus, averaging Eq (3 |
1 | 2569-2572 | The average value of ti then is t
(known as relaxation time) Thus, averaging Eq (3 16) over the
N-electrons at any given time t gives us for the average velocity vd
(
)
(
)
( )
≡
=
−
E
v
V
v
d
i
i
i
average
average
average
e
t
m
0 –
τ
τ
=
= −
E
E
e
e
m
m
(3 |
1 | 2570-2573 | Thus, averaging Eq (3 16) over the
N-electrons at any given time t gives us for the average velocity vd
(
)
(
)
( )
≡
=
−
E
v
V
v
d
i
i
i
average
average
average
e
t
m
0 –
τ
τ
=
= −
E
E
e
e
m
m
(3 17)
This last result is surprising |
1 | 2571-2574 | (3 16) over the
N-electrons at any given time t gives us for the average velocity vd
(
)
(
)
( )
≡
=
−
E
v
V
v
d
i
i
i
average
average
average
e
t
m
0 –
τ
τ
=
= −
E
E
e
e
m
m
(3 17)
This last result is surprising It tells us that the
electrons move with an average velocity which is
independent of time, although electrons are
accelerated |
1 | 2572-2575 | 16) over the
N-electrons at any given time t gives us for the average velocity vd
(
)
(
)
( )
≡
=
−
E
v
V
v
d
i
i
i
average
average
average
e
t
m
0 –
τ
τ
=
= −
E
E
e
e
m
m
(3 17)
This last result is surprising It tells us that the
electrons move with an average velocity which is
independent of time, although electrons are
accelerated This is the phenomenon of drift and the
velocity vd in Eq |
1 | 2573-2576 | 17)
This last result is surprising It tells us that the
electrons move with an average velocity which is
independent of time, although electrons are
accelerated This is the phenomenon of drift and the
velocity vd in Eq (3 |
1 | 2574-2577 | It tells us that the
electrons move with an average velocity which is
independent of time, although electrons are
accelerated This is the phenomenon of drift and the
velocity vd in Eq (3 17) is called the drift velocity |
1 | 2575-2578 | This is the phenomenon of drift and the
velocity vd in Eq (3 17) is called the drift velocity Because of the drift, there will be net transport of
charges across any area perpendicular to E |
1 | 2576-2579 | (3 17) is called the drift velocity Because of the drift, there will be net transport of
charges across any area perpendicular to E Consider
a planar area A, located inside the conductor such that
the normal to the area is parallel to E (Fig |
1 | 2577-2580 | 17) is called the drift velocity Because of the drift, there will be net transport of
charges across any area perpendicular to E Consider
a planar area A, located inside the conductor such that
the normal to the area is parallel to E (Fig 3 |
1 | 2578-2581 | Because of the drift, there will be net transport of
charges across any area perpendicular to E Consider
a planar area A, located inside the conductor such that
the normal to the area is parallel to E (Fig 3 4) |
1 | 2579-2582 | Consider
a planar area A, located inside the conductor such that
the normal to the area is parallel to E (Fig 3 4) Then
because of the drift, in an infinitesimal amount of time
Dt, all electrons to the left of the area at distances upto
|vd|Dt would have crossed the area |
1 | 2580-2583 | 3 4) Then
because of the drift, in an infinitesimal amount of time
Dt, all electrons to the left of the area at distances upto
|vd|Dt would have crossed the area If n is the number
of free electrons per unit volume in the metal, then
there are n Dt |vd|A such electrons |
1 | 2581-2584 | 4) Then
because of the drift, in an infinitesimal amount of time
Dt, all electrons to the left of the area at distances upto
|vd|Dt would have crossed the area If n is the number
of free electrons per unit volume in the metal, then
there are n Dt |vd|A such electrons Since each
electron carries a charge –e, the total charge transported across this area
A to the right in time Dt is –ne A|vd|Dt |
1 | 2582-2585 | Then
because of the drift, in an infinitesimal amount of time
Dt, all electrons to the left of the area at distances upto
|vd|Dt would have crossed the area If n is the number
of free electrons per unit volume in the metal, then
there are n Dt |vd|A such electrons Since each
electron carries a charge –e, the total charge transported across this area
A to the right in time Dt is –ne A|vd|Dt E is directed towards the left and
hence the total charge transported along E across the area is negative of
this |
1 | 2583-2586 | If n is the number
of free electrons per unit volume in the metal, then
there are n Dt |vd|A such electrons Since each
electron carries a charge –e, the total charge transported across this area
A to the right in time Dt is –ne A|vd|Dt E is directed towards the left and
hence the total charge transported along E across the area is negative of
this The amount of charge crossing the area A in time Dt is by definition
[Eq |
1 | 2584-2587 | Since each
electron carries a charge –e, the total charge transported across this area
A to the right in time Dt is –ne A|vd|Dt E is directed towards the left and
hence the total charge transported along E across the area is negative of
this The amount of charge crossing the area A in time Dt is by definition
[Eq (3 |
1 | 2585-2588 | E is directed towards the left and
hence the total charge transported along E across the area is negative of
this The amount of charge crossing the area A in time Dt is by definition
[Eq (3 2)] I Dt, where I is the magnitude of the current |
1 | 2586-2589 | The amount of charge crossing the area A in time Dt is by definition
[Eq (3 2)] I Dt, where I is the magnitude of the current Hence,
v
∆ = +
∆
d
I
t
n e A
t
(3 |
1 | 2587-2590 | (3 2)] I Dt, where I is the magnitude of the current Hence,
v
∆ = +
∆
d
I
t
n e A
t
(3 18)
Substituting the value of |vd| from Eq |
1 | 2588-2591 | 2)] I Dt, where I is the magnitude of the current Hence,
v
∆ = +
∆
d
I
t
n e A
t
(3 18)
Substituting the value of |vd| from Eq (3 |
1 | 2589-2592 | Hence,
v
∆ = +
∆
d
I
t
n e A
t
(3 18)
Substituting the value of |vd| from Eq (3 17)
2
E
τ
∆ =
∆
e A
I t
n
t
m
(3 |
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