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1 | 3390-3393 | 4), we find that there is an upward force F, of
magnitude IlB, For mid-air suspension, this must be balanced by
the force due to gravity:
m g = I lB
m g
B
I l
=
0 2
9 8
0 |
1 | 3391-3394 | For mid-air suspension, this must be balanced by
the force due to gravity:
m g = I lB
m g
B
I l
=
0 2
9 8
0 65 T
2
×1 |
1 | 3392-3395 | 2
9 8
0 65 T
2
×1 5
=
=
×
Note that it would have been sufficient to specify m/l, the mass per
unit length of the wire |
1 | 3393-3396 | 8
0 65 T
2
×1 5
=
=
×
Note that it would have been sufficient to specify m/l, the mass per
unit length of the wire The earth’s magnetic field is approximately
4 × 10–5 T and we have ignored it |
1 | 3394-3397 | 65 T
2
×1 5
=
=
×
Note that it would have been sufficient to specify m/l, the mass per
unit length of the wire The earth’s magnetic field is approximately
4 × 10–5 T and we have ignored it Example 4 |
1 | 3395-3398 | 5
=
=
×
Note that it would have been sufficient to specify m/l, the mass per
unit length of the wire The earth’s magnetic field is approximately
4 × 10–5 T and we have ignored it Example 4 2 If the magnetic field is parallel to the positive y-axis
and the charged particle is moving along the positive x-axis (Fig |
1 | 3396-3399 | The earth’s magnetic field is approximately
4 × 10–5 T and we have ignored it Example 4 2 If the magnetic field is parallel to the positive y-axis
and the charged particle is moving along the positive x-axis (Fig 4 |
1 | 3397-3400 | Example 4 2 If the magnetic field is parallel to the positive y-axis
and the charged particle is moving along the positive x-axis (Fig 4 4),
which way would the Lorentz force be for (a) an electron (negative
charge), (b) a proton (positive charge) |
1 | 3398-3401 | 2 If the magnetic field is parallel to the positive y-axis
and the charged particle is moving along the positive x-axis (Fig 4 4),
which way would the Lorentz force be for (a) an electron (negative
charge), (b) a proton (positive charge) FIGURE 4 |
1 | 3399-3402 | 4 4),
which way would the Lorentz force be for (a) an electron (negative
charge), (b) a proton (positive charge) FIGURE 4 4
Solution The velocity v of particle is along the x-axis, while B, the
magnetic field is along the y-axis, so v × B is along the z-axis (screw
rule or right-hand thumb rule) |
1 | 3400-3403 | 4),
which way would the Lorentz force be for (a) an electron (negative
charge), (b) a proton (positive charge) FIGURE 4 4
Solution The velocity v of particle is along the x-axis, while B, the
magnetic field is along the y-axis, so v × B is along the z-axis (screw
rule or right-hand thumb rule) So, (a) for electron it will be along –z
axis |
1 | 3401-3404 | FIGURE 4 4
Solution The velocity v of particle is along the x-axis, while B, the
magnetic field is along the y-axis, so v × B is along the z-axis (screw
rule or right-hand thumb rule) So, (a) for electron it will be along –z
axis (b) for a positive charge (proton) the force is along +z axis |
1 | 3402-3405 | 4
Solution The velocity v of particle is along the x-axis, while B, the
magnetic field is along the y-axis, so v × B is along the z-axis (screw
rule or right-hand thumb rule) So, (a) for electron it will be along –z
axis (b) for a positive charge (proton) the force is along +z axis EXAMPLE 4 |
1 | 3403-3406 | So, (a) for electron it will be along –z
axis (b) for a positive charge (proton) the force is along +z axis EXAMPLE 4 2
Charged particles moving in a magnetic field |
1 | 3404-3407 | (b) for a positive charge (proton) the force is along +z axis EXAMPLE 4 2
Charged particles moving in a magnetic field Interactive demonstration:
http://www |
1 | 3405-3408 | EXAMPLE 4 2
Charged particles moving in a magnetic field Interactive demonstration:
http://www phys |
1 | 3406-3409 | 2
Charged particles moving in a magnetic field Interactive demonstration:
http://www phys hawaii |
1 | 3407-3410 | Interactive demonstration:
http://www phys hawaii edu/~teb/optics/java/partmagn/index |
1 | 3408-3411 | phys hawaii edu/~teb/optics/java/partmagn/index html
Rationalised 2023-24
Physics
112
4 |
1 | 3409-3412 | hawaii edu/~teb/optics/java/partmagn/index html
Rationalised 2023-24
Physics
112
4 3 MOTION IN A MAGNETIC FIELD
We will now consider, in greater detail, the motion of a charge moving in
a magnetic field |
1 | 3410-3413 | edu/~teb/optics/java/partmagn/index html
Rationalised 2023-24
Physics
112
4 3 MOTION IN A MAGNETIC FIELD
We will now consider, in greater detail, the motion of a charge moving in
a magnetic field We have learnt in Mechanics (see Class XI book, Chapter
6) that a force on a particle does work if the force has a component along
(or opposed to) the direction of motion of the particle |
1 | 3411-3414 | html
Rationalised 2023-24
Physics
112
4 3 MOTION IN A MAGNETIC FIELD
We will now consider, in greater detail, the motion of a charge moving in
a magnetic field We have learnt in Mechanics (see Class XI book, Chapter
6) that a force on a particle does work if the force has a component along
(or opposed to) the direction of motion of the particle In the case of motion
of a charge in a magnetic field, the magnetic force is perpendicular to the
velocity of the particle |
1 | 3412-3415 | 3 MOTION IN A MAGNETIC FIELD
We will now consider, in greater detail, the motion of a charge moving in
a magnetic field We have learnt in Mechanics (see Class XI book, Chapter
6) that a force on a particle does work if the force has a component along
(or opposed to) the direction of motion of the particle In the case of motion
of a charge in a magnetic field, the magnetic force is perpendicular to the
velocity of the particle So no work is done and no change in the magnitude
of the velocity is produced (though the direction of momentum may be
changed) |
1 | 3413-3416 | We have learnt in Mechanics (see Class XI book, Chapter
6) that a force on a particle does work if the force has a component along
(or opposed to) the direction of motion of the particle In the case of motion
of a charge in a magnetic field, the magnetic force is perpendicular to the
velocity of the particle So no work is done and no change in the magnitude
of the velocity is produced (though the direction of momentum may be
changed) [Notice that this is unlike the force due to an electric field, qE,
which can have a component parallel (or antiparallel) to motion and thus
can transfer energy in addition to momentum |
1 | 3414-3417 | In the case of motion
of a charge in a magnetic field, the magnetic force is perpendicular to the
velocity of the particle So no work is done and no change in the magnitude
of the velocity is produced (though the direction of momentum may be
changed) [Notice that this is unlike the force due to an electric field, qE,
which can have a component parallel (or antiparallel) to motion and thus
can transfer energy in addition to momentum ]
We shall consider motion of a charged particle in a uniform magnetic
field |
1 | 3415-3418 | So no work is done and no change in the magnitude
of the velocity is produced (though the direction of momentum may be
changed) [Notice that this is unlike the force due to an electric field, qE,
which can have a component parallel (or antiparallel) to motion and thus
can transfer energy in addition to momentum ]
We shall consider motion of a charged particle in a uniform magnetic
field First consider the case of v perpendicular to B |
1 | 3416-3419 | [Notice that this is unlike the force due to an electric field, qE,
which can have a component parallel (or antiparallel) to motion and thus
can transfer energy in addition to momentum ]
We shall consider motion of a charged particle in a uniform magnetic
field First consider the case of v perpendicular to B The
perpendicular force, q v × B, acts as a centripetal force and
produces a circular motion perpendicular to the magnetic field |
1 | 3417-3420 | ]
We shall consider motion of a charged particle in a uniform magnetic
field First consider the case of v perpendicular to B The
perpendicular force, q v × B, acts as a centripetal force and
produces a circular motion perpendicular to the magnetic field The particle will describe a circle if v and B are perpendicular
to each other (Fig |
1 | 3418-3421 | First consider the case of v perpendicular to B The
perpendicular force, q v × B, acts as a centripetal force and
produces a circular motion perpendicular to the magnetic field The particle will describe a circle if v and B are perpendicular
to each other (Fig 4 |
1 | 3419-3422 | The
perpendicular force, q v × B, acts as a centripetal force and
produces a circular motion perpendicular to the magnetic field The particle will describe a circle if v and B are perpendicular
to each other (Fig 4 5) |
1 | 3420-3423 | The particle will describe a circle if v and B are perpendicular
to each other (Fig 4 5) If velocity has a component along B, this component
remains unchanged as the motion along the magnetic field will
not be affected by the magnetic field |
1 | 3421-3424 | 4 5) If velocity has a component along B, this component
remains unchanged as the motion along the magnetic field will
not be affected by the magnetic field The motion in a plane
perpendicular to B is as before a circular one, thereby producing
a helical motion (Fig |
1 | 3422-3425 | 5) If velocity has a component along B, this component
remains unchanged as the motion along the magnetic field will
not be affected by the magnetic field The motion in a plane
perpendicular to B is as before a circular one, thereby producing
a helical motion (Fig 4 |
1 | 3423-3426 | If velocity has a component along B, this component
remains unchanged as the motion along the magnetic field will
not be affected by the magnetic field The motion in a plane
perpendicular to B is as before a circular one, thereby producing
a helical motion (Fig 4 6) |
1 | 3424-3427 | The motion in a plane
perpendicular to B is as before a circular one, thereby producing
a helical motion (Fig 4 6) You have already learnt in earlier classes (See Class XI,
Chapter 4) that if r is the radius of the circular path of a particle,
then a force of m v2 / r, acts perpendicular to the path towards
the centre of the circle, and is called the centripetal force |
1 | 3425-3428 | 4 6) You have already learnt in earlier classes (See Class XI,
Chapter 4) that if r is the radius of the circular path of a particle,
then a force of m v2 / r, acts perpendicular to the path towards
the centre of the circle, and is called the centripetal force If the
velocity v is perpendicular to the magnetic field B, the magnetic
force is perpendicular to both v and B and acts
like a centripetal force |
1 | 3426-3429 | 6) You have already learnt in earlier classes (See Class XI,
Chapter 4) that if r is the radius of the circular path of a particle,
then a force of m v2 / r, acts perpendicular to the path towards
the centre of the circle, and is called the centripetal force If the
velocity v is perpendicular to the magnetic field B, the magnetic
force is perpendicular to both v and B and acts
like a centripetal force It has a magnitude q v
B |
1 | 3427-3430 | You have already learnt in earlier classes (See Class XI,
Chapter 4) that if r is the radius of the circular path of a particle,
then a force of m v2 / r, acts perpendicular to the path towards
the centre of the circle, and is called the centripetal force If the
velocity v is perpendicular to the magnetic field B, the magnetic
force is perpendicular to both v and B and acts
like a centripetal force It has a magnitude q v
B Equating the two expressions for centripetal
force,
m v 2/r = q v B, which gives
r = m v / qB
(4 |
1 | 3428-3431 | If the
velocity v is perpendicular to the magnetic field B, the magnetic
force is perpendicular to both v and B and acts
like a centripetal force It has a magnitude q v
B Equating the two expressions for centripetal
force,
m v 2/r = q v B, which gives
r = m v / qB
(4 5)
for the radius of the circle described by the
charged particle |
1 | 3429-3432 | It has a magnitude q v
B Equating the two expressions for centripetal
force,
m v 2/r = q v B, which gives
r = m v / qB
(4 5)
for the radius of the circle described by the
charged particle The larger the momentum, the
larger is the radius and bigger the circle
described |
1 | 3430-3433 | Equating the two expressions for centripetal
force,
m v 2/r = q v B, which gives
r = m v / qB
(4 5)
for the radius of the circle described by the
charged particle The larger the momentum, the
larger is the radius and bigger the circle
described If w is the angular frequency, then v
= w r |
1 | 3431-3434 | 5)
for the radius of the circle described by the
charged particle The larger the momentum, the
larger is the radius and bigger the circle
described If w is the angular frequency, then v
= w r So,
w = 2p n = q B/ m
[4 |
1 | 3432-3435 | The larger the momentum, the
larger is the radius and bigger the circle
described If w is the angular frequency, then v
= w r So,
w = 2p n = q B/ m
[4 6(a)]
which is independent of the velocity or energy |
1 | 3433-3436 | If w is the angular frequency, then v
= w r So,
w = 2p n = q B/ m
[4 6(a)]
which is independent of the velocity or energy Here n is the frequency of rotation |
1 | 3434-3437 | So,
w = 2p n = q B/ m
[4 6(a)]
which is independent of the velocity or energy Here n is the frequency of rotation The
independence of n from energy has important
application in the design of a cyclotron (see
Section 4 |
1 | 3435-3438 | 6(a)]
which is independent of the velocity or energy Here n is the frequency of rotation The
independence of n from energy has important
application in the design of a cyclotron (see
Section 4 4 |
1 | 3436-3439 | Here n is the frequency of rotation The
independence of n from energy has important
application in the design of a cyclotron (see
Section 4 4 2) |
1 | 3437-3440 | The
independence of n from energy has important
application in the design of a cyclotron (see
Section 4 4 2) The time taken for one revolution is T= 2p/w
º 1/n |
1 | 3438-3441 | 4 2) The time taken for one revolution is T= 2p/w
º 1/n If there is a component of the velocity
parallel to the magnetic field (denoted by v||), it will make the particle
FIGURE 4 |
1 | 3439-3442 | 2) The time taken for one revolution is T= 2p/w
º 1/n If there is a component of the velocity
parallel to the magnetic field (denoted by v||), it will make the particle
FIGURE 4 5 Circular motion
FIGURE 4 |
1 | 3440-3443 | The time taken for one revolution is T= 2p/w
º 1/n If there is a component of the velocity
parallel to the magnetic field (denoted by v||), it will make the particle
FIGURE 4 5 Circular motion
FIGURE 4 6 Helical motion
Rationalised 2023-24
113
Moving Charges and
Magnetism
EXAMPLE 4 |
1 | 3441-3444 | If there is a component of the velocity
parallel to the magnetic field (denoted by v||), it will make the particle
FIGURE 4 5 Circular motion
FIGURE 4 6 Helical motion
Rationalised 2023-24
113
Moving Charges and
Magnetism
EXAMPLE 4 3
Example 4 |
1 | 3442-3445 | 5 Circular motion
FIGURE 4 6 Helical motion
Rationalised 2023-24
113
Moving Charges and
Magnetism
EXAMPLE 4 3
Example 4 3 What is the radius of the path of an electron (mass
9 × 10-31 kg and charge 1 |
1 | 3443-3446 | 6 Helical motion
Rationalised 2023-24
113
Moving Charges and
Magnetism
EXAMPLE 4 3
Example 4 3 What is the radius of the path of an electron (mass
9 × 10-31 kg and charge 1 6 × 10–19 C) moving at a speed of 3 ×107 m/s in
a magnetic field of 6 × 10–4 T perpendicular to it |
1 | 3444-3447 | 3
Example 4 3 What is the radius of the path of an electron (mass
9 × 10-31 kg and charge 1 6 × 10–19 C) moving at a speed of 3 ×107 m/s in
a magnetic field of 6 × 10–4 T perpendicular to it What is its
frequency |
1 | 3445-3448 | 3 What is the radius of the path of an electron (mass
9 × 10-31 kg and charge 1 6 × 10–19 C) moving at a speed of 3 ×107 m/s in
a magnetic field of 6 × 10–4 T perpendicular to it What is its
frequency Calculate its energy in keV |
1 | 3446-3449 | 6 × 10–19 C) moving at a speed of 3 ×107 m/s in
a magnetic field of 6 × 10–4 T perpendicular to it What is its
frequency Calculate its energy in keV ( 1 eV = 1 |
1 | 3447-3450 | What is its
frequency Calculate its energy in keV ( 1 eV = 1 6 × 10–19 J) |
1 | 3448-3451 | Calculate its energy in keV ( 1 eV = 1 6 × 10–19 J) Solution Using Eq |
1 | 3449-3452 | ( 1 eV = 1 6 × 10–19 J) Solution Using Eq (4 |
1 | 3450-3453 | 6 × 10–19 J) Solution Using Eq (4 5) we find
r = m v / (qB) = 9 ×10–31 kg × 3 × 107 m s–1 / ( 1 |
1 | 3451-3454 | Solution Using Eq (4 5) we find
r = m v / (qB) = 9 ×10–31 kg × 3 × 107 m s–1 / ( 1 6 × 10–19 C × 6 × 10–4 T)
= 28 × 10–2 m = 28 cm
n = v / (2 pr) = 17×106 s–1 = 17×106 Hz =17 MHz |
1 | 3452-3455 | (4 5) we find
r = m v / (qB) = 9 ×10–31 kg × 3 × 107 m s–1 / ( 1 6 × 10–19 C × 6 × 10–4 T)
= 28 × 10–2 m = 28 cm
n = v / (2 pr) = 17×106 s–1 = 17×106 Hz =17 MHz E = (½ )mv 2 = (½ ) 9 × 10–31 kg × 9 × 1014 m2/s2 = 40 |
1 | 3453-3456 | 5) we find
r = m v / (qB) = 9 ×10–31 kg × 3 × 107 m s–1 / ( 1 6 × 10–19 C × 6 × 10–4 T)
= 28 × 10–2 m = 28 cm
n = v / (2 pr) = 17×106 s–1 = 17×106 Hz =17 MHz E = (½ )mv 2 = (½ ) 9 × 10–31 kg × 9 × 1014 m2/s2 = 40 5 ×10–17 J
≈ 4×10–16 J = 2 |
1 | 3454-3457 | 6 × 10–19 C × 6 × 10–4 T)
= 28 × 10–2 m = 28 cm
n = v / (2 pr) = 17×106 s–1 = 17×106 Hz =17 MHz E = (½ )mv 2 = (½ ) 9 × 10–31 kg × 9 × 1014 m2/s2 = 40 5 ×10–17 J
≈ 4×10–16 J = 2 5 keV |
1 | 3455-3458 | E = (½ )mv 2 = (½ ) 9 × 10–31 kg × 9 × 1014 m2/s2 = 40 5 ×10–17 J
≈ 4×10–16 J = 2 5 keV move along the field and the path of the particle would be a helical one
(Fig |
1 | 3456-3459 | 5 ×10–17 J
≈ 4×10–16 J = 2 5 keV move along the field and the path of the particle would be a helical one
(Fig 4 |
1 | 3457-3460 | 5 keV move along the field and the path of the particle would be a helical one
(Fig 4 6) |
1 | 3458-3461 | move along the field and the path of the particle would be a helical one
(Fig 4 6) The distance moved along the magnetic field in one rotation is
called pitch p |
1 | 3459-3462 | 4 6) The distance moved along the magnetic field in one rotation is
called pitch p Using Eq |
1 | 3460-3463 | 6) The distance moved along the magnetic field in one rotation is
called pitch p Using Eq [4 |
1 | 3461-3464 | The distance moved along the magnetic field in one rotation is
called pitch p Using Eq [4 6 (a)], we have
p = v||T = 2pm v|| / q B
[4 |
1 | 3462-3465 | Using Eq [4 6 (a)], we have
p = v||T = 2pm v|| / q B
[4 6(b)]
The radius of the circular component of motion is called the radius of
the helix |
1 | 3463-3466 | [4 6 (a)], we have
p = v||T = 2pm v|| / q B
[4 6(b)]
The radius of the circular component of motion is called the radius of
the helix 4 |
1 | 3464-3467 | 6 (a)], we have
p = v||T = 2pm v|| / q B
[4 6(b)]
The radius of the circular component of motion is called the radius of
the helix 4 4
MAGNETIC FIELD DUE TO A CURRENT ELEMENT,
BIOT-SAVART LAW
All magnetic fields that we know are due to currents (or moving
charges) and due to intrinsic magnetic moments of particles |
1 | 3465-3468 | 6(b)]
The radius of the circular component of motion is called the radius of
the helix 4 4
MAGNETIC FIELD DUE TO A CURRENT ELEMENT,
BIOT-SAVART LAW
All magnetic fields that we know are due to currents (or moving
charges) and due to intrinsic magnetic moments of particles Here, we shall study the relation between current and the
magnetic field it produces |
1 | 3466-3469 | 4 4
MAGNETIC FIELD DUE TO A CURRENT ELEMENT,
BIOT-SAVART LAW
All magnetic fields that we know are due to currents (or moving
charges) and due to intrinsic magnetic moments of particles Here, we shall study the relation between current and the
magnetic field it produces It is given by the Biot-Savart’s law |
1 | 3467-3470 | 4
MAGNETIC FIELD DUE TO A CURRENT ELEMENT,
BIOT-SAVART LAW
All magnetic fields that we know are due to currents (or moving
charges) and due to intrinsic magnetic moments of particles Here, we shall study the relation between current and the
magnetic field it produces It is given by the Biot-Savart’s law Fig |
1 | 3468-3471 | Here, we shall study the relation between current and the
magnetic field it produces It is given by the Biot-Savart’s law Fig 4 |
1 | 3469-3472 | It is given by the Biot-Savart’s law Fig 4 7 shows a finite conductor XY carrying current I |
1 | 3470-3473 | Fig 4 7 shows a finite conductor XY carrying current I Consider
an infinitesimal element dl of the conductor |
1 | 3471-3474 | 4 7 shows a finite conductor XY carrying current I Consider
an infinitesimal element dl of the conductor The magnetic field
dB due to this element is to be determined at a point P which is at
a distance r from it |
1 | 3472-3475 | 7 shows a finite conductor XY carrying current I Consider
an infinitesimal element dl of the conductor The magnetic field
dB due to this element is to be determined at a point P which is at
a distance r from it Let q be the angle between dl and the
displacement vector r |
1 | 3473-3476 | Consider
an infinitesimal element dl of the conductor The magnetic field
dB due to this element is to be determined at a point P which is at
a distance r from it Let q be the angle between dl and the
displacement vector r According to Biot-Savart’s law, the
magnitude of the magnetic field dB is proportional to the current
I, the element length |dl|, and inversely proportional to the square
of the distance r |
1 | 3474-3477 | The magnetic field
dB due to this element is to be determined at a point P which is at
a distance r from it Let q be the angle between dl and the
displacement vector r According to Biot-Savart’s law, the
magnitude of the magnetic field dB is proportional to the current
I, the element length |dl|, and inversely proportional to the square
of the distance r Its direction* is perpendicular to the plane
containing dl and r |
1 | 3475-3478 | Let q be the angle between dl and the
displacement vector r According to Biot-Savart’s law, the
magnitude of the magnetic field dB is proportional to the current
I, the element length |dl|, and inversely proportional to the square
of the distance r Its direction* is perpendicular to the plane
containing dl and r Thus, in vector notation,
d
I d
r
B
r
∝
×
l
3
=
×
µ0
3
4π
I d
lr
r
[4 |
1 | 3476-3479 | According to Biot-Savart’s law, the
magnitude of the magnetic field dB is proportional to the current
I, the element length |dl|, and inversely proportional to the square
of the distance r Its direction* is perpendicular to the plane
containing dl and r Thus, in vector notation,
d
I d
r
B
r
∝
×
l
3
=
×
µ0
3
4π
I d
lr
r
[4 11(a)]
where m0/4p is a constant of proportionality |
1 | 3477-3480 | Its direction* is perpendicular to the plane
containing dl and r Thus, in vector notation,
d
I d
r
B
r
∝
×
l
3
=
×
µ0
3
4π
I d
lr
r
[4 11(a)]
where m0/4p is a constant of proportionality The above expression
holds when the medium is vacuum |
1 | 3478-3481 | Thus, in vector notation,
d
I d
r
B
r
∝
×
l
3
=
×
µ0
3
4π
I d
lr
r
[4 11(a)]
where m0/4p is a constant of proportionality The above expression
holds when the medium is vacuum FIGURE 4 |
1 | 3479-3482 | 11(a)]
where m0/4p is a constant of proportionality The above expression
holds when the medium is vacuum FIGURE 4 7 Illustration of
the Biot-Savart law |
1 | 3480-3483 | The above expression
holds when the medium is vacuum FIGURE 4 7 Illustration of
the Biot-Savart law The
current element I dl
produces a field dB at a
distance r |
1 | 3481-3484 | FIGURE 4 7 Illustration of
the Biot-Savart law The
current element I dl
produces a field dB at a
distance r The Ä sign
indicates that the
field is perpendicular
to the plane of this
page and directed
into it |
1 | 3482-3485 | 7 Illustration of
the Biot-Savart law The
current element I dl
produces a field dB at a
distance r The Ä sign
indicates that the
field is perpendicular
to the plane of this
page and directed
into it *
The sense of dl × r is also given by the Right Hand Screw rule : Look at the
plane containing vectors dl and r |
1 | 3483-3486 | The
current element I dl
produces a field dB at a
distance r The Ä sign
indicates that the
field is perpendicular
to the plane of this
page and directed
into it *
The sense of dl × r is also given by the Right Hand Screw rule : Look at the
plane containing vectors dl and r Imagine moving from the first vector towards
second vector |
1 | 3484-3487 | The Ä sign
indicates that the
field is perpendicular
to the plane of this
page and directed
into it *
The sense of dl × r is also given by the Right Hand Screw rule : Look at the
plane containing vectors dl and r Imagine moving from the first vector towards
second vector If the movement is anticlockwise, the resultant is towards you |
1 | 3485-3488 | *
The sense of dl × r is also given by the Right Hand Screw rule : Look at the
plane containing vectors dl and r Imagine moving from the first vector towards
second vector If the movement is anticlockwise, the resultant is towards you If it is clockwise, the resultant is away from you |
1 | 3486-3489 | Imagine moving from the first vector towards
second vector If the movement is anticlockwise, the resultant is towards you If it is clockwise, the resultant is away from you Rationalised 2023-24
Physics
114
EXAMPLE 4 |
1 | 3487-3490 | If the movement is anticlockwise, the resultant is towards you If it is clockwise, the resultant is away from you Rationalised 2023-24
Physics
114
EXAMPLE 4 5
The magnitude of this field is,
0
2
d sin
d
4
I
l
r
µ
θ
=
π
B
[4 |
1 | 3488-3491 | If it is clockwise, the resultant is away from you Rationalised 2023-24
Physics
114
EXAMPLE 4 5
The magnitude of this field is,
0
2
d sin
d
4
I
l
r
µ
θ
=
π
B
[4 11(b)]
where we have used the property of cross-product |
1 | 3489-3492 | Rationalised 2023-24
Physics
114
EXAMPLE 4 5
The magnitude of this field is,
0
2
d sin
d
4
I
l
r
µ
θ
=
π
B
[4 11(b)]
where we have used the property of cross-product Equation [4 |
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