Contents
Physics Topics can also be used to explain the behavior of complex systems, such as the stock market or the dynamics of traffic flow.
What is the Path of Charge in Uniform Magnetic Field?
Let \(\vec{B}\) = magnetic field at a point,
q = electric charge of a particle,
v = velocity of the particle at the given point.
Magnetic force on the charged particle due to magnetic field at that point,
\(\vec{F}\) = q\(\vec{v}\) × \(\vec{B}\) …….. (1)
If the angle between \(\vec{v}\) and \(\vec{B}\) be θ, from the cross product of two vectors,
F = |\(\vec{F}\)| = qvBsinθ ….. (2)
Naturally, if v = 0, F = 0, i.e., if a charged particle is at rest then no magnetic force acts on it.
Definition of magnetic field \(\vec{B}\):
i) Direction of \(\vec{B}\): If θ = 0° or θ = 180°, F = 0 . Thus no magnetic force acts on a charged particle which is moving parallel or antiparallel to the magnetic field. Hence, in a magnetic field, the direction (or its opposite direction) of a moving charged particle for which no magnetic force acts on it, defines the direction of \(\vec{B}\).
ii) Magnitude of \(\vec{B}\): If \(\vec{v}\) and \(\vec{B}\) remain perpendicular to each other, θ = 90°. Thus, the magnitude of the magnetic force \(\vec{F}\) becomes maximum. Expressing this maximum force by Fm, equation (2) can be written as,
Fm = qvB …….. (3)
If q = 1 and v = 1, then B = Fm. The maximum possible force exerted by a magnetic field on a particle of unit charge moving with a unit velocity through the field, defines this magnitude of the magnetic field.
Unit of B: From equation (3), [B] = \(\frac{[F]}{[q][v]}\)
Hence the unit of B is
\(\frac{\mathrm{N}}{\mathrm{A} \cdot \mathrm{s} \cdot\left(\mathrm{m} \cdot \mathrm{s}^{-1}\right)}\) or N ᐧ A-1 ᐧ m-1
This unit is known as tesla (T) or weber/metre2 (Wb ᐧ m-2).
Significance of the cross product: Applying the rule of the cross product of two vectors in equation (1) we can conclude that \(\vec{F}\) is always perpendicular to the plane containing \(\vec{v}\) and \(\vec{B}\) [Fig.]. If a right-handed corkscrew be rotated from the direc-tion of \(\vec{v}\) to \(\vec{B}\), the direction of advancement of the screw-head indicates the direction of \(\vec{F}\).
Work done by the magnetic force is zero: \(\vec{F}\) and \(\vec{v}\) are always perpendicular to each other. Since the displacement of a particle is taken along the direction of its velocity, at any point in the magnetic field, the force \(\vec{F}\) and a small displacement d\(\vec{s}\) of the particle are perpendicular to each other. Then, for the magnetic force \(\vec{F}\) and its displacement d\(\vec{s}\), work done,
dW = \(\vec{F} \cdot d \vec{s}\) = Fdscos90° = 0
Hence, the work done by the magnetic force on a moving charged particle in a magnetic field is zero. In other words, the magnetic force is a no work force.
Again, we know that, work done on a free particle = change in kinetic energy of the particle. Since magnetic force is a no-work force, the kinetic energy of a charged particle is not affected.
∴ ΔK = W = 0
or, \(\frac{1}{2} m v^2\) = constant (m = mass of the particle)
or, v = constant
So, when a magnetic force acts on a charged particle moving in a magnetic field, its speed as well as its kinetic energy remains unaffected.
Magnetic force in CGS system: in this system, charge q is expressed in esu [see the chapter : ‘Electric Field’]. But to indicate the magnetic force, the current as well as the charge must be measured in emu.
1 emu of charge = 3 × 1010 esu of charge = c esu
[c = velocity of light in vacuum = 3 × 1010 cm ᐧ s-1]
So, if the magnitude of an electric charge be q esu, in electro-magnetic unit, it will be \(\frac{q}{c}\) emu.
Hence, the above mentioned SI equation (1) can be expressed in CGS system as \(\vec{F}\) = \(\frac{q}{c} \vec{v} \times \vec{B}\).
In this equation, magnetic field \(\vec{B}\) is used but not the magnetic intensity \(\vec{H}\). Here c = 3 × 1010 cm ᐧ s-1 ᐧ F, q, v, and B are measured in dyn, esu, cm ᐧ s-1, and gauss or G, respectively.
Note that, 1 C = 0.1 emu of charge = 3 × 109 esu of charge.
Fleming’s left hand rule: If a charged particle moves at right angles to the magnetic field, that is if the angle between \(\vec{v}\) and \(\vec{B}\) be θ = 90°, then from the Fig. it can be concluded that the three vectors \(\vec{v}\), \(\vec{B}\) and \(\vec{F}\) are mutually perpendicular. This special case can be easily explained by Fleming’s left hand rule [Fig.].
Path of a Moving Charge in a Uniform Magnetic Field
Uniform magnetic field: A magnetic field is said to be uniform if its magnitude and direction remain constant in a region. We know that a magnetic field is represented by mag-netic lines of force. For a uniform magnetic field
1. magnetic lines of force are parallel to each other because the direction of magnetic field remains constant;
2. the magnitude of the mag-netic field also remains unchanged and hence the number den-sity of the lines of force at different points are equal, i.e., the lines of force are equispaced [Fig.(a)]. Moreover, to represent a uniform normal and upward magnetic field with respect to the plane of the page, equispaced marked points are used [Fig.(b)]; on the other hand, to denote a downward uniform magnetic field similar marked points are used [Fig.(c)].
In the region adjacent to the earth’s surface, the terrestrial mag-netic field is assumed to be uniform. Only in the vicinity of mag-nets or magnetic materials, are these lines of force distorted a little. A uniform magnetic field is also generated between two strong opposite magnetic poles kept very close to each other [Fig(d)],
Determination of the path of a charged particle:
i) The charged particle is at rest: Since \(\vec{F}\) = \(\overrightarrow{q v} \times \vec{B}\), if \(\vec{v}\) = 0, \(\vec{F}\) = 0. Hence, in this case no magnetic force acts on the particle and the charged particle remains at rest.
ii) The charged particle enters with a velocity \(\vec{v}\) parallel to the magnetic field: If the velocity of the particle and the magnetic field are parallel to each other, then \(\vec{v}\) × \(\vec{B}\) = 0.
So, the magnetic force, \(\vec{F}\) = q\(\vec{v}\) × \(\vec{B}\) = 0. Since no magnetic force acts in this case, the particle continues to move along a straight line.
iii) The charged particle enters with a velocity v perpendic-ular to the magnetic field: Let a uniform magnetic field B be acting upward, perpendicular to the plane of the paper [Fig.]. A particle having charge +q enters that mag-netic field at P with a velocity \(\vec{v}\) parallel to the plane of the paper. Applying cross product rule, we see that the magnetic force \(\vec{F}\) act-ing on the particle will be normal to \(\vec{v}\) and along the direction \(\overrightarrow{P O}\).
As a result, the particle will be accel-erated towards \(\overrightarrow{P O}\) and hence it will tend to deviate from its path. When the particle reaches another point Q, the magnetic force \(\vec{F}\) will still act normal to \(\vec{v}\) and along \(\overrightarrow{Q O}\). In this way, magnetic force acting on the charged particle at every point of its path is always directed towards the point O.
This force acts as a centripetal force on the particle which therefore keeps revolving along a circular path centred at O and of radius r (r = PO = QO). Since the magnitude of velocity does not change under the influence of the mag-netic field, the charged particle will have a uniform circular motion.
Radius of the circular path: Magnetic force, \(\vec{F}\) = \(q \vec{v} \times \vec{B}\). Since the angle between \(\vec{v}\) and \(\vec{B}\) is 90°, the magnitude of the magnetic force,
F = |\(\vec{F}\)| = qvBsin90° = qvB
If the mass of the charged particle be m,
So, qvB = \(\frac{m v^2}{r}\) or, r = \(\frac{m v}{q B}\) …. (1)
We see from the equation (1) that—
1. For a given charged particle (q = constant) and for a definite magnetic field (B = constant), the radius of the circular path is directly proportional to the momentum (mv) of the particle. This property is utilised in the mea-surement of mass of a charged particle in a mass spectro-scope.
2. If a given charged particle (q – constant) enters a magnetic field with a definite momentum (mv= constant) then the radius of the circular path is inversely propor-tional to the applied magnetic field (B).
3. The radius of the circular path is inversely proportional to its specific charge \(\frac{q}{m}\). For example, the charge of a proton and electron is the same but the mass of a proton is 1836 times that of an electron. Hence, the value of \(\frac{q}{m}\) for an electron is 1836 times higher. It means that if a proton and an electron enter a magnetic field with equal velocity, the electron revolves in a circular path of much smaller radius.
4. It is clear that if the charge be -q instead of +q, the direction of uniform circular motion will be reversed.
Period of revolution and cyclotron frequency: Circumference of the circular path = 2πr.
Since it is a uniform circular motion, time period,
T = \(\frac{2 \pi r}{v}\) = \(2 \pi \frac{m}{q B}\) ……..(2)
The number of complete revolutions made in unit time, i.e., frequency of the circular motion,
n = \(\frac{\mathbf{1}}{T}\) = \(\frac{1}{2 \pi}\left(\frac{q}{m}\right) B\) …. (3)
This frequency n is called cyclotron frequency.
Evidently, both T and n are independent of the radius of the path as well as of the velocity. From equation (1), it is clear that, \(\frac{r}{v}\) = constant . This property for the motion of a charged particle in a magnetic field is utilised in particle accelerators like cyclotrons.
iv) The charged particle enters the magnetic field at an inclined path: Let the z-axis be chosen along the direction of a uniform magnetic field acting at a place [Fig.].
A particle of charge + q and mass m enters that magnetic field with velocity \(\vec{v}\) at the point P on the xy -plane. If this velocity \(\vec{v}\) is inclined at an angle θ with the magnetic field \(\vec{B}\),the component of velocity along \(\vec{B}\), i.e., along z-axis = vcosθ and the component of velocity perpendicular to \(\vec{B}\), i.e., on the xy-plane = vsinθ.
Naturally, no magnetic force acts on the particle along the component vcosθ and hence this component remains unchanged. So, the charged particle performs a uniform linear motion along the magnetic field.
Again, the component vsinθ produces a uniform circular motion. The radius of this uniform circular motion can be obtained by substituting vsinθ for v in equation (1),
r = \(\frac{m v \sin \theta}{q B}\) …. (4)
Time period and frequency of the particle in uniform circu-lar motion do not depend on the velocity of the particle; hence just like equations (2) and (3), it can be written as,
T = \(\frac{2 \pi m}{q B}\) and n = \(\frac{1}{T}\) = \(\frac{q B}{2 \pi m}\) ….. (5)
Due to the combination of linear motion parallel to z -axis and uniform circular motion on the xy-plane, the charged particle follows a spiral or helical path.
The axis of this helical path is the z -axis. For each complete revolution of the particle, the distance covered along z -axis, i.e., along the direction of magnetic field [from P to Q in Fig.] is called pitch of this helical motion.
So, pitch = time period × linear velocity
= \(\frac{2 \pi m}{q B} \cdot v \cos \theta\)
= \(\frac{2 \pi m}{q B} \cdot \frac{q B r}{m \sin \theta} \cdot \cos \theta\)
= 2πrcotθ
The property of the helical motion of a charged particle in a magnetic field is utilised in magnetic focussing of different equipments.