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Definition, explanation, Formula, Application, Solved Example Problems - Lorentz Force | 12th Physics : Magnetism and Magnetic Effects of Electric Current

Chapter: 12th Physics : Magnetism and Magnetic Effects of Electric Current

Lorentz Force

When an electric charge q is kept at rest in a magnetic field, no force acts on it. At the same time, if the charge moves in the magnetic field, it experiences a force.

LORENTZ FORCE

When an electric charge q is kept at rest in a magnetic field, no force acts on it. At the same time, if the charge moves in the magnetic field, it experiences a force. This force is different from Coulomb force, studied in unit 1. This force is known as magnetic force. It is given by the equation


In general, if the charge is moving in both the electric and magnetic fields, the total force experienced by the charge is given by . It is known as Lorentz force.

 

1. Force on a moving charge in a magnetic field

When an electric charge q is moving with velocity  in the magnetic field  , it experiences a force, called magnetic force  . After careful experiments, Lorentz deduced the force experienced by a moving charge in the magnetic field 


The equations (3.54) and (3.55) imply

1.   is directly proportional to the magnetic field 

2.   is directly proportional to the velocity 

3.   is directly proportional to sine of the angle between the velocity and magnetic field

4.   is directly proportional to the magnitude of the charge q

5. The direction of  is always perpendicular to  and as  is the cross product of  and 


6. The direction of  on negative charge is opposite to the direction of  on positive charge provided other factors are identical as shown Figure 3.49

7. If velocity  of the charge q is along magnetic field  then,  is zero


Definition of tesla

The strength of the magnetic field is one tesla if unit charge moving in it with unit velocity experiences unit force.


 

EXAMPLE 3.20

A particle of charge q moves with along positive y - direction invelocity  a magnetic field . Compute the Lorentz force experienced by the particle (a) when magnetic field is along positive y-direction (b) when magnetic field points in positive z - direction (c) when magnetic field is in zy - plane and making an angle θ with velocity of the particle. Mark the direction of magnetic force in each case.

Solution

Velocity of the particle is 

(a) Magnetic field is along positive  y - direction, this implies, 


From Lorentz force, 

So, no force acts on the particle when it moves along the direction of magnetic field.

(b) Magnetic field points in positive z - direction, this implies, 


From Lorentz force, 


Therefore, the magnitude of the Lorentz force is qvB and direction is along positive x - direction.

(c) Magnetic field is in zy - plane and making an angle θ with the velocity of the particle, which implies 


From Lorentz force,


 

EXAMPLE 3.21

Compute the work done and power delivered by the Lorentz force on the particle of charge q moving with velocity  . Calculate the angle between Lorentz force and velocity of the charged particle and also interpret the result.

Solution

For a charged particle moving on a magnetic field,


The work done by the magnetic field is


Since  is perpendicular to  and hence  This means that Lorentz force do no work on the particle. From work kinetic energy theorem, (Refer section 4th chapter, XI th standard Volume I)


Since,  and  are perpendicular to each other. The angle between Lorentz force and velocity of the charged particle is 90º. Thus Lorentz force changes the direction of the velocity but not the magnitude of the velocity. Hence Lorentz force does no work and also does not alter kinetic energy of the particle.

 

2. Motion of a charged particle in a uniform magnetic field


Consider a charged particle of charge q having mass m enters into a region of uniform magnetic field  with velocity  such that velocity is perpendicular to the magnetic field. As soon as the particle enters into the field, Lorentz force acts on it in a direction perpendicular to both magnetic field  and velocity .

As a result, the charged particle moves in a circular orbit as shown in Figure 3.50.

The Lorentz force on the charged particle is given by


Since Lorentz force alone acts on the particle, the magnitude of the net force on the particle is


This Lorentz force acts as centripetal force for the particle to execute circular motion. Therefore,


The radius of the circular path is


where p = mv is the magnitude of the linear momentum of the particle. Let T be the time taken by the particle to finish one complete circular motion, then


Hence substituting (3.56) in (3.57), we get


Equation (3.58) is called the cyclotron period. The reciprocal of time period is the frequency f, which is


In terms of angular frequency ω,


Equations (3.59) and (3.60) are called as cyclotron frequency or gyro-frequency.

From equations (3.58), (3.59) and (3.60), we infer that time period and frequency depend only on charge-to-mass ratio (specific charge) but not velocity or the radius of the circular path.

If a charged particle moves in a region of uniform magnetic field such that its velocity is not perpendicular to the magnetic field, then the velocity of the particle is split up into two components; one component is parallel to the field while the other perpendicular to the field. The component of velocity parallel to field remains unchanged and the component perpendicular to field keeps changing due to the Lorentz force. Hence the path of the particle is not a circle; it is a helix around the field lines as shown in Figure 3.51.


For an example, the helical path of an electron when it moves in a magnetic field is shown in Figure 3.52. Inside the particle detector called cloud chamber, the path is made visible by the condensation of water droplets.


 

EXAMPLE 3.22

An electron moving perpendicular to a uniform magnetic field 0.500 T undergoes circular motion of radius 2.80 mm. What is the speed of electron?

Solution

Charge of an electron q = -1.60 × 10-19 C

|q| = 1.60 ×10−19 C

Magnitude of magnetic field B = 0.500 T

Mass of the electron, m = 9.11 × 10-31 kg

Radius of the orbit, r = 2.50 mm = 2.50 × 10-3 m

Velocity of the electron, v = |q| rB/m


v = 2.195 ×108 m s−1

 

EXAMPLE 3.23

A proton moves in a uniform magnetic field of strength 0.500 T magnetic field is directed along the x-axis. At initial time, t = 0 s, the proton has velocity . Find

(a) At initial time, what is the acceleration of the proton.

(b) Is the path circular or helical?. If helical, calculate the radius of helical trajectory and also calculate the pitch of the helix (Note: Pitch of the helix is the distance travelled along the helix axis per revolution).

Solution


Pitch of the helix is the distance travelled along x-axis in a time T, which is P = vx T

But time,


The proton experiences appreciable acceleration in the magnetic field, hence the pitch of the helix is almost six times greater than the radius of the helix.

 

EXAMPLE 3.24

Two singly ionized isotopes of uranium 23592U and 23892U (isotopes have same atomic number but different mass number) are sent with velocity 1.00 × 105 m s-1 into a magnetic field of strength 0.500 T normally. Compute the distance between the two isotopes after they complete a semi-circle. Also compute the time taken by each isotope to complete one semi-circular path. (Given: masses of the isotopes: m235 = 3.90 x 10-25 kg and m238 = 3.95 x 10-25 kg)


Solution

Since isotopes are singly ionized, they have equal charge which is equal to the charge of an electron, q = - 1.6 × 10-19 C. Mass of uranium 23592U and 23892U are 3.90 × 10-25 kg and 3.95 × 10-25 kg respectively. Magnetic field applied, B = 0.500 T. Velocity of the electron is 1.00 × 105 m s-1, then

(a) the radius of the path of 23592U is r235


The diameter of the semi-circle due to 23892U is d238 = 2r238 = 98.8 cm

Therefore the separation distance between the isotopes is Δd = d238 − d235 = 1.2cm

(b) The time taken by each isotope to complete one semi-circular path are

 

Note that even though the difference between mass of two isotopes are very small, this arrangement helps us to convert this small difference into an easily measurable distance of separation. This arrangement is known as mass spectrometer. A mass spectrometer is used in many areas in sciences, especially in medicine, in space science, in geology etc. For example, in medicine, anaesthesiologists use it to measure the respiratory gases and biologist use it to determine the reaction mechanisms in photosynthesis.

 

3. Motion of a charged particle under crossed electric and magnetic field (velocity selector)


Consider an electric charge q of mass m which enters into a region of uniform magnetic field  with velocity  such that velocity is not perpendicular to the magnetic field. Then the path of the particle is a helix. The Lorentz force on the charged particle moving in a uniform magnetic field can be balanced by Coulomb force by proper arrangement of electric and magnetic fields.

The Coulomb force acts along the direction of electric field (for a positive charge q) whereas the Lorentz force is perpendicular to the direction of magnetic field. Therefore in order to balance these forces, both electric and magnetic fields must be perpendicular to each other. Such an arrangement of perpendicular electric and magnetic fields are known as cross fields.

 For illustration, consider an experimental arrangement as shown in Figure 3.53. In the region of space between parallel plates of a capacitor (which produces uniform electric field), uniform magnetic field is maintained perpendicular to the direction of electric field. Suppose a charged particle enters this space from the left side as shown, the net force on the particle is


For a positive charge, the electric force on the charge acts in downward direction whereas the Lorentz force acts upwards. When these two forces balance one another, then


This means, for a given magnitude of - field and - field, the forces act only for the particle moving with particular speed v0 = E/B . This speed is independent of mass and charge.

If the charge enters into the crossed fields with velocity v, other than vo, it results in any of the following possibilities (Table 3.4).


So by proper choice of electric and magnetic fields, the particle with particular speed can be selected. Such an arrangement of fields is called a velocity selector.

 

EXAMPLE 3.25

Let E be the electric field of magnitude 6.0 × 106 N C-1 and B be the magnetic field magnitude 0.83 T. Suppose an electron is accelerated with a potential of 200 V, will it show zero deflection?. If not, at what potential will it show zero deflection.

Solution:

Electric field, E = 6.0 × 106 N C-1 and magnetic field, B = 0.83 T.

Then


When an electron goes with this velocity, it shows null deflection. Since the accelerating potential is 200 V, the electron acquires kinetic energy because of this accelerating potential. Hence,


Since the mass of the electron, m = 9 .1×10−31 kg and charge of an electron, |q| = e = 1.6 ×10−19 C. The velocity due to accelerating potential 200 V


Since the speed v200 > v, the electron is deflected towards direction of Lorentz force. So, in order to have null deflection, the potential, we have to supply is


V =148 65 V

 

4. Cyclotron

Cyclotron (Figure 3.54) is a device used to accelerate the charged particles to gain large kinetic energy. It is also called as high energy accelerator. It was invented by Lawrence and Livingston in 1934.

Principle

When a charged particle moves normal to the magnetic field, it experiences magnetic Lorentz force.


Construction

The schematic diagram of a cyclotron is shown in Figure 3.55. The particles are allowed to move in between two semi-circular metal containers called Dees (hollow D - shaped objects). Dees are enclosed in an evacuated chamber and it is kept in a region with uniform magnetic field controlled by an electromagnet. The direction of magnetic field is normal to the plane of the Dees. The two Dees are kept separated with a gap and the source S (which ejects the particle to be accelerated) is placed at the center in the gap between the Dees. Dees are connected to high frequency alternating potential difference.


Working

Let us assume that the ion ejected from source S is positively charged. As soon as ion is ejected, it is accelerated towards a Dee (say, Dee – 1) which has negative potential at that time. Since the magnetic field is normal to the plane of the Dees, the ion undergoes circular path. After one semi-circular path in Dee-1, the ion reaches the gap between Dees. At this time, the polarities of the Dees are reversed so that the ion is now accelerated towards Dee-2 with a greater velocity. For this circular motion, the centripetal force of the charged particle q is provided by Lorentz force.


From the equation (3.62), the increase in velocity increases the radius of circular path. This process continues and hence the particle undergoes spiral path of increasing radius. Once it reaches near the edge, it is taken out with the help of deflector plate and allowed to hit the target T.

Very important condition in cyclotron operation is the resonance condition. It happens when the frequency f at which the positive ion circulates in the magnetic field must be equal to the constant frequency of the electrical oscillator fosc

From equation (3.59), we have


The time period of oscillation is


The kinetic energy of the charged particle is


Limitations of cyclotron

a)     the speed of the ion is limited

b)    electron cannot be accelerated

c)     uncharged particles cannot be accelerated

 

EXAMPLE 3.26

Suppose a cyclotron is operated to accelerate protons with a magnetic field of strength 1 T. Calculate the frequency in which the electric field between two Dees could be reversed.

Solution

Magnetic field B = 1 T

Mass of the proton, mp = 1.67 ×10−27 kg

Charge of the proton, q = 1.60 ×10−19 C


 

5. Force on a current carrying conductor placed in a magnetic field


When a current carrying conductor is placed in a magnetic field, the force experienced by the wire is equal to the sum of Lorentz forces on the individual charge carriers in the wire. Consider a small segment of wire of length dl, with cross-sectional area A and current I as shown in Figure 3.56. The free electrons drift opposite to the direction of current. So the relation between current I and magnitude of drift velocity vd (Refer Unit 2) is


If the wire is kept in a magnetic field , then average force experienced by the charge (here, electron) in the wire is


Let n be the number of free electrons per unit volume, therefore


where N is the number of free electrons in the small element of volume V = A dl.

Hence Lorentz force on the wire of length dl is the product of the number of the electrons

(N = nA dl) and the force acting on an electron.


The length dl is along the length of the wire and hence the current element in the wire is 

Therefore the force on the wire is


The force in a straight current carrying conducting wire of length l placed in a uniform magnetic field is


In magnitude,

F = BIl sinθ

(a) If the conductor is placed along the direction of the magnetic field, the angle between them is θ = 0º. Hence, the force experienced by the conductor is zero.

(b) If the conductor is placed perpendicular to the magnetic field, the angle between them is θ =90º Hence, the force experienced by the conductor is maximum, which is F = BIl.

Fleming’s left hand rule (mnemonic)

When a current carrying conductor is placed in a magnetic field, the direction of the force experienced by it is given by Fleming’s Left Hand Rule (FLHR) as shown in Figure 3.57.


Stretch forefinger, the middle finger and the thumb of the left hand such that they are in mutually perpendicular directions. If forefinger points the direction of magnetic field, the middle finger points the direction of the electric current, then thumb will point the direction of the force experienced by the conductor.

 

EXAMPLE 3.27

A metallic rod of linear density is 0.25 kg m-1 is lying horizontally on a smooth inclined plane which makes an angle of 45º with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of strength 0.25 T is acting on it in the vertical direction. Calculate the electric current flowing in the rod to keep it stationary.


Solution

The linear density of the rod i.e., mass per unit length of the rod is 0.25 kg m-1

m/l = 0.25 kg m−1

Let I be the current flowing in the metallic rod. The direction of electric current is into the paper. The direction of magnetic force IBl is given by Fleming’s left hand rule.


For equilibrium,

mg sin 45º = IBl cos 45 º

 ⇒ I = I/B m/l g tan 45 º

 ⇒ I = 9.8 A

So, we need to supply current of 9.8 A to keep the metallic rod stationary.

 

6. Force between two long parallel current carrying conductors

Two long straight parallel current carrying conductors separated by a distance r are kept in air as shown in Figure 3.58. Let I1 and I2 be the electric currents passing through the conductors A and B in same direction (i.e. along z - direction) respectively. The net magnetic field at a distance r due to current I1 in conductor A is



From thumb rule, the direction of magnetic field is perpendicular to the plane of the paper and inwards (arrow into the page ) i.e. along negative i ^ direction.

Let us consider a small elemental length dl in conductor B at which the magnetic field B1 is present. From equation 3.65, Lorentz force on the element dl of conductor B is


Therefore the force on dl of the wire B is directed towards the wire W1. So the length dl is attracted towards the conductor A. The force per unit length of the conductor B due to the wire conductor A is


In the same manner, we compute the magnitude of net magnetic induction due to current I2 (in conductor A) at a distance r in the elemental length dl of conductor A is


From the thumb rule, direction of magnetic field is perpendicular to the plane of the paper and outwards (arrow out of the page ʘ) i.e., along positive i ^ direction.

Hence, the magnetic force at element dl of the wire is W1 is


Therefore the force on dl of conductor A is directed towards the conductor B. So the length dl is attracted towards the conductor B as shown in Figure (3.59).


The force per unit length of the conductor A due to the conductor B is


Thus the force experienced by two parallel current carrying conductors is attractive if the direction of electric current passing through them is same as shown in Figure 3.60.


Thus the force experienced by two parallel current carrying conductors is repulsive if they carry current in the opposite directions as shown in Figure 3.61.

Definition of ampère

One ampère is defined as that current when it is passed through each of the two infinitely long parallel straight conductors kept at a distance of one meter apart in vacuum causes each conductor to experience a force of 2 × 10−7 newton per meter length of conductor. 

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