Home | | Physics 12th Std | Lorentz Force: Solved Example Problems

# Lorentz Force: Solved Example Problems

Physics : Magnetism and Magnetic Effects of Electric Current: Lorentz Force: Solved Example Problems with Answers, Solution and Explanation

## 1. Force on a moving charge in a magnetic field: Solved Example Problems

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: Solved Example Problems

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): Solved Example Problems

### 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: Solved Example Problems

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: Solved Example Problems

### 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.

Tags : Physics , 12th Physics : Magnetism and Magnetic Effects of Electric Current
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
12th Physics : Magnetism and Magnetic Effects of Electric Current : Lorentz Force: Solved Example Problems | Physics