Informative line

### Various Applications

Learn about motion of charged particle in electric field and magnetic field and cyclotron frequency, velocity & mass filter.

# Velocity Filter

• In various applications, there is a need that particles should move with essential velocity .
• It can be achieved by applying specific combination of electric and magnetic field.
• In this combination , Electric field $$\vec E$$ is in upward direction .
• Magnetic field $$\vec B$$ is into the  page.
• Velocity $$\vec v$$ is from right to left (perpendicular to both positive charges).
• $$\vec F_E$$ is in upward direction .
• $$\vec F_B$$ is in downward direction .

• Magnitude of $$\vec E$$ and $$\vec B$$ is chosen such that

$$qE = q\,v\,B$$

So, velocity $$v = \dfrac {E}{B}$$

Such , particles move undeviated .

• If particle has velocity higher than $$v$$

$$q\,v\, B > qE$$

$$F_B > F_E$$

The particle is deviated downwards.

• If particle has velocity lower than $$v$$

$$q \,v\,B< qE$$

$$F_B < F_E$$

So, The particle is deviated upwards .

#### If a particle goes undeviated in the region containing electric and magnetic field, which option is correct regarding it?

A Electric field and magnetic field are parallel to each other and velocity vector is perpendicular to magnetic field

B All velocity vector, magnetic field and electric field are perpendicular to each other

C Electric field is parallel to magnetic field and velocity vector is not parallel to $$\vec E$$

D $$\vec B$$ is parallel to $$\vec E$$ and $$\vec v$$ is perpendicular to $$\vec E$$

×

When $$\vec v$$ is perpendicular to  $$\vec E$$ , it will cause deviation of particle.

Hence, Option (A) is incorrect.

Since, $$\vec v$$ is perpendicular to both $$\vec E$$ and $$\vec B$$ , the force due to $$\vec E$$ nullifies the force due to $$\vec B$$ .

Hence, Option (B) is correct.

Since, $$\vec v$$ is not parallel to  $$\vec E$$ so, it will cause deviation of particle.

Hence, Option (C) is incorrect.

Since, $$\vec v$$ is perpendicular to  $$\vec E$$ , it will cause deviation of particle.

Hence, Option (D) is incorrect.

### If a particle goes undeviated in the region containing electric and magnetic field, which option is correct regarding it?

A

Electric field and magnetic field are parallel to each other and velocity vector is perpendicular to magnetic field

.

B

All velocity vector, magnetic field and electric field are perpendicular to each other

C

Electric field is parallel to magnetic field and velocity vector is not parallel to $$\vec E$$

D

$$\vec B$$ is parallel to $$\vec E$$ and $$\vec v$$ is perpendicular to $$\vec E$$

Option B is Correct

# Mass Filter

• Mass Filter or mass spectrometer is a device in which ions are separated according to their mass to change ratio.
• The ions pass through velocity selector having magnetic field B1 and Then enter a second uniform magnetic field B.
• The ions move in a second region in a semicircle of radius r.
• The radius of semicircle can be calculated as

$$r = \dfrac{mv}{qB_2}$$

$$\Rightarrow\dfrac{r\,B_2}{v} = \dfrac{m}{q}$$

• Since, electron passing through velocity selector possess velocity

$$v = \dfrac{E}{B_1}$$

• So, mass to charge ratio will be

$$\dfrac{m}{q} = \dfrac{r\, B_2\, B_1}{E}$$

#### A charge of $$q= 1$$ $$\mu C$$ and mass $$m=1\,\mu g$$ , enters in a region of magnetic field at point P at  velocity v = 1 m / sec.  Find the combination of radius of semicircular region (r) and magnetic field (B) so, that it completes semicircular path and exits at point Q.

A $$r = 5\,m ,\, B =0.2\,T$$

B $$r= 10\,m ,\, B = 2\, T$$

C $$r = 5\,m ,\, B = 4\,T$$

D $$r = 3\, m ,\, B =0.2\,T$$

×

For the combination of r and B

$$r = \dfrac{mv}{qB}$$

$$\Rightarrow r= \dfrac{1× 10^{-6} × 1}{ 1× 10^{-6} × B}$$

$$\Rightarrow r= \dfrac{1}{B}$$       ....(1)

Since , equation (1) is satisfied only for option (A).

Hence, Option (A) is correct.

### A charge of $$q= 1$$ $$\mu C$$ and mass $$m=1\,\mu g$$ , enters in a region of magnetic field at point P at  velocity v = 1 m / sec.  Find the combination of radius of semicircular region (r) and magnetic field (B) so, that it completes semicircular path and exits at point Q.

A

$$r = 5\,m ,\, B =0.2\,T$$

.

B

$$r= 10\,m ,\, B = 2\, T$$

C

$$r = 5\,m ,\, B = 4\,T$$

D

$$r = 3\, m ,\, B =0.2\,T$$

Option A is Correct

# Calculation of  Radius and Time Period of Charged Particle in Uniform Circular Motion

• Consider a charged particle is projected perpendicular to the magnetic field with velocity  $$\vec v$$ , as shown in figure.
• The velocity of charged particle  is perpendicular to the magnetic field .
• Path of charged particle when $$\vec F_B$$ is perpendicular to $$\vec v$$

When magnetic forces is perpendicular to velocity ,then the path of particle is a circle .

• Magnetic force acting on charged particle when $$\vec v$$ is perpendicular to $$\vec B$$

Since, $$\vec v$$ is perpendicular to $$\vec B$$ .

Therefore , $$\theta = 90º$$

Magnetic force will be

$$F_B = q\,v\,B \,sin90º$$

$$F_B = q\,v\,B$$  .......(1)

• Radius of the circular motion :

For uniform circular motion ,

$$F = \dfrac{mv^2}{r}$$      ........(2)

Comparing equation (1) and (2)

$$q\,v\,B = \dfrac{mv^2}{r}$$         ......(3)

$$\Rightarrow r = \dfrac{mv}{qB}$$

where r is the radius of circular motion.

• Time period of circular motion :

Time period is the time taken by the particle to complete one revolution.

It is given by

Time period = $$\dfrac{circumference}{speed}$$

$$T = \dfrac{2\pi r}{v}$$

from equation (3)   $$r = \dfrac{mv}{qB}$$

$$T = \dfrac{2 \,\pi \,m\,v }{v\,q\,B}$$

$$T = \dfrac{2 \,\pi \,m }{q\,B}$$         ........(4)

where T is the time period of circular motion.

It is clear from equation (4) that the time period of circular motion is independent of speed of the charged particle and radius of the circular motion.

#### A charge of mass m and charge q , enters in uniform magnetic field B with velocity $$v$$ , if the speed of particle is doubled ,what will happen on the radius of curvature (r) ?

A Halved

B Doubled

C No Change

D None of These

×

The radius of curvature is given as

$$r = \dfrac{mv}{qB}$$

When velocity is doubled $$v\,' = 2v$$

$$r\,' = \dfrac{m(2v)}{qB} = 2\dfrac{mv}{qB}$$

$$r\,' = 2r$$

The radius of curvature gets doubled .

### A charge of mass m and charge q , enters in uniform magnetic field B with velocity $$v$$ , if the speed of particle is doubled ,what will happen on the radius of curvature (r) ?

A

Halved

.

B

Doubled

C

No Change

D

None of These

Option B is Correct

# Motion of Charged Particle in Electric Field and Magnetic field

• Consider a charged particle moving in a region where electric field $$(\vec E)$$ and magnetic field $$(\vec B)$$ are present.
• Due to magnetic field , force acts on the particle in the direction perpendicular to direction of velocity .
• Since, there is no displacement in the direction of force. So, due to this force work done is zero.

From work  energy theorem ,

$$W = \Delta K\cdot E =0$$

• Since, change in kinetic energy is also zero . So, change in speed is also zero.
• Only change in direction of particle takes place.
• The change in speed will be only due to the force of electric field because this force is dependent only on the direction of electric field for positive charge and opposite to direction of electric field for negative charge .

#### A particle of mass $$m$$ and charge $$q$$ , is moving in a region of magnetic field $$\vec B$$  and electric field  $$\vec E$$ , what will be the work done by $$\vec B$$ on the particle?

A Same as work done by Electric Field ($$W_E$$)

B $$\dfrac{1}{W_E}$$

C Zero

D $$2W_E$$

×

Since, there is no displacement in the direction of force due to magnetic field.

So, work done WB = 0

### A particle of mass $$m$$ and charge $$q$$ , is moving in a region of magnetic field $$\vec B$$  and electric field  $$\vec E$$ , what will be the work done by $$\vec B$$ on the particle?

A

Same as work done by Electric Field ($$W_E$$)

.

B

$$\dfrac{1}{W_E}$$

C

Zero

D

$$2W_E$$

Option C is Correct

# Cyclotron

• Cyclotron is a charged particle accelerator .
• The magnetic force bends the moving charge into a semicircular  path.
• The electric field accelerates the charges between two semicircular charge containers of D shape.

Note :- The containers of D shape should be hollow so that electric field inside remain zero as T remain constant.

• The accelerating electric field reverses its direction at the time the electrons finish their half cycle, so that they can be accelerated across the gap.

Cyclotron frequency

The time period to complete one orbit.

$$T = \dfrac{2\pi m}{qB}\,=\,constant$$

where, m = mass of charge q

We know   $$\omega = \dfrac{2\pi}{T}$$

$$\omega_{cyclotron} = \dfrac{qB}{m}$$

This frequency is independent of speed and radius.

• If the time taken by charged particle to describe a semicircle is equal to the time during which D1 and D2 change their polarity, the charged particle gets accelerated when it arrives in between the gaps.
• The electric field accelerates the charged particle further as its direction is along the velocity of charged particle.
• Once the charged particle is inside the dee D2, it now describes a greater semicircle in same time with increased speed due to the magnetic field. This process continues and the ion goes on describing a circular path of greater radius and finally acquires high energy.

#### Determine the cyclotron frequency of a charged particle having mass $$m\,=\,1\,\mu g$$  and charge $$q = 2$$ $$\mu C$$. Given the value of magnetic field $$B = 0·2 \,T$$.

×

The frequency of cyclotron  is given as

$$\omega_{cyclotron} = \dfrac{qB}{m}$$

Where, B = magnetic field

m = mass of charge q

Given : $$m\,=\,1\,\mu g$$ ,  $$q=2\,\mu C$$ , $$B = 0·2\,T$$

$$\omega _{cyclotron} = \dfrac{qB}{m}$$

$$=$$ $$\dfrac{2× 10^{-6}}{1× 10^{-6}} × 0.2$$

$$=$$ $$0·4\, rad/ sec$$

### Determine the cyclotron frequency of a charged particle having mass $$m\,=\,1\,\mu g$$  and charge $$q = 2$$ $$\mu C$$. Given the value of magnetic field $$B = 0·2 \,T$$.

A

.

B

C

D

Option B is Correct

# Maximum Kinetic Energy of the Ion in Cyclotron

• Cyclotron is a charge particle accelerator .
• The magnetic force bends the moving charge  into a semicircular  path.
• The electric field accelerates the charges between two semicircular charge containers of D shape.
• The accelerating electric field reverses its direction at the time the electrons finish their half cycle, so that they can be accelerated across the gap.

Cyclotron frequency:

• The time period to complete one orbit.

$$T = \dfrac{2\pi m}{qB}$$   where m is mass of charge q.

•   We know  $$\omega = \dfrac{2\pi}{T}$$

$$\omega_{cyclotron} = \dfrac{qB}{m}$$

This frequency is independent of speed and radius.

• If the time taken by charged particle to describe a semicircle is equal to the time during which D1 and D2 change their polarity, the charged particle gets accelerated when it arrives in between the gaps.
• The electric field accelerates the charged particle further as its direction is along the velocity of charged particle.
• Once the charged particle is inside the dee D2, it now describes a greater semicircle in same time with increase speed due to the magnetic field. This process continues and the ion goes on describing a circular path of greater radius and finally acquires high energy.

• Radius of curvature $$r= \dfrac{mv}{qB}$$
• Velocity of charged particle $$v = \dfrac{qBr}{m}$$
• Kinetic energy is given by

$$K = \dfrac{1}{2} mv^2$$

$$K = \dfrac{q^2B^2r^2}{2m}$$

• At the time of exit of particle,

$$\ r=R$$

where $$R$$ is the radius of semicircle .

So,    $$K_{max} = \dfrac{q^2B^2R^2}{2m}$$

#### Find the maximum kinetic energy of the ion having charge $$q= 1\,\mu C$$ and mass $$m = 1\,\mu g$$ in a cyclotron having magnetic field $$B = 0·2\,T$$ and radius of dees is $$R = 20\,cm$$.

A $$100\,pJ$$

B  $$300\,pJ$$

C  $$800\,pJ$$

D $$600 \,pJ$$

×

Maximum Kinetic energy $$K_{max} = \dfrac{q^2B^2R^2}{2m}$$

where B = Magnetic field

R = Radius of dees

q = Charge

m = Mass of charge

Given : $$m = 1\,\mu g$$ , $$q= 1\,\mu C$$$$B = 0.2\,T$$$$R= 20 \,cm$$

$$K _{max} = \dfrac{(0.2)^2 × (0.2)^2 × 1×10^{-6} × 1×10^{-6} }{2× 1×10^{-6} }$$

$$K_{max} = 8× 10^{-10} J$$

$$K_{max} = 800× 10^{-12} J$$

$$K_{max} = 800\,pJ$$

### Find the maximum kinetic energy of the ion having charge $$q= 1\,\mu C$$ and mass $$m = 1\,\mu g$$ in a cyclotron having magnetic field $$B = 0·2\,T$$ and radius of dees is $$R = 20\,cm$$.

A

$$100\,pJ$$

.

B

$$300\,pJ$$

C

$$800\,pJ$$

D

$$600 \,pJ$$

Option C is Correct