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

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 \(\vec E\)

- 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 B
_{1}and Then enter a second uniform magnetic field B_{2 }. - 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 \(r = 5\,m ,\, B =0.2\,T\)

B \(r= 10\,m ,\, B = 2\, T\)

C \(r = 5\,m ,\, B = 4\,T \)

D \(r = 3\, m ,\, B =0.2\,T\)

- 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 D
_{1}and D_{2}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 D
_{2,}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.

A 0·8 rad/sec

B 0·4 rad/sec

C 1 rad/sec

D 0·7 rad/sec

- 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 D
_{1}and D_{2}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 D
_{2,}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, \(K_{max} = \dfrac{q^2B^2R^2}{2m}\)

A \(100\,pJ\)

B \(300\,pJ\)

C \(800\,pJ\)

D \(600 \,pJ\)

- 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 Halved

B Doubled

C No Change

D None of These

- 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 Same as work done by Electric Field (\(W_E\))

B \(\dfrac{1}{W_E}\)

C Zero

D \(2W_E\)