A charged particle enters at right angles into a uniform magnetic field as shown:What should be the nature of charge on the particle if it begins to move in a direction pointing vertically out of the page due to its interaction with the magnetic field?

According to the Fleming's left-hand rule, the charged particle should be a positively charged in order to experience the force in the direction out of the page after interaction with the magnetic field.

When a proton is released from rest in room, it starts with an initial acceleration \${ a }_{ 0 }\$ towards west When it is projected towards north with a speed \${ V }_{ 0 }\$ it moves with an initial acceleration \${ 3a }_{ 0 }\$ towards west. The electric and magnetic fields in the room are :

Initial acceleration \$={ a }_{ 0 }\$when a proton is released from rest in a room it starts with an initial acceleration \${ a }_{ 0 }\$ towards west.\${ F }_{ 1 }=\$ electric force acting on the proton in west direction\$q=\$ charge on proton\$E=\$ electric field in the region in west direction\$m=\$ mass of protonelectric force on the proton can be given on \${ F }_{ 1 }=qE\$Using Newton's second law\${ F }_{ 1 }={ ma }_{ 0 }\$hence,\${ ma }_{ 0 }=qE\$\$E={ ma }_{ 0 }/q\$ Along westCase -2\${ V }_{ 0 }=\$ initial velocity of the particle\${ F }_{ 2 }=\$ magnetic force acting on the proton in west direction electric force on proton \$=3{ ma }_{ 0 }\$\${ F }_{ 1 }+{ F }_{ 2 }=3{ ma }_{ 0 }\$\${ ma }_{ 0 }+{ F }_{ 2 }=3{ ma }_{ 0 }\$maximum magnetic force is given by \${ F }_{ 2 }=q{ V }_{ 0 }B\$So, \$q{ V }_{ 0 }B=2{ ma }_{ 0 }\$\$B=2{ ma }_{ 0 }/q{ V }_{ 0 }\$ (down).Option C

A proton and an \$\alpha \$ -particle enter a uniform magnetic field perpendicularly with the same speed. If proton takes 25 \$\mu\$ second to make 5 revolutions, then times period for the \$\alpha\$-particle would be

Time taken by proton to make one revolution \$= \displaystyle \frac{25}{5} = 5 \mu sec\$.As, \$\displaystyle T = \frac{2 \pi m}{qB} \$So, \$\displaystyle \frac{T_2}{T_1} = \frac{m_2}{m_1} \times \frac{q_1}{q_2}\$\$\displaystyle T_2 = T_1 \frac{m_2 q_1}{m_1 q_2} = \frac{5 \times 4 m_1}{m_1} \times \frac{q}{2q} = 10 \mu sec.\$

A particle of charge \$q\$ and mass \$m\$ starts moving from the origin under the action of an electric field \$\overrightarrow{E}=E_o\hat{i}\$ and \$\overrightarrow{B}=B_0\hat{i}\$ with a velocity \$\overrightarrow{v}=v_0\hat{j}\$. The speed of the particle will become \$2v_0\$ after a time

\$\overrightarrow{E}=E_0\hat{i} \parallel \overrightarrow{B}=B_0\hat{i}\$\$\overrightarrow{v}=v_0\hat{j} \perp \overrightarrow{E}\$ \$\overrightarrow{v} \perp \overrightarrow{B}\$Hence, the path of the particle is a helix with speed remaining constant for the circular path in the \$yz\$ plane.Magnitude of velocity of particle at any time \$t:\$ \$v= \sqrt{v_x^2 + v_y^2 + v_z^2}=\sqrt{ (a_xt)^2 +v_0^2}\$ where \$v_y^2 + v_z^2 = v_0^2\$Given \$v= 2v_0\$\$ \Rightarrow (2v_0)^2 = v_x^2 + v_0^2\$\$ \Rightarrow (v_x)^2 = 3(v_0)^2\$\$ \Rightarrow v_x = \sqrt {3} v_0\$\$ \Rightarrow a_xt = \sqrt{3}v_0\$\$ \Rightarrow (\dfrac{qE}{m})t= \sqrt{3}v_0\$\$ \Rightarrow t= \dfrac{\sqrt{3}v_0m}{qE}\$

A beam of protons with a velocity \$4 \times 10^5 ms^{-1}\$ enters a uniform magnetic field of 0.3 T at an angle of \$60^o\$ to the magnetic field. Find the pitch of the helix (which is the distance travelled by a proton in the beam parallel to the magnetic field during one period of the rotation). Mass of the proton \$= 1.67 \times 10^{-27} kg\$

When a charged particle is projected at an angle \$\theta\$ to a magnetic field, the component of velocity parallel to the field is \$v\cos \theta\$ while perpendicular to the field is \$v\sin \theta\$, so the particle will move in a circle of radius\$r = \dfrac{m(v\sin \theta)}{qB}=\dfrac{( 1.67 \times 10^{-27}) \times \left( 4 \times 10^{5} \times \sin 60^0 \right)}{1.6 \times 10^{-19} \times 0.3}=\dfrac{ \left( 1.67 \times 10^{-27} \right) \times \left( 4 \times 10^{5} \times \dfrac{\sqrt{3}}{2} \right)}{1.6 \times 10^{-19} \times 0.3}=\dfrac{2 \times 10^{-2}}{\sqrt{3}}\$Time period: \$T = \dfrac{2\pi r}{vsin\theta}= \dfrac{2\pi \times \dfrac{2 \times 10^{-2}}{\sqrt{3}}}{4 \times 10^5 \times \sin 60^0}= \dfrac{2\pi \times \dfrac{2 \times 10^{-2}}{\sqrt{3}}}{4 \times 10^5 \times \dfrac{\sqrt{3}}{2}}=\dfrac{2\pi}{3} \times 10^{-7}\$Pitch: \$P = v \cos \theta T= (4\times 10^5) \times \cos 60^0 \times \dfrac{2\pi}{3} \times 10^{-7}=\dfrac{4\pi}{3}\times 10^{-2}= 4.35 \times 10^{-2}\: m\$

A charged particle enters into a uniform magnetic field with velocity vector at an angle of \$45^o\$ with the magnetic field. The pitch of the helical path followed by the particle is \$p\$. The radius of the helix will be

Pitch of the helix is product of velocity component along the magnetic field and the time period \$T\$.Pitch: \$p= \dfrac{2\pi m}{qB}v \cos \theta=\dfrac{2\pi P\cos \theta}{qB}=\dfrac{2\pi P\cos 45^0}{qB}\$Here \$P\$ denotes momentumRadius of the helix\$:\ r =\dfrac{mv\sin \theta}{qB}=\dfrac{mv\sin 45^0}{qB}=\dfrac{P\sin 45^0}{qB}\$\$ \therefore p =2\pi r\$\$\Rightarrow r =\dfrac{p}{2\pi}\$

The time period of the charged particle in cyclotron is \$T\$ and maximum speed is \$V\$, then

Solution: As we know that,frequency of revolution of particle in cyclotron is,\$\nu_c=\dfrac{qB}{2\pi m}\$and time period is,\$T=\dfrac{1}{\nu_c}\$So,\$T=\dfrac{2\pi m}{qB}\$ \$...(1)\$\$eq^n(1)\$ is clearly depict thattime period is independent of velocity of particle.hence,The correct opt: D

An electron is moving in a circular path under the influence of a transverse magnetic field of \$3 \cdot 57 \times 10^{-2}T\$. If the value of e/m is \$1 \cdot 76 \times 10^{11}\$ C/kgg. the frequency of revolution of the electron is.

Given : \$\dfrac{q}{m} = 1.76 \times 10^{11}\$ \$C/kg\$ \$B = 3.57 \times 10^{-2}\$ \$T\$ Frequency of revolution \$\nu = \dfrac{Bq}{2\pi m}\$\$\therefore\$ \$\nu = \dfrac{3.57 \times 10^{-2} \times 1.76 \times 10^{11}}{2\pi} = 1\times 10^9\$ Hz \$ =1\$ GHz

This section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d ) out of which ONE OR MORE may be correct.A proton moving with a constant velocity passes through a region of space without any change in its velocity. If E and B represent the electric and magnetic fields respectively, Identify which of the following options can be true for the space.

A proton moving through a space contain electric and magnetic field will experience a force \$F=q(\vec {E} +\vec {v}\times\vec {B})\$If the proton passes through the space without changing its velocity then two cases can be possible:i) either \$E= 0\$ and \$B= 0\$. (Trivial case)ii) or, \$E\neq 0\$ and \$B \neq 0 \$ (motion of the proton (v) is perpendicular to the B)

In a cyclotron, a charged particle

In a cyclotron, charged particle experiences coulombic force(force due to electric field) between the Dees and magnetic force ( force due to magnetic field ) while circulating inside the Dees. In total, it always experiences a force.

Cyclotron | Definition, Examples, Diagrams

definition

Describe the limitations of cyclotron

Limitations of Cyclotron are as follows:

Cyclotron cannot accelerate uncharged particles like neutrons.
Cyclotron cannot accelerate electrons because of its small mass.
It cannot accelerate positively charged particles with large mass due to relativistic effect.

definition

Cyclotron

The cyclotron is a machine to accelerate charged particles or ions to high energies.The cyclotron uses both electric and magnetic fields in combination to increase the energy of charged particles.As the fields are perpendicular to each other they are called crossed fields. Cyclotron uses the fact that the frequency of revolution of the charged particle in a magnetic field is independent of its energy. The particles move most of the time inside two semicircular disc-like metal containers, D1 and D2, which are called dees as they look like the letter D.Inside the metal boxes the particle is shielded and is not acted on by the electric field. The magnetic field, however, acts on the particle and makes it go round in a circular path inside a dee. Every time the particle moves from one dee to another it is acted upon by the electric field. The sign of the electric field is changed alternately in tune with the circular motion of the particle.This ensures that the particle is always accelerated by the electric field.Each time the acceleration increases the energy of the particle. As energy increases, the radius of the circular path increases. So the path is a spiral one.

definition

Cyclotron frequency

The cyclotron frequency or gyrofrequency is the frequency of a charged particle moving perpendicular to the direction of a uniform magnetic field $B$ (constant magnitude and direction).

v_{c} = \frac{q B}{2 π m}

T = \frac{1}{v _{c}}

where

m

is the particle mass,

q

its charge and

B

the magnetic field,

T

is the time period and

v_{c}

is the cyclotron frequency.

definition

Uses of cyclotron

The cyclotron is used to bombard nuclei with energetic particles, so accelerated by it, and study the resulting nuclear reactions. It is also used to implant ions into solids and modify their properties or even synthesise new materials. It is used in hospitals to produce radioactive substances which can be used in diagnosis and treatment.

Cyclotron

definition

Describe the limitations of cyclotron

definition

Cyclotron

definition

Cyclotron frequency

definition

Uses of cyclotron

REVISE WITH CONCEPTS

Motion in the Presence of Electric and Magnetic Field

Quick Summary With Stories

Application and Limitations of a Cyclotron

A charged particle enters at right angles into a uniform magnetic field as shown:
What should be the nature of charge on the particle if it begins to move in a direction pointing vertically out of the page due to its interaction with the magnetic field?

When a proton is released from rest in room, it starts with an initial acceleration $a_{0}$ towards west When it is projected towards north with a speed $V_{0}$ it moves with an initial acceleration $3 a_{0}$ towards west. The electric and magnetic fields in the room are :

A proton and an $α$ -particle enter a uniform magnetic field perpendicularly with the same speed. If proton takes 25 $μ$ second to make 5 revolutions, then times period for the $α$ -particle would be

A particle of charge $q$ and mass $m$ starts moving from the origin under the action of an electric field $E = E_{o} \hat{i}$ and $B = B_{0} \hat{i}$ with a velocity $v = v_{0} \hat{j}$ . The speed of the particle will become $2 v_{0}$ after a time

A charged particle enters into a uniform magnetic field with velocity vector at an angle of $4 5^{o}$ with the magnetic field. The pitch of the helical path followed by the particle is $p$ . The radius of the helix will be

The time period of the charged particle in cyclotron is $T$ and maximum speed is $V$ , then

An electron is moving in a circular path under the influence of a transverse magnetic field of $3 \cdot 57 \times 1 0^{- 2} T$ . If the value of e/m is $1 \cdot 76 \times 1 0^{11}$ C/kgg. the frequency of revolution of the electron is.

In a cyclotron, a charged particle

definition

Describe the limitations of cyclotron

definition

Cyclotron

definition

Cyclotron frequency

definition

Uses of cyclotron

REVISE WITH CONCEPTS

Motion in the Presence of Electric and Magnetic Field

Quick Summary With Stories

Application and Limitations of a Cyclotron

A charged particle enters at right angles into a uniform magnetic field as shown: What should be the nature of charge on the particle if it begins to move in a direction pointing vertically out of the page due to its interaction with the magnetic field?

When a proton is released from rest in room, it starts with an initial acceleration a0​ towards west When it is projected towards north with a speed V0​ it moves with an initial acceleration 3a0​ towards west. The electric and magnetic fields in the room are :

A proton and an α -particle enter a uniform magnetic field perpendicularly with the same speed. If proton takes 25 μ second to make 5 revolutions, then times period for the α-particle would be

A particle of charge q and mass m starts moving from the origin under the action of an electric field E=Eo​i^ and B=B0​i^ with a velocity v=v0​j^​. The speed of the particle will become 2v0​ after a time

A charged particle enters into a uniform magnetic field with velocity vector at an angle of 45o with the magnetic field. The pitch of the helical path followed by the particle is p. The radius of the helix will be

The time period of the charged particle in cyclotron is T and maximum speed is V, then

An electron is moving in a circular path under the influence of a transverse magnetic field of 3⋅57×10−2T. If the value of e/m is 1⋅76×1011 C/kgg. the frequency of revolution of the electron is.

In a cyclotron, a charged particle

A charged particle enters at right angles into a uniform magnetic field as shown:
What should be the nature of charge on the particle if it begins to move in a direction pointing vertically out of the page due to its interaction with the magnetic field?

When a proton is released from rest in room, it starts with an initial acceleration $a_{0}$ towards west When it is projected towards north with a speed $V_{0}$ it moves with an initial acceleration $3 a_{0}$ towards west. The electric and magnetic fields in the room are :

A proton and an $α$ -particle enter a uniform magnetic field perpendicularly with the same speed. If proton takes 25 $μ$ second to make 5 revolutions, then times period for the $α$ -particle would be

A particle of charge $q$ and mass $m$ starts moving from the origin under the action of an electric field $E = E_{o} \hat{i}$ and $B = B_{0} \hat{i}$ with a velocity $v = v_{0} \hat{j}$ . The speed of the particle will become $2 v_{0}$ after a time

A charged particle enters into a uniform magnetic field with velocity vector at an angle of $4 5^{o}$ with the magnetic field. The pitch of the helical path followed by the particle is $p$ . The radius of the helix will be

The time period of the charged particle in cyclotron is $T$ and maximum speed is $V$ , then

An electron is moving in a circular path under the influence of a transverse magnetic field of $3 \cdot 57 \times 1 0^{- 2} T$ . If the value of e/m is $1 \cdot 76 \times 1 0^{11}$ C/kgg. the frequency of revolution of the electron is.