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}\$

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>>Class 12
>>Physics
>>Moving Charges and Magnetism
>>Motion of Charged Particle Combined in Magnetic Field and Electric Fields
>> $Force on a moving charge in magnetic and...$

Force on a moving charge in magnetic and electric fields

Q: 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.

Q: 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

Q: 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\$

Q: 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}\$

Q: 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

Q: 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

Q: 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)

Q: 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.

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Important Questions

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?

Medium

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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 :

Hard

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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

Medium

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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

Hard

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A beam of protons with a velocity $4 \times 1 0^{5} m s^{- 1}$ enters a uniform magnetic field of 0.3 T at an angle of $6 0^{o}$ 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 $= 1.67 \times 1 0^{- 27} k g$

Medium

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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

Medium

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

Medium

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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.

Medium

NEET

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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.

This question has multiple correct options

Medium

View solution

In a cyclotron, a charged particle

Medium

View solution

Force on a moving charge in magnetic and electric fields

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REVISE WITH CONCEPTS

Motion in the Presence of Electric and Magnetic Field

QUICK SUMMARY WITH STORIES

Motion in the Presence of Electric and Magnetic Field

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.