
Under the influence of a uniform magnetic field, a charged particle moves with constant speed v in a circle of radius R. The time period of rotation of the particle
(A) Depends on R and not on v
(B) Is independent of both v and R
(C) Depends on both v and R
(D) Depends on v and not on R
Answer
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Hint:A cyclotron is a type of tiny particle accelerator that generates radioactive isotopes for imaging purposes. Stable, non-radioactive isotopes are fed into the cyclotron, which uses a magnetic field to accelerate charged particles (protons) to high energies. When stable isotopes interact with the particle beam, a nuclear reaction occurs between the protons and the target atoms, resulting in radioactive isotopes for use in nuclear medicine and other applications.
Formula used
\[\mathbf{T}=\dfrac{\mathbf{2\pi m}}{qB}\]
where B is the magnetic field intensity, q is the charged particle's electric charge, and m is the charged particle's relativistic mass.
Complete step by step solution:
A cyclotron uses a high-frequency alternating voltage delivered between two hollow "D"-shaped sheet metal electrodes called "dees" within a vacuum chamber to accelerate a charged particle beam. The dees are arranged face to face with a little gap between them, forming a cylindrical area within which the particles can travel. The particles are injected into this space's centre. The dees are situated between the poles of a huge electromagnet that generates a perpendicular static magnetic field B across the electrode plane. Due to the Lorentz force perpendicular to the particles' direction of travel, the magnetic field forces their path to bend in a circle.
The time it takes for one complete cycle of vibration to pass a specific location is called a time period (abbreviated as 'T'). The time period of a wave reduces as the frequency of the wave rises. 'Seconds' is the unit of time measurement. The reciprocal connection between frequency and time period can be represented mathematically as \[T=\dfrac{1}{f}\text{ or }f=\dfrac{1}{T}\].
\[\mathbf{T}=\dfrac{\mathbf{2\pi m}}{qB}\] is the period of an electron travelling on a helical orbit with an axis perpendicular to the magnetic field, where m is the mass of the electron, q is the charge of the electron, and B is the magnetic field in the region.
When a particle moves with a speed of v on the plane of paper and moves in a magnetic field of strength B pointing downwards into the page:
$\mathrm{F}=\mathrm{qvB}$ (This is the Force of a charged particle that is present in a magnetic field)
we know that $\mathrm{F}=\dfrac{\mathrm{mv}^{2}}{\mathrm{r}} \quad(\mathrm{r}$ is the radius of motion and $\mathrm{m}$ is mass of particle)
$\Rightarrow \mathrm{qvB}=\dfrac{\mathrm{mv}^{2}}{\mathrm{r}}$
$\Rightarrow \mathrm{r}=\dfrac{\mathrm{mv}}{\mathrm{Bq}}$
as we know that $\omega=\dfrac{v}{r}$
$\Rightarrow \omega=\dfrac{\mathrm{Bq}}{\mathrm{m}}$
Time period, $\mathrm{T}=\dfrac{2 \pi}{ \omega}$
$\Rightarrow \mathrm{T}=\dfrac{2 \pi \mathrm{m}}{\mathrm{Bq}}$
And this shows that it is independent of both radius and velocity.
Hence option B is correct.
Note:
Under the influence of the magnetic field, the particles would travel in a circular route within the dees if their speeds were constant. Between the dees, however, a radio frequency (RF) alternating voltage of several thousand volts is supplied. The voltage accelerates the particles by creating an oscillating electric field in the space between the dees. The frequency is chosen so that the particles form a single circuit during each voltage cycle. The frequency must match the particle's cyclotron resonance frequency to do this.
Formula used
\[\mathbf{T}=\dfrac{\mathbf{2\pi m}}{qB}\]
where B is the magnetic field intensity, q is the charged particle's electric charge, and m is the charged particle's relativistic mass.
Complete step by step solution:
A cyclotron uses a high-frequency alternating voltage delivered between two hollow "D"-shaped sheet metal electrodes called "dees" within a vacuum chamber to accelerate a charged particle beam. The dees are arranged face to face with a little gap between them, forming a cylindrical area within which the particles can travel. The particles are injected into this space's centre. The dees are situated between the poles of a huge electromagnet that generates a perpendicular static magnetic field B across the electrode plane. Due to the Lorentz force perpendicular to the particles' direction of travel, the magnetic field forces their path to bend in a circle.
The time it takes for one complete cycle of vibration to pass a specific location is called a time period (abbreviated as 'T'). The time period of a wave reduces as the frequency of the wave rises. 'Seconds' is the unit of time measurement. The reciprocal connection between frequency and time period can be represented mathematically as \[T=\dfrac{1}{f}\text{ or }f=\dfrac{1}{T}\].
\[\mathbf{T}=\dfrac{\mathbf{2\pi m}}{qB}\] is the period of an electron travelling on a helical orbit with an axis perpendicular to the magnetic field, where m is the mass of the electron, q is the charge of the electron, and B is the magnetic field in the region.
When a particle moves with a speed of v on the plane of paper and moves in a magnetic field of strength B pointing downwards into the page:
$\mathrm{F}=\mathrm{qvB}$ (This is the Force of a charged particle that is present in a magnetic field)
we know that $\mathrm{F}=\dfrac{\mathrm{mv}^{2}}{\mathrm{r}} \quad(\mathrm{r}$ is the radius of motion and $\mathrm{m}$ is mass of particle)
$\Rightarrow \mathrm{qvB}=\dfrac{\mathrm{mv}^{2}}{\mathrm{r}}$
$\Rightarrow \mathrm{r}=\dfrac{\mathrm{mv}}{\mathrm{Bq}}$
as we know that $\omega=\dfrac{v}{r}$
$\Rightarrow \omega=\dfrac{\mathrm{Bq}}{\mathrm{m}}$
Time period, $\mathrm{T}=\dfrac{2 \pi}{ \omega}$
$\Rightarrow \mathrm{T}=\dfrac{2 \pi \mathrm{m}}{\mathrm{Bq}}$
And this shows that it is independent of both radius and velocity.
Hence option B is correct.
Note:
Under the influence of the magnetic field, the particles would travel in a circular route within the dees if their speeds were constant. Between the dees, however, a radio frequency (RF) alternating voltage of several thousand volts is supplied. The voltage accelerates the particles by creating an oscillating electric field in the space between the dees. The frequency is chosen so that the particles form a single circuit during each voltage cycle. The frequency must match the particle's cyclotron resonance frequency to do this.
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