Cyclotron
Trending Questions
Q. A cyclotron is used to
- Charge subatomic particles
- Accelerate charged particles
- Heat up subatomic particles.
- Increase the mass of particles
Q. In cyclotron we apply electric field(→E) & magnetic field(→B) to accelerate the charged particles or ions, where →E & →B
- are parallel
- are applied at some angle other than 90∘
- are applied in perpendicular directions
- none of the above
Q.
Fe+ ions are accelerated through a potential difference of 500 V and are injected normally into a homogeneous magnetic field B of strength 20.0 mT. Find the radius of the circular paths followed by the isotopes with mass numbers 57 and 58. Take the mass of an ion = A (1.6×10−27) kg where A is the mass number.
Q. (a) A monoenergetic electron beam with electron speed of 5.20 × 106m/s is subject to a magnetic field of 1.30 × 10–4 Tnormal to the beam velocity. What is the radius of the circle tracedby the beam, given e/m for electron equals 1.76 × 1011C /kg (b) Is the formula you employ in (a) valid for calculating radius ofthe path of a 20 MeV electron beam? If not, in what way is itmodified?[Note: Exercises 11.20(b) and 11.21 (b) take you to relativisticmechanics which is beyond the scope of this book. They have beeninserted here simply to emphasise the point that the formulas youuse in part (a) of the exercises are not valid at very high speeds orenergies. See answers at the end to know what ‘very high speed orenergy’ means.]
Q. (a) Estimate the speed with which electrons emitted from a heatedemitter of an evacuated tube impinge on the collector maintainedat a potential difference of 500 V with respect to the emitter.Ignore the small initial speeds of the electrons. Thespecific charge of the electron, i.e., its e/m is given to be1.76 × 1011 C kg–1. (b) Use the same formula you employ in (a) to obtain electron speedfor an collector potential of 10 MV. Do you see what is wrong ? Inwhat way is the formula to be modified?
Q. The cyclotron frequency of an electrons gyrating in a magnetic field of 1T is approximately
- 28MHz
- 280MHz
- 2.8MHz
- 28GHz
Q. Assertion :If a proton and an α-particle enter a uniform magnetic field perpendicularly with the same speed, the time period of revolution of α-particle is double than that of proton. Reason: In a magnetic field, the period of revolution of a charged particle is directly proportional to the mass of the particle and inversely proportional to the charge of particle.
- Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
- Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
- Assertion is correct but Reason is incorrect
- Assertion is incorrect but Reason is correct
Q. The region between x=0 and x=L is filled with uniform, steady magnetic field B0^k. A particle of mass m, positive charge q and velocity v0^i travels along x-axis and enters the region of magnetic field. Neglect gravity throughout the question. The field now extends up to x=2.1L If the time spent by the particle in the magnetic field is T=πmX×qB0. Find X?
Q. Find the time required for it to make half of a revolution.
Q. A proton is projected with a velocity 107ms−1 , at right angles to a uniform magnetic field of induction 100 mT. The time (in seconds ) taken by the proton to traverse 900 arc is
(Mass of proton =1.65×10−27 kg and charge of proton =1.6×10−19C )
(Mass of proton =1.65×10−27 kg and charge of proton =1.6×10−19C )
- 2.43×10−7
- 1.62×10−7
- 3.24×10−7
- 0.81×10−7