Question

Notify Me Bookmark

Electrons (mass m, charge −e) are accelerated from rest through a potential difference V and are then deflected by a magnetic field B→ that is perpendicular to their velocity. The radius of the resulting electron trajectory is: (2eV/m)/B B2eV/m (2mV/e)/B B2mV/e none of these

Electrons (mass m, charge −e) are accelerated from rest through a potential difference V and are then deflected by a magnetic field B→ that is perpendicular to their velocity. The radius of the resulting electron trajectory is: (2eV/m)/B B2eV/m (2mV/e)/B B2mV/e none of these

Image text
Electrons (mass m , charge − e ) are accelerated from rest through a potential difference V and are then deflected by a magnetic field B → that is perpendicular to their velocity. The radius of the resulting electron trajectory is: ( 2 e V / m ) / B B 2 e V / m ( 2 m V / e ) / B B 2 m V / e none of these

Detailed Answer

0 0

Please enter your email here. You will get email when this question is answered by our tutor.

Doubtrix Logo

Problem is not yet solved!

Click on Notify me to get email when question is answered.

Join Doubtrix with 7-days free trial

  • Icon Correct & Verified Answers
  • Icon No AI-Generated Answers
  • Icon Video Solution for Difficult Questions
  • Icon Work With Experts to Reach at Correct Answers
Notify me Login Sign Up

Similar/Related Questions

An a-particle has a charge of +2 e and a mass of 6.64×10−27 kg. It is accelerated from rest through a potential difference that has a value of 1.93×106 V and then enters a uniform magnetic field whose magnitude is 2.29 T. The a-particle moves perpendicular to the magnetic field at all times. What is (a) the speed of the a-particle, (b) the magnitude of the magnetic force on it, and (c) the radius of its circular path? (a) Number Units (b) Number Units (c) Number Units
Solution
0 0

A 150 V battery is connected across two parallel metal plates of area 28.5 cm2 and separation 9.20 mm. A beam of alpha particles (charge +2e, mass 6.64×10−27 kg ) is accelerated from rest through a potential difference of 1.75 kV and enters the region between the plates perpendicular to the electric field, as shown in (Figure 1). Figure 1 of 1 Part A What magnitude of magnetic field is needed so that the alpha particles emerge undeflected from between the plates? Express your answer with the appropriate units. Submit Request Answer Part B What is the direction of this magnetic field? The magnetic field is directed into the page. The magnetic field is directed out of the page. The magnetic field is directed upward. The magnetic field is directed downward. Submit Request Answer Request Answer

Solution
0 0

An ion with a mass m and a magnitude of charge of 1 e is initially at rest when it is accelerated by a potential difference of magnitude ΔV. After exiting the accelerating potential, it enters a region with a uniform magnetic field perpendicular to the ion's velocity. The ion moves in a semicircle of radius R. Next, an ion with a mass m′ and a magnitude of charge of 4e is accelerated by the same potential difference and deflected by the same magnetic field. The radius of its semicircular path is R′ = 3R. Find the ratio of the ion masses, m′/m. m′/m =
Solution
0 0