Home
Class 12
PHYSICS
In Thomson's experiment for determining ...

In Thomson's experiment for determining `e//m` the potential different between the cathod and the anod (in accelerating coloumns ) is same as that between the deflecting plates (in the regions of crossed field ) . If the`p.d` is doubled , by what factor should the magnetic field be increased to ensure that the electron beam remains undeflected

Text Solution

Verified by Experts

`1/2m upsilon^(2)=V.q, 1/2m (E/B)^(2)= V.q,1/2mV^(2)/(d^(2)B^(2))=V.q`
`(q)/(m) = (V)/(2d^(2)B^(2)), B^(2) alpha V rarr B alpha sqrt(V), B^(1) alpha sqrt(2V)`
`B^(1) = sqrt(2)B`
Promotional Banner

Topper's Solved these Questions

Similar Questions

Explore conceptually related problems

Both, the electric and magnetic fields can deflect a moving electron. What is the difference between these deflections?

The figure shows the schematic of an absolute electrometer. The potential difference that is to be measured is applied between the plates 1 and 2 , with the upper plate connected to one arm of a balance beam.* The pan connected to the other arm is loaded with weights until balance is achieved, that is, when the upper plate begins to move upward. In this way the force acting between the charged plates is measured, and this enables one to determine the magnitude of the potential di fference between the plates. It the aquiltbrium in the electrometer stable or unstable? • The figure does not show the protecting rings around plates 1 and 2 with the same potentials. Theso are used to ensure that the field is as uniform as possible

In an experiment to measure the charge on an electrons , the electrons travelling at a speed of 1.825 xx 10^(7) m/s are subjected to a deflecting electric field 3.2 xx 10^(4) V/m . The intensity of magnetic field applied so that the beam remains undeflected is

In photoelectric effect experiment potential difference between plates increases keeping incident light on cathode plate remains unchanged which of the following is correct about saturation current :

Following experiment was performed by J.J. Thomson in order to measure ratio of charge e and mass m of electron. Electrons emitted from a hot filament are accelerated by a potential difference V. As the electrons pass through deflecting plates, they encounter both electric and magnetic fields. the entire region in which electrons leave the plates they enters a field free region that extends to fluorescent screen. The entire region in which electrons travel is evacuated. Firstly, electric and magnetic fields were made zero and position of undeflected electron beam on the screen was noted. The electric field was turned on and resulting deflection was noted. Deflection is given by d_(1) = (eEL^(2))/(2mV^(2)) where L = length of deflecting plate and v = speed of electron. In second part of experiment, magnetic field was adjusted so as to exactly cancel the electric force leaving the electron beam undeflected. This gives eE = evB . Using expression for d_(1) we can find out (e)/(m) = (2d_(1)E)/(B^(2)L^(2)) A beam of electron with velocity 3 xx 10^(7) m s^(-1) is deflected 2 mm while passing through 10 cm in an electric field of 1800 V//m perpendicular to its path. e//m for electron is

Following experiment was performed by J.J. Thomson in order to measure ratio of charge e and mass m of electron. Electrons emitted from a hot filament are accelerated by a potential difference V. As the electrons pass through deflecting plates, they encounter both electric and magnetic fields. the entire region in which electrons leave the plates they enters a field free region that extends to fluorescent screen. The entire region in which electrons travel is evacuated. Firstly, electric and magnetic fields were made zero and position of undeflected electron beam on the screen was noted. The electric field was turned on and resulting deflection was noted. Deflection is given by d_(1) = (eEL^(2))/(2mV^(2)) where L = length of deflecting plate and v = speed of electron. In second part of experiment, magnetic field was adjusted so as to exactly cancel the electric force leaving the electron beam undeflected. This gives eE = evB. Using expression for d_(1) we can find out (e)/(m) = (2d_(1)E)/(B^(2)L^(2)) If the electron is deflected downward when only electric field is turned on, in what direction do the electric and magnetic fields point in second part of experiment

Following experiment was performed by J.J. Thomson in order to measure ratio of charge e and mass m of electron. Electrons emitted from a hot filament are accelerated by a potential difference V. As the electrons pass through deflecting plates, they encounter both electric and magnetic fields. the entire region in which electrons leave the plates they enters a field free region that extends to fluorescent screen. The entire region in which electrons travel is evacuated. Firstly, electric and magnetic fields were made zero and position of undeflected electron beam on the screen was noted. The electric field was turned on and resulting deflection was noted. Deflection is given by d_(1) = (eEL^(2))/(2mV^(2)) where L = length of deflecting plate and v = speed of electron. In second part of experiment, magnetic field was adjusted so as to exactly cancel the electric force leaving the electron beam undeflected. This gives eE = evB . Using expression for d_(1) we can find out (e)/(m) = (2d_(1)E)/(B^(2)L^(2)) If the electron speed were doubled by increasing the potential differece V, which of the following would be true in order to correctly measure e//m

When an electron is accelerated through a potential difference V , it experience a fromce F through as uniform transverse magnetic field. If the potential difference is increased to 2 V, the force experoenced by the electron in the same magnetic field is