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

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)) 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

The mass of an electron is m, charge is e and it is accelerated from rest through a potential difference of V volts. The velocity acquired by electron will be

Electron at rest are accelerated by a potential of V volt. These electrons enter the region of space having a uniform, perpendicular magnetic induction field B. the radius of the path of the electrons inside the magnetic field is

If an electron is not deflected in passing through a certain region of space can we be sure that there is no magnetic field in that region?

If an electron is not deflected in passing through a certain region of space can we be sure that there is no magnetic field in that region?