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

Options

A. The electric field point to the top, while the magnetic field point into the page
B. The electric field point to the top, while the magnetic field point out of page
C. Electric field points to the bottom, while the magnetic field point out of pate
D. Electric field points to the bottom, while the magnetic field points into the page

Answer

B

Explanation

Since electrons are deflected downward, hence electric field must be upward. The magnetic force in order to cancel the electric force must point upward. From right hand rule, magnetic field must point out of page. Hence choice (b) is correct.