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`

Options

A. Magnetic field would have to be halved
B. Magnetic field would have to be doubled
C. Length L of the plates would have to be doubled
D. Length L of the plates would have to be halved

Answer

A

Explanation

In the passage, we have
`d_(1) = (eEL^2)/(2mv^2) and e/m = (2d_(1)E)/(B^(2)L^(2)`
By increasing speed of electron to two times, `(d_1)` will become `1//4th`. For this magnetic field should be halved, so that we can correctly meaure `e//m`. Hence choice (a) is correct.