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Both, the electric and magnetic fields c...

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

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To understand the difference between the deflections of a moving electron in an electric field and a magnetic field, we can analyze the forces acting on the electron in both scenarios. ### Step-by-Step Solution: 1. **Understanding the Forces**: - When an electron moves in an electric field, it experiences a force given by \( F = qE \), where \( q \) is the charge of the electron and \( E \) is the electric field strength. - In a magnetic field, the force on a moving electron is given by \( F = q(v \times B) \), where \( v \) is the velocity of the electron and \( B \) is the magnetic field strength. ...
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Knowledge Check

  • An electric field can deflect

    A
    X-rays
    B
    `gamma`-rays
    C
    `alpha`-rays
    D
    All of these

  • An Electric field can deflect

    A
    `alpha`-particles
    B
    `X`-rays
    C
    Neutrons
    D
    `gamma`-rays

  • Assertion (A),If electrons travel undefeccted by the electric field E and magnetic field B then the velocity of electrons is given by V=e/b . Reason (R ) : When both electric and magnetic field are applied simultaneously on electron beam. If force due to electric field is equal and opposite to force due to magnetic field then they travel undeflected.

    A
    `A` and `R` are true and `R` is the correct explaination of `A`
    B
    `A` and `R` are true and `R`is not the correct explaination of `A`
    C
    `A` is true and `R` is false.
    D
    `A` is false but `R` is true

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    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)) 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 speed were doubled by increasing the potential differece V, which of the following would be true in order to correctly measure e//m

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