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An electron passes undeflected through p...

An electron passes undeflected through perpendicular electric and magnetic filed of intensity `3.4xx10^(3) V//m` and `2xx10^(-3) Wb//m^(2)` respectively . Then its velocity un `m//s` is

A

`1.7xx10^(6)`

B

`6.8xx10^(6)`

C

`6.8`

D

`1.7xx10^(8)`

Text Solution

Verified by Experts

The correct Answer is:
A
`V=E/B`
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Knowledge Check

  • Speed of an electron passing undeviated through a region of cross electric and magnetic fields of magnitude 4 xx 10^(5) V/m and 0.02 Wb//m^(2) respectively in meter per second is

    A
    `2 xx 10^(6)`
    B
    `8 xx 10^(7)`
    C
    `8 xx 10^(6)`
    D
    `2 xx 10^(7)`

  • In a Thomson set-up for the demination of e/m, electrons accelerated by 2.5 kV enter the region of crossed electric and magnetic fields of strengths 3.6 xx 10^(4) Vm^(-1) and 1.2 xx 10^(-3) T respectively and through undeflected. The meaured value of e//m of the electron is equal to

    A
    `1.0 xx 10^(11) C-kg^(-1)`
    B
    `1.76 xx 10^(11) C-kg^(-1)`
    C
    `1.80 xx 10^(11) C-kg^(-1)`
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  • A narrow electron beam passes undeviated through an electric field E = 3 xx 10^(4) "volt"//m and an overlapping magnetic field B = 2 xx 10^(-3) Weber//m^(2) . If electric field and magnetic field are mutually perpendicular. The speed of the electron is

    A
    `60 m//s`
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