When a proton is released from rest in a room, it starts with an initial acceleration a0 towards west. When it is projected towards north with a speed \nu0 it moves with an initial acceleration 3a0 toward west. The electric and magnetic fields in the room are:

  • Option 1)

    \frac{\text{ma}_{0}}{\text{e}} \text{east}, \frac{3\text{ma}_{0}}{\text{e}\nu_{0}}\text{down}

  • Option 2)

    \frac{\text{ma}_{0}}{\text{e}} \text{west}, \frac{2\text{ma}_{0}}{\text{e}\nu_{0}}\text{up}

  • Option 3)

    \frac{\text{ma}_{0}}{\text{e}} \text{west}, \frac{2\text{ma}_{0}}{\text{e}\nu_{0}}\text{down}

  • Option 4)

    \frac{\text{ma}_{0}}{\text{e}} \text{east}, \frac{3\text{ma}_{0}}{\text{e}\nu_{0}}\text{up}

 

Answers (1)

As we learned in @8252 and @8250

Acceleration of charged particle, 

\vec{a} = \frac{q}{m}\left ( \vec{E}+\vec{V}\times \vec{B} \right )

Release from west = \vec{a} = \frac{q}{w}\vec{E} = a_{o}\:\left ( west \right )

                                \Rightarrow \vec{E} = \frac{ma_{o}}{e}\:\left ( west \right )

When it is projected towards north, acceleration due to magnetic force = 2ao

\therefore Magnetic field, B = \frac{2ma_{o}}{ev_{o}}\:\left ( down \right )


Option 1)

\frac{\text{ma}_{0}}{\text{e}} \text{east}, \frac{3\text{ma}_{0}}{\text{e}\nu_{0}}\text{down}

This option is incorrect

Option 2)

\frac{\text{ma}_{0}}{\text{e}} \text{west}, \frac{2\text{ma}_{0}}{\text{e}\nu_{0}}\text{up}

This option is incorrect

Option 3)

\frac{\text{ma}_{0}}{\text{e}} \text{west}, \frac{2\text{ma}_{0}}{\text{e}\nu_{0}}\text{down}

This option is correct

Option 4)

\frac{\text{ma}_{0}}{\text{e}} \text{east}, \frac{3\text{ma}_{0}}{\text{e}\nu_{0}}\text{up}

This option is incorrect

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