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When a proton is released from rest in a room, it starts with an initial acceleration \mathrm{a}_0  towards west. When it is projected towards north with a speed \mathrm{v}_0  it moves with an initial acceleration 3 \mathrm{a}_0  towards west. The electric and magnetic fields in the room are respectively

Option: 1

\mathrm{\frac{\mathrm{ma}_0}{\mathrm{e}} ~west~, \frac{2 \mathrm{ma}_0}{\mathrm{ev}_0}~ down }


Option: 2

\mathrm{\frac{\mathrm{ma}_0}{\mathrm{e}}~ east, ~\frac{3 \mathrm{ma}_0}{\mathrm{ev}_0}~up}


Option: 3

\mathrm{\frac{\mathrm{ma}_0}{\mathrm{e}}~ east, ~\frac{3 \mathrm{ma}_0}{\mathrm{ev}_0}~ down }


Option: 4

\mathrm{\frac{\mathrm{ma}_0}{\mathrm{e}}~ west, ~\frac{2 \mathrm{ma}_0}{\mathrm{ev}}~ up}


Answers (1)

best_answer

When moves with an acceleration \mathrm{a}_0  towards west, electric field
\mathrm{ \mathrm{E}=\frac{\mathrm{F}}{\mathrm{q}}=\frac{\mathrm{ma}_0}{\mathrm{e}} \text { (West) } }


When moves with an acceleration \mathrm{ 3 \mathrm{a}_0 }  towards east, magnetic field
\mathrm{ \mathrm{B}=\frac{2 \mathrm{ma}_0}{\mathrm{ev}_0}(\text { downward }) }

Posted by

Ritika Kankaria

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