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>>Motion of Charged Particle Combined in Magnetic Field and Electric Fields
>> $When a proton is released from the rest$

Question

When a proton is released from the rest in a room, it starts with an initial acceleration $a_{0}$ towards north with a speed $v_{0}$ it moves with an initial acceleration $3 a_{0}$ towards west. The electric and magnetic fields in the room are

A

$\frac{m a _{0}}{e}$ west, $\frac{2 m a _{0}}{e v _{0}}$ up

B

$\frac{m a _{0}}{e}$ west, $\frac{2 m a _{0}}{e v _{0}}$ down

C

$\frac{m a _{0}}{e}$ west, $\frac{3 m a _{0}}{e v _{0}}$ up

D

$\frac{m a _{0}}{e}$ east, $\frac{3 m a _{0}}{e v _{0}}$ down

Medium

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Updated on : 2022-09-05

Solution

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Correct option is B)

$a_{0} = \frac{e E}{m} \Rightarrow E = \frac{m a _{0}}{e}$

$\frac{e v _{0} B + e E}{m} = 3 a_{0}$

$e v_{0} B = 3 m a_{0} - e E$

$= 3 m a_{0} - m a_{0}$

$= 2 m a_{0}$

$B = \frac{2 m a _{0}}{e v _{0}}$

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