When a proton is released from rest in room, it starts with an initial acceleration $a_{0}$ towards west When it is projected 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 :

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Answer 1

Initial acceleration $= a_{0}$
when a proton is released from rest in a room it starts with an initial acceleration $a_{0}$ towards west.
$F_{1} =$ electric force acting on the proton in west direction

$q =$ charge on proton
$E =$ electric field in the region in west direction
$m =$ mass of proton
electric force on the proton can be given on $F_{1} = q E$
Using Newton's second law
$F_{1} = m a_{0}$

hence,
$m a_{0} = q E$

$E = m a_{0} / q$ Along west

Case -2
$V_{0} =$ initial velocity of the particle

$F_{2} =$ magnetic force acting on the proton in west direction electric force on proton $= 3 m a_{0}$

$F_{1} + F_{2} = 3 m a_{0}$

$m a_{0} + F_{2} = 3 m a_{0}$

maximum magnetic force is given by
$F_{2} = q V_{0} B$

So, $q V_{0} B = 2 m a_{0}$
$B = 2 m a_{0} / q V_{0}$ (down).

Option C

Answer 2

Initial acceleration

= a_{0}

when a proton is released from rest in a room it starts with an initial acceleration

a_{0}

towards west.

F_{1} =

electric force acting on the proton in west direction

q =

charge on proton

E =

electric field in the region in west direction

m =

mass of proton

electric force on the proton can be given on

F_{1} = q E

Using Newton's second law

F_{1} = m a_{0}

hence,

m a_{0} = q E

E = m a_{0} / q

Along west

Case -2

V_{0} =

initial velocity of the particle

F_{2} =

magnetic force acting on the proton in west direction electric force on proton

= 3 m a_{0}

F_{1} + F_{2} = 3 m a_{0}

m a_{0} + F_{2} = 3 m a_{0}

maximum magnetic force is given by

F_{2} = q V_{0} B

So,

q V_{0} B = 2 m a_{0}

B = 2 m a_{0} / q V_{0}

(down).

Option C

When a proton is released from rest in room, it starts with an initial acceleration $a_{0}$ towards west When it is projected 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 :

$\frac{m a _{0}}{e} e a s t, \frac{3 m a _{0}}{e v _{0}} d o w n$

$\frac{m a _{0}}{e} w e s t, \frac{3 m a _{0}}{e v _{0}} u p$

$\frac{m a _{0}}{e} w e s t, \frac{2 m a _{0}}{e v _{0}} d o w n$

$\frac{m a _{0}}{e} e a s t, \frac{3 m a _{0}}{e v _{0}} u p$

A charged particle is released from rest in a region of uniform electric and magnetic fields which are parallel to each other. The particle will move on a

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A charged particle is released from rest in a region of steady and uniform electric and magnetic fields which are parallel to each other. The particle will move in a

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When a proton is released from rest in room, it starts with an initial acceleration a0​ towards west When it is projected towards north with a speed V0​ it moves with an initial acceleration 3a0​ towards west. The electric and magnetic fields in the room are :

ema0​​east,ev0​3ma0​​down

ema0​​west,ev0​3ma0​​up

ema0​​west,ev0​2ma0​​down

ema0​​east,ev0​3ma0​​up

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When a proton is released from the rest in a room, it starts with an initial acceleration a0​ towards north with a speed v0​ it moves with an initial acceleration 3a0​ towards west. The electric and magnetic fields in the room are

A charged particle is released from the rest in a region of steady and uniform electric and magnetic fields which are parallel to each other. The particle will move in a

A charged particle is released from rest in a region of steady and uniform electric and magnetic fields which are parallel to each other. The particle will move in a

Revise with Concepts

Motion in the Presence of Electric and Magnetic Field

Cyclotron

When a proton is released from rest in room, it starts with an initial acceleration $a_{0}$ towards west When it is projected 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 :

$\frac{m a _{0}}{e} e a s t, \frac{3 m a _{0}}{e v _{0}} d o w n$

$\frac{m a _{0}}{e} w e s t, \frac{3 m a _{0}}{e v _{0}} u p$

$\frac{m a _{0}}{e} w e s t, \frac{2 m a _{0}}{e v _{0}} d o w n$

$\frac{m a _{0}}{e} e a s t, \frac{3 m a _{0}}{e v _{0}} u p$

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