First slide

Force on a charged particle moving in a magnetic field

Question

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

Difficult

A

$\frac{m a_{0}}{e} e a s t, \frac{3 m a_{0}}{e υ_{0}} u p$
B

$\frac{m a_{0}}{e} e a s t, \frac{3 m a_{0}}{e υ_{0}} d o w n$
C

$\frac{m a_{0}}{e} w e s t, \frac{2 m a_{0}}{e υ_{0}} u p$
D

$\frac{m a_{0}}{e} w e s t, \frac{2 m a_{0}}{e υ_{0}} d o w n$

Solution

when proton is stationary force is only due to electric field i.e, $\overset{⇀}{E}$ = $\frac{m a_{0}}{e} a l o n g w e s t$
force due to magnetic field must be 2m $a_{0}$ along west
its velocity is along north (j) force is along west (-i)
$\vec{F}$ =q( $\overset{⇀}{V}$ x $\overset{⇀}{B}$ ) (-i=j x-k) hence magnetic field is vertically downwards direction and having a magnitude $\frac{2 m a_{0}}{e}$

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