When a proton is released from rest in a room | vedclass

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(B) $1$. When the proton is at rest,the only force acting on it is the electric force $F_E = qE = eE$. Given the acceleration is $a_0$ towards west,we

Vedclass · Accepted Answer

$\frac{ma_0}{e}$ west,$\frac{2ma_0}{ev_0}$ down

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(B) $1$. When the proton is at rest,the only force acting on it is the electric force $F_E = qE = eE$. Given the acceleration is $a_0$ towards west,we have $ma_0 = eE$,which implies $E = \frac{ma_0}{e}$ towards west.$2$. When the proton is projected towards north with speed $v_0$,the total force is the vector sum of the electric force and the magnetic force: $F_{net} = F_E + F_B = ma_{net}$.$3$. The net acceleration is $3a_0$ towards west. Since $F_E$ is $ma_0$ towards west,the magnetic force $F_B$ must be $2ma_0$ towards west to result in a total force of $3ma_0$ towards west $(F_B = F_{net} - F_E = 3ma_0 - ma_0 = 2ma_0)$.$4$. The magnetic force is given by $F_B = q(v 	imes B) = ev_0B \sin(	heta)$. For the force to be towards west,with velocity towards north,the magnetic field $B$ must be directed downwards (using Fleming's Left-Hand Rule or the cross product $v 	imes B$).$5$. Setting the magnitudes equal: $ev_0B = 2ma_0$,which gives $B = \frac{2ma_0}{ev_0}$ downwards.

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 $v_0$,it moves with an initial acceleration $3a_0$ towards west. The electric and magnetic fields in the room are

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