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

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 $v_{0}$ , it moves with an initial acceleration $3 a_{0}$ towards west. The electric and magnetic fields in the room are

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(3)

$Acceleration of charged particle$ $v_{0}$

$Released from rest \Rightarrow \vec{a} = \frac{q}{m} \vec{E} = a_{0} (west) \Rightarrow \vec{E} = \frac{m a_{0}}{e} (west)$

$When it is projected towards north, acceleration due to magnetic force = 2 a_{0}$

$Therefore magnetic field = \frac{2 m a_{0}}{e v_{0}} (down),$

$Direction of magnetic field {\vec{F}}_{m} \to (\vec{v} \times \vec{B})$ $- \hat{i} \to (\hat{j} \times - \hat{k}),$

$The direction of magnetic force is towards west, for that the direction of$ $magnetic field is vertically downward (- \hat{k})$

Solution

Q.
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 $3 a_{0}$ towards west. The electric and magnetic fields in the room are

1 $\frac{m a_{0}}{e} east, \frac{3 m a_{0}}{e v_{0}} down$

2 $\frac{m a_{0}}{e} west, \frac{2 m a_{0}}{e v_{0}} up$

3 $\frac{m a_{0}}{e} west, \frac{2 m a_{0}}{e v_{0}} down$

4 $\frac{m a_{0}}{e} east, \frac{3 m a_{0}}{e v_{0}} up$

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Solution

Q. When a proton is released from rest in a 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

1 ma0e east, 3ma0ev0 down

2 ma0e west, 2ma0ev0 up

3 ma0e west, 2ma0ev0 down