Single Choice

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 $${ 3a }_{ 0 }$$ towards west. The electric and magnetic fields in the room are :

A$$\dfrac { { ma }_{ 0 } }{ e } east,\dfrac { { 3ma }_{ 0 } }{ { ev }_{ 0 } } down$$
B$$\dfrac { { ma }_{ 0 } }{ e } west,\dfrac { { 3ma }_{ 0 } }{ { ev }_{ 0 } } up$$
C$$\dfrac { { ma }_{ 0 } }{ e } west,\dfrac { { 2ma }_{ 0 } }{ { ev }_{ 0 } } down$$
Correct Answer
D$$\dfrac { { ma }_{ 0 } }{ e } east,\dfrac { { 3ma }_{ 0 } }{ { ev }_{ 0 } } up$$

Solution

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 $${ F }_{ 1 }=qE$$
Using Newton's second law
$${ F }_{ 1 }={ ma }_{ 0 }$$
hence,
$${ ma }_{ 0 }=qE$$
$$E={ ma }_{ 0 }/q$$ 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 $$=3{ ma }_{ 0 }$$
$${ F }_{ 1 }+{ F }_{ 2 }=3{ ma }_{ 0 }$$
$${ ma }_{ 0 }+{ F }_{ 2 }=3{ ma }_{ 0 }$$
maximum magnetic force is given by
$${ F }_{ 2 }=q{ V }_{ 0 }B$$
So, $$q{ V }_{ 0 }B=2{ ma }_{ 0 }$$
$$B=2{ ma }_{ 0 }/q{ V }_{ 0 }$$ (down).
Option C


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