Question

When a potion is released from in a room, it starts with an initial acceleration $v_0$ towards west. When it is projected towards North with a speed $3a_0$ it moves with an initial acceleration towards West. The electric and magnetic fields in the room are

• A. $\frac{m{a}_0}{e}West,\frac{2m{a}_0}{e{v}_0}up$
• B. $\frac{m{a}_0}{e}West,\frac{2m{a}_0}{e{v}_0}down$
• C. \$\frac{m{a}_0}{e}East,\frac{3m{a}_0}{e{v}_0}up$
• D. $\frac{m{a}_0}{e}East,\frac{2m{a}_0}{e{v}_0}down$

Answer 1

The correct option is B $\frac{m{a}_0}{e}West,\frac{2m{a}_0}{e{v}_0}down$

Given:

Initial acceleration = $a_0$

When a proton is released from rest in a room, it starts with an initial acceleration towards the west.

$F_1=Electricforceactingontheprotoninthewestdirection.$

Q= Charge on proton

E = Electric field in the region in west direction.

M= mass of proton.

Solution:

Electric force on proton,$F_1=QE$

Using Newton's second law,

$F_1=M{a}_0$

Therefore, $M{a}_0=QE$

$\Rightarrow E=\frac{M{a}_0}{Q}$ along west.

Case:b

$V_0=velocityoftheparticleinitially$

$F_2=Magneticforceactingontheprotoninwestdirection$

Electric force on proton =$3M{a}_0$

$F_1+F_2=3M{a}_0$

$\Rightarrow M{a}_0+Q{V}_{0}B=3M{a}_0$

$\Rightarrow Q{V}_{0}B=2M{a}_0$

$\Rightarrow B=\frac{2M{a}_0}{Q{V}_0}$(down)

Hence,Option B is correct.