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Question

A uniform electric field (E) and magnetic field (B0) exist in a region as shown in figure. Both the fields are mutually perpendicular.

A positively charged particle having charge q and mass m is projected vertically upward, and it's crossed the line AB after time t0. Find the speed of projection, if the particle moves with constant velocity after time t0.
(Given, g is acceleration due to gravity, and qE=mg) ( Here, the electric field and gravity exist everywhere but the magnetic field starts only after crossing the line AB)

A
gt0
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B
2gt0
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C
gt04
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D
gt02
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Solution

The correct option is B 2gt0
Let the velocity of the particle be v, when it crosses the line AB.

The value of magnetic force is,

Fm=qvB0sin90=qvB0


Now resolving the magnetic force in vertical & horizontal direction and applying the equilibrium condition (v=constant, or a=0) for particle at shown position.

Fmcosθ=mg ........(i)

And, Fmsinθ=qE=FE

Fmsinθ=qE .........(ii)

From equations (i) & (ii);

tanθ=qEmg

tanθ=1 (given, qE=mg)

θ=45

After time t0, the velocity of the particle along x axis is given by

vx=vx+axt

vx=0+(qEm)t0

vcosθ=(qEm)t0=gt0 .....(iii)

Similarly, vy=uy+ayt

vsinθ=ugt0 .....(iv)

Dividing equations (iv) to (iii), we get

ugt0gt0=tanθ=tan45=1

u=2gt0

Hence, option (b) is correct.
Why this question ?
Tip: The net force on a particle must be zero when it achieves constant velocity after time t0.
Concept : Draw FBD of particle and balance the acting external forces (mg,FE,Fm) to achieve equilibrium.

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