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Question

A particle of specific charge (charge/mass) α starts moving from the origin from rest under the action of an electric field E=E0 ^i and magnetic field B=B0 ^k. Its velocity at (x0, y0, 0) is (4^i3^j). The value of x0 is :
(All parameters are in SI unit.)

A
13αE02B0
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B
16αB0E0
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C
252αE0
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D
5α2B0
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Solution

The correct option is C 252αE0
Given:
E=E0 ^i and B=B0 ^k

Initially the charged particle is at rest. Hence, at t=0, it will start to accelerate along (+ve) xdirection due to presence of electric field. So initial velocity of charge will be along +ve x direction.

Given,

Final velocity, Vf=4^i3^j

|Vf|=Vf=42+32=5 m/s

Now using work energy theorem;

Welect+Wmag=ΔK.E

As we know that work done by magnetic force (Wmag)=0.

Welec+0=12×m(52)0

Welec=25m2 ........(1)

The position of particle is (x0, y0, 0)
Here, Δx=x00=x0

Workdone by electric force:

Welec=(qE0)Δx=(qE0)x0 .....(2)

From Eqs (1) & (2) we get;

25m2=qE0x0

According to problem, α=q/m

x0=25m2qE0=25α12E0=252αE0

Hence, option (c) is the correct answer.

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