A particle of charge +q and mass m moving under the influence of a uniform electric field Ei^ and a uniform magnetic field Bk^ follows trajectory from P to Q as shown in figure. The velocities at P and Q are vi^ and −2vj^ respectively. Which of the following statement(s) is/are correct?

The work is only done by electric field, and so Eq×2a=12m[(2v)2−v2]Or E=34mv2qa .Rate of work done at P, =Fvcos00 =Eqv =34mv2qa×qv=34mv3a Rate of work done at Q , =F×2v×cos900=0

A particle of charge +q and mass m moving under the influence of a uniform electric field Ei^ and a uniform magnetic field Bk^ follows trajectory from P to Q as shown in figure. The velocities at P and Q are vi^ and −2vj^ respectively. Which of the following statement(s) is/are correct?

The work is only done by electric field, and so Eq×2a=12m[(2v)2−v2]Or E=34mv2qa .Rate of work done at P, =Fvcos00 =Eqv =34mv2qa×qv=34mv3a Rate of work done at Q , =F×2v×cos900=0

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A particle of charge $+ q$ and mass m moving under the influence of a uniform electric field $E \hat{i}$ and a uniform magnetic field $B \hat{k}$ follows trajectory from P to Q as shown in figure. The velocities at P and Q are $v \hat{i}$ and $- 2 v \hat{j}$ respectively. Which of the following statement(s) is/are correct?

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$E = \frac{3}{4} \frac{m v^{2}}{q a}$

Rate of wok done by electric field at P is $\frac{3}{4} \frac{m v^{3}}{a}$

Rate of work done by electric field at P is zero

Rate of work done by both the fields at Q is zero

answer is A, B, D.

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Detailed Solution

The work is only done by electric field, and so
$E q \times 2 a = \frac{1}{2} m [{(2 v)}^{2} - v^{2}]$
Or $E = \frac{3}{4} \frac{m v^{2}}{q a}$ .
Rate of work done at $P$ ,
$= F v \cos 0^{0} = E q v = \frac{3}{4} \frac{m v^{2}}{q a} \times q v = \frac{3}{4} \frac{m v^{3}}{a}$
Rate of work done at $Q$ ,
$= F \times 2 v \times \cos 90^{0} = 0$