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  • A particle of charge Q, mass m is moving under the influence of uniform electric field <img alt="\mathrm{E \hat{i}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BE%20%5Chat%7Bi%

A particle of charge Q, mass m is moving under the influence of uniform electric field \mathrm{E \hat{i}} and a uniform magnetic field\mathrm{{B} \hat{k}} follows a trajectory from P to Q as shown in figure. The velocity at P and Q are\mathrm{v \hat{i} \text { and } 2 v \hat{j}} respectively. Find the rate of work done by electric field at P

Option: 1

\mathrm{\frac{3}{4}\left(\frac{m v^2}{q a}\right)}


Option: 2

\mathrm{\frac{3}{4}\left(\frac{m v^3}{ a}\right)}


Option: 3

\mathrm{\frac{3}{4}\left(\frac{m v^3}{q a}\right)}


Option: 4

none of these


Answers (1)

best_answer

Work done = change in KE

\mathrm{\begin{aligned} & W_{P O}+W_{O Q}=1 / 2 m(2 v)^2-1 / 2 m(v)^2, q E\left(a \cos 90^{\circ}\right)+q E(2 a \cos \theta)=3 / 2 m v^2 \\ & 0+q E(2 a)=3 / 2 m v^2, \quad E=\frac{3}{4}\left(\frac{m v^2}{q a}\right) \end{aligned}}

\mathrm{\text { Rate of work done } F . v=q E . v=\frac{3}{4}\left(\frac{m v^3}{a}\right)}

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Anam Khan

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