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  • A charged particle of mass 'm' and charge 'q' moving under the influence of uniform electric field  and a uniform magnetic field

A charged particle of mass 'm' and charge 'q' moving under the influence of uniform electric field E\hat{i} and a uniform magnetic field B\hat{k} follows a trajectory from point P to Q as shown in figure. The velocities at P and Q are respectively, v\hat{i} and -2v\hat{j}. Then which of the following statements (A, B, C, D) are the correct? (trajectory shown is schematic and not to scale) (A) E = \frac{3}{4}\left(\frac{mv^2}{qa} \right ) (B) Rate of work done by the electric field at P is \frac{3}{4}\left(\frac{mv^2}{a} \right ) (C) Rate of work done by both the fields at Q is Zero. (D) The difference between the magnitude of angular momentum of the particle at P and Q is 2mav
Option: 1  (A), (C), (D)
Option: 2  (A), (B), (C)
Option: 3  (A), (B), (C), (D)
Option: 4  (B), (C), (D)
 

Answers (1)

best_answer

A) by work-energy theorem

W_{mag} + W_{ele} = \frac{1}{2} m(2v)^2 - \frac{1}{2}m(v)^2

0 + qE_02a = \frac{3}{2}mv^2

E_0 = \frac{3}{4} \frac{mv^2}{qa}

(B) Rate of change of work done at A = power of electric force

                                                                   = qE_0V

                                                                    =\frac{3}{4}\frac{mv^3}{a}

(C) at Q, \frac{dw}{dt} = 0 for both forces

(D) \\\Delta\vec{L} = (-m(2v)(2a)\hat{k}) - (-mva\hat{k})\\|\Delta \vec{L} |= 3mva

 

Hence, the correct option is (2).

Posted by

avinash.dongre

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