InterviewSolution
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InterviewSolution
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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 uniform magnetic field ` B hat(k)` follows a trajectory from ` P to Q` as shown in fig. The velocities at `P and Q` are ` v hat(i)` and ` - 2v hat(j)` . which of the following statement(s) is/are correct ? A. `E=3/4((mv^2)/(qa))`B. Rate of work done by the electric field at P is `3/4[(mv^3)/(a)]`C. Rate of work done by the electric field at `P=0`D. Rate of work done by both fields at Q is zero |
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Answer» Correct Answer - A::B::D Increase in K.E=work done by electric field `1/2m(2v)^2-1/2mv^2=qE(2a)` or `E=3/4((mv^2)/(qa))` (b) Rate of work done by electric field at P =Electric force x Velocity at P `qExxv=q[3/4(mv^2)/(qa)]v`, `=3/4(mv^3)/(a)` (d) At Q, force acting on particle due to electric field is zero (as `2vhatj`) is perpendicular to both `Ehati` and `Bhatk`) and hence rate of work done by both fields is zero. |
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