A particle of mass m and charge q is thrown from origin at t=0 with velocity $2 \overset{\land}{i} + 3 \overset{\land}{j} + 4 \overset{\land}{k}$ units in a region with uniform magnetic field $\vec{B} = 2 \overset{\land}{i}$ units. After time $t = π m / q B$ , an electric field $\vec{E}$ is switched on such that particle moves on a straight line with constant speed. $\vec{E}$ may be;

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$6 \overset{\land}{k} + 8 \overset{\land}{j}$ units

$- 6 \overset{\land}{k} - 8 \overset{\land}{j}$ units

$5 \overset{\land}{k} - 10 \overset{\land}{j}$ units

$- 6 \overset{\land}{k} - 9 \overset{\land}{j}$ units

answer is C.

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Detailed Solution

In time $\frac{T}{2} = \frac{π m}{q B}$ Which is half of revolution time

Velocity will get reversed in yz plane which is given as

$\vec{v} = 2 \overset{\land}{i} - 3 \overset{\land}{j} - 4 \overset{\land}{k}$

Net force on the particle is

$q (\vec{E} + \vec{V} \times \vec{B}) = 0$

$\Rightarrow \vec{E} = - (\vec{V \times} \vec{B}) = - (2 \overset{\land}{i} - 3 \overset{\land}{j} - 4 \overset{\land}{k}) \times 2 \overset{\land}{i} = - 6 \overset{\land}{k} + 8 \overset{\land}{j}$

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