A) \[\overrightarrow{E}=0;\,\overrightarrow{B}=b\hat{i}\,+c\hat{k}\]
B) \[\overrightarrow{E}=ai;\,\overrightarrow{B}=c\hat{k}\,+a\hat{i}\]
C) \[\overrightarrow{E}=0;\,\overrightarrow{B}=c\hat{j}\,+b\hat{k}\]
D) \[\overrightarrow{E}=ai;\,\overrightarrow{B}=c\hat{k}\,+b\hat{j}\]
Correct Answer: B
Solution :
Electric field can deviate the path of the particle in the shown direction only when it is along negative y-direction. In the given options \[\overrightarrow{E}\] is either zero or along x-direction. Hence it is the magnetic field which is really responsible for its curved path. Options (a) and (c) can?t be accepted as the path will be helix in that case (when the velocity vector makes an angle other than 0°, 180° or 90° with the magnetic field, path is a helix) option (d) is wrong because in that case component of net force on the particle also comes in k direction which is not acceptable as the particle is moving in x-y plane. Only in option (b) the particle can move in x-y plane. In option (d): \[{{\overrightarrow{F}}_{net}}=q\overrightarrow{E}+q(\overrightarrow{v}\times \overrightarrow{B})\] Initial velocity is along x-direction. So let \[\overrightarrow{v}=v\hat{i}\] \ \[{{\overrightarrow{F}}_{net}}=qa\hat{i}+q[(v\hat{i})\times (c\hat{k}+b\hat{j}]=qa\hat{i}-qvc\hat{j}+qvb\hat{k}\] In option (b) \[{{\overrightarrow{F}}_{net}}=q(a\hat{i})+q[(v\hat{i})\times (c\hat{k}+a\overrightarrow{i})=qa\hat{i}-qvc\hat{j}\]
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