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For a positively charged particle moving in a $x-y$ plane initially along the $x$-axis, there is a sudden change in its path due to the presence of electric and/or magnetic fields beyond $P$. The curved path is shown in the $x-y$ plane and is found to be non-circular. Which one of the following combinations is possible
$\overrightarrow E = 0;\,\overrightarrow B = b\hat i\, + c\hat k$
$\overrightarrow E = ai;\,\overrightarrow B = c\hat k\, + a\hat i$
$\overrightarrow E = 0;\,\overrightarrow B = c\hat j\, + b\hat k$
$\overrightarrow E = ai;\,\overrightarrow B = c\hat k\, + b\hat j$
Solution
(b) 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 $(b)$ ${\overrightarrow F _{net}} = q(a\hat i) + q[(v\hat i) \times (c\hat k + a\overrightarrow i ) = qa\hat i – qvc\hat j$
