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Question

(B) The particle moves in the $x-y$ plane. For the motion to remain in the $x-y$ plane,the net force $\overrightarrow F_{net} = q(\overrightarrow E +

Vedclass · Accepted Answer

$\overrightarrow E = a\hat i; \overrightarrow B = c\hat k + a\hat i$

Vedclass · Answer

(B) The particle moves in the $x-y$ plane. For the motion to remain in the $x-y$ plane,the net force $\overrightarrow F_{net} = q(\overrightarrow E + \overrightarrow v 	imes \overrightarrow B)$ must have no component along the $z$-axis. Initial velocity is $\overrightarrow v = v\hat i$. In option $(d)$,$\overrightarrow F_{net} = q(a\hat i) + q(v\hat i 	imes (c\hat k + b\hat j)) = qa\hat i - qvc\hat j + qvb\hat k$. The presence of a $z$-component $(qvb\hat k)$ forces the particle out of the $x-y$ plane,so $(d)$ is incorrect. In option $(b)$,$\overrightarrow F_{net} = q(a\hat i) + q(v\hat i 	imes (c\hat k + a\hat i)) = qa\hat i - qvc\hat j$. Here,the force has only $x$ and $y$ components,keeping the particle in the $x-y$ plane. The path is non-circular because the electric field changes the speed of the particle. Thus,option $(b)$ is the correct combination.

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