A charged particle passes through a region that could have electric field only or magnetic field only or both electric and magnetic fields or none of the fields. Match Column I with Column II

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Solution

i. Kinetic energy of the particle can remain constant, if both the fields are present. This is possible if the force due to both fields cancel each other.
Kinetic energy of the particle can also remain constant if only magnetic field is present, because magnetic field does not do any work.
Obviously KE will remain constant if no field is present.
ii. This is possible if either both the fields are present or no field is present. This is also possible if only magnetic field is present and the particle is at rest or moving in the direction of field.
iii. This is possible if electric field is present, magnetic field may or may not be present.
iv. This is possible if only electric field is present and velocity and electric field are along the parallel lines. Magnetic field may also be present if it is parallel to velocity.

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\begin{matrix} Colum I & Colum II i. & \to E = 0, \to B \neq 0 and initial velocity may & a. & Straight line be at any angle with \to B . ii. & \to E = 0, \to B = 0 and initial velocity may & b. & Parabola be at any angle with \to E . iii & \to E \neq 0, \to B \neq 0, \to E | | \to B and initial velocity is & c. & Circular ⊥ to both. d. & Helical path \end{matrix}

\begin{matrix} Colum I & Colum II i. & \to E = 0, \to B \neq 0 and initial velocity may & a. & Straight line be at any angle with \to B . ii. & \to E = 0, \to B = 0 and initial velocity may & b. & Parabola be at any angle with \to E . iii & \to E \neq 0, \to B \neq 0, \to E | | \to B and initial velocity is & c. & Circular ⊥ to both. d. & Helical path \end{matrix}

v

a

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