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Two uncharged parallel plates A and B are at the same electrical potential, and separated by a distance d. A uniform magnetic induction field B acts perpendicular to plane of paper (and parallel to plates) between the plates. Particles, each with charge q, mass m and kinetic energy
K are shot through a hole in A directly towards B, particle entering the hole perpendicular to plate A. An ammeter A_m connected to B and measures the current collected by the plate B, as the particles hit it. The limiting value of the field B for which the ammeter will stop showing a current is

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$\frac{\sqrt{3 mK}}{qd}$

$\frac{\sqrt{mK}}{2 qd}$

$\frac{\sqrt{2 mK}}{qd}$

$\frac{2 \sqrt{mK}}{qd}$

answer is A.

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Detailed Solution

Obviously for the limiting field, the path of the particles will be semi-circular arcs of radius equal to the distance between the plates.

i.e., radius r of circular path is r = d.

Now, for a particle of charge q, mass m and kinetic energy K, where velocity $v = \sqrt{\frac{2 k}{m}}$ , the radius r is

$\begin{array}{l} r = d = \frac{mv}{qB} = \frac{m}{qB} \sqrt{\frac{2 k}{m}} \\ Hence, B = \frac{\sqrt{2 Km}}{q d} \end{array}$

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