The diagram shows a circular non-conducting smooth fixed track kept concentric to a magnetic field of magnitude B which exists in a cylindrical region of radius r (r > R). There is small bead of mass m and charge q rests on the track. A bead can move freely along track. If the magnetic field is switched off then

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If $q = \frac{m}{B} \sqrt{\frac{g}{R}}$ then bead will complete vertical circular motion

Impulse imparted to bead is qRB

If $q = \frac{4 m}{B} \sqrt{\frac{g}{R}}$ then bead will complete vertical circular motion

Impulse imparted to bead is $\frac{q R B}{2}$

answer is A, C.

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Detailed Solution

$\begin{array}{l} Sudden removal of magnetic field results in induced emf, which produces non conservative eletric field. \\ This electric field applies impulsive force on charge. \\ \oint \vec{E} . \vec{dr} = ξ = - \frac{d ϕ}{d t} \\ \Rightarrow E 2 π R = π R^{2} \frac{d B}{d t} \\ \Rightarrow E d t = \frac{R d B}{2} \end{array}$

$\begin{array}{l} \Rightarrow Impulse = \int F d t = \int q E d t = \int \frac{q R}{2} d B = \frac{q R B}{2} \\ \Rightarrow m v = \frac{q R B}{2} \end{array}$

$\begin{array}{l} \Rightarrow Minimum velocity to complete a circle = \sqrt{4 g R} \\ \Rightarrow q = \frac{2 m v}{R B} = \frac{2 m}{R B} \sqrt{4 g R} = \frac{4 m}{B} \sqrt{\frac{g}{R}} \end{array}$