A particle of mass m and charge q is released from rest in a vertical plane from height H. A uniform magnetic field also exist in horizontal direction. Magnitude of magnetic field is B. Then

Charge particle will never reach ground if

H > \frac{2 m^{2} g}{(q B)^{2}}

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Vertical motion of charge particle will be SHM.

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Maximum speed attained by particle is

\frac{2 m}{q B} \sqrt{g}

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Motion of particle is oscillatory

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Solution

The correct option is B Vertical motion of charge particle will be SHM.

$\to F =_{B} + \to W \Rightarrow \to W = - m g^j$
$\to V = V_{x}^i - V_{y}^j$

$\to F = (q V_{x} B - m g)^j + q V_{y} B^i m \Rightarrow \to a = \frac{\to F}{m}$
$a_{x} = \frac{q B}{m} V_{y} & a_{y} = (\frac{q B V_{x}}{m} - g)$
$\frac{d}{d t} a_{y} = \frac{q B}{m} a_{x} = {(\frac{q B}{m})}^{2} V_{y} \Rightarrow \frac{d^{2} V_{y}}{d t^{2}} = {(\frac{q B}{m})}^{2} V_{y}$
which means that the particle is having an SHM motion along y -axis.

$V_{y} = V_{0 y} s i n (w t + ϕ)$
here, $w = \frac{q B}{m} & ϕ = 0$
$\Rightarrow a_{y} = \frac{d}{d t} V_{y} = V_{0 y} w c o s w t$

As we know that at $t = 0, a_{y} = g$
$g = V_{0 y} w \Rightarrow V_{0 y} = \frac{g}{w} = \frac{g m}{q B}$
$V_{y} = \frac{d y}{d t} = \frac{m g}{q B} s i n w t = \int_{0}^{y} d y = \frac{m g}{q B} \int_{0}^{t} s i n w t d t$
$y = \frac{m^{2} g}{(q B)^{2}} (1 - c o s w t) \Rightarrow y_{m a x} = \frac{2 m^{2} g}{(q B)^{2}}$
which means that the path is cycloid.

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