A charged particle with specific charge s moves undeflected through a region of space containing mutually perpendicular and uniform electric and magnetic fields E and B.When the electric field is switched off,the particle will move in a circular path of radius

\frac{E}{B s}

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\frac{E s}{B}

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\frac{E s}{B^{2}}

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\frac{E}{B^{2} s}

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Solution

The correct option is D $\frac{E}{B^{2} s}$
$q E = B q v$
$∴ v = \frac{E}{B}$
$r = \frac{m v}{B q} = \frac{m (E / B)}{B (q)}$
$= \frac{E}{B^{2} S}$ (where S=q/m)

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A charged particle with specific charge s moves undeflected through a region of space containing mutually perpendicular and uniform electric and magnetic fields E and B.When the electric field is switched off,the particle will move in a circular path of radius

The correct option is D EB2sqE=Bqv∴v=EB r=mvBq=m(E/B)B(q) =EB2S (where S=q/m)

The correct option is D $\frac{E}{B^{2} s}$
$q E = B q v$
$∴ v = \frac{E}{B}$
$r = \frac{m v}{B q} = \frac{m (E / B)}{B (q)}$
$= \frac{E}{B^{2} S}$ (where S=q/m)