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A charged particle having charge $10^{- 6} C$ and mass of $10^{- 10} k g$ is fired from the middle of the plate making an angle $30 °$ with plane of the plate. Length of the plate is 0.17m and it is separated by $0.1 m$ . Electric field $E = 10^{- 3} \frac{N}{C}$ is present between the plates. Just outside the plates magnetic field is present. Find the velocity of the projection of charged particle and magnitude of the magnetic field perpendicular to the plane of the figure, if it has to graze the plate at $C and A$ parallel to the surface of the plane.(negative gravity)
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$3.0 m / s, 2.46 m T$

$2.0 m / s, 3.46 m T$

$2.0 m / s, 2.46 m T$

$3.0 m / s, 3.46 m T$

answer is B.

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Detailed Solution

To graze at C using equation of trajectory of parabola,

$y = x \tan θ - \frac{a x^{2}}{2 v^{2} \cos^{2} θ}$ …(i)

$here, a = \frac{q E}{m} = \frac{10^{- 6} \times 10^{- 3}}{10^{- 10}} = 10 m / s^{2}$

Substituting in Eq. (i), we have

$0.05 = 0.17 \tan 30 ° - \frac{10 \times {(0.17)}^{2}}{2 v^{2} \times {(\frac{\sqrt{3}}{2})}^{2}}$

Solving this equation, we have $v = 2 m / s$

In magnetic field, $A C = 2 r$

$or 0.1 = 2 r or r = 0.05 m = \frac{m v \cos 30 °}{B q} ∴ B = \frac{m v \cos 30 °}{(0.05) q} = \frac{(10^{- 10}) (2) (\frac{\sqrt{3}}{2})}{(0.05) (10^{- 6})} = 3.46 \times 10^{- 3} T$

$= 3.46 m T$ Ans.

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