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A potential difference of 600 V is applied across the plates of a parallel plate capacitor. The separation between the plates is 3 mm. An electron projected vertically, parallel to the plates, with a velocity of 2 x 10⁶ m s^-1 moves un-deflected between the plates. The magnitude of the magnetic field between the capacitor plates is $(\frac{α}{100}) T .$ The value of $α$ is ____.

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Detailed Solution

Electric field $E = \frac{V}{d}$

where V is the potential difference between the plates and d, the separation between them.

$d = 3 mm = 3 \times 10^{- 3} m;$

$E = \frac{V}{d} = \frac{600}{3 \times 10^{- 3}} = 2 \times 10^{5} V m^{- 1}$

Since the electron moves un-deflected between the plates, the force due to magnetic field must balance the force due to electric field. Thus:

$Bev = eE$

$\Rightarrow B = \frac{E}{v} = \frac{2 \times 10^{5}}{2 \times 10^{6}} = 0 .1 = \frac{10}{100} T$

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