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 $2 \times 10^{6} {ms}^{- 1}$ moves undeflected between the plates. What is the magnitude of the magnetic field between the capacitor plates?

Moderate

Solution

Electric field $E = \frac{V}{d}$
where V is the potential difference between the plates and d, the separation between them.
$\begin{array}{l} d = 3 mm = 3 \times 10^{- 3} m \\ E = \frac{V}{d} = \frac{600}{3 \times 10^{- 3}} = 2 \times 10^{5} {Vm}^{- 1} \end{array}$
Since the electron moves undeflected between the plates, the force due to magnetic field must balance the force due to electric field. Thus
$\begin{array}{l} Bev = eE \\ B = \frac{E}{v} = \frac{2 \times 10^{5}}{2 \times 10^{6}} = 0 .1 T \end{array}$

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