Question

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

• A. 0.1 T
• B. 0.2 T
• C. 0.3 T
• D. 0.4 T

Answer 1

The correct option is A

0.1 T

Electric intensity $E=\frac{V}{d}$
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 undeflected between the plates, the force clue to magnetic field must balance the force due to electric field.
$Be\nu =eEorB=\frac{E}{\nu }=\frac{2\times {10}^5}{2\times {10}^6}=0.1T$