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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,

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} \mathrm{~m} / \mathrm{s}  moves undeflected between the plates. What is the magnitude of the magnetic field between the capacitor plates?

Option: 1

0.1 T


Option: 2

0.2 T


Option: 3

0.3 T


Option: 4

0.4 T


Answers (1)

best_answer

Electric field  \mathrm{E=\frac{v}{d}}
where V is the potential difference between the plates and d, the separation between them.
\mathrm{ d=3 \mathrm{~mm}=3 \times 10^{-3} \mathrm{~m} \\ }

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

Since the electron moves undeflected between the plates, the force due to magnetic field must balance the force due to electric field. Thus

\mathrm{ B e v=e E }

of   \mathrm{ \quad B=\frac{E}{v}=\frac{2 \times 10^5}{2 \times 10^6}=0.1 \mathrm{~T} }

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