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  • A potential difference of 100V is applied across the plates of a parallel plate capacitor. The separation bey the plates is 1mm. An electron projected Vertically p

A potential difference of 100V is applied across the plates of a parallel plate capacitor. The separation bey the plates is 1mm. An electron projected Vertically parallel to the plates, with a velocity of 2 \times 10^6 \mathrm{~m} / \mathrm{sec}. moves undeflected between the plates. What is the magnitude of the magnetic field. between the capacitor plates.

Option: 1

0.01T


Option: 2

0.05T


Option: 3

0.5T


Option: 4

10 T


Answers (1)

best_answer

V=100 \mathrm{~V}, \quad d= 100m=-1 \times 10^{-3} \mathrm{~m} \\

E=\frac{V}{d}=\frac{100 \mathrm{~V}}{1 \times 10^{-3}}=2 \times 10^5 \mathrm{~V} / \mathrm{m}

Since the electron moves under fleeted between plates, the force due to magnetic field must balance the force due to electric field

BeV=eE

B=\frac{E}{V}=\frac{1 \times 10^5}{2 \times 10^6}=\frac{1}{20}=0.05 \mathrm{~T}

Posted by

Ritika Kankaria

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