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  • A proton is moving with a uniform velocity of  along

A proton is moving with a uniform velocity of 10^6 \mathrm{~m} s^{-1} along the Y-axis, under the joint action of a magnetic field along Z-axis and an electric field of magnitude 2 \times 10^4 \mathrm{~V} \mathrm{m}^{-1} along the negative x-axis. If the electric field is switched off, the proton starts moving in a circle. The radius of the circle is nearly ( given: \frac{e}{m} ratio for proton \approx 10^8 \mathrm{e} kg^{-1} )

Option: 1

0.5m


Option: 2

0.2m


Option: 3

0.1m


Option: 4

0.05m


Answers (1)

best_answer

Initially the lorentz force acting on the proton,

\vec{F}=\vec{F}_e+\vec{F}_m=\vec{E}_q+\overrightarrow{q v} \times \vec{B}=-E q i+q v B \hat{i}

Since the proton has no acceleration,

\therefore \vec{F}=0 \quad \therefore E q=q V B

or B=\frac{E}{V}=\frac{2 \times 10^4}{10^6}=2 \times 10^{-2} \mathrm{T}

When the electric field is switched off then the radius of the circular path,

R=\frac{m V}{q B}=\frac{10^6}{10^8 \times 2 \times 10^{-2}}=0.5 \mathrm{~m}

 

Posted by

Divya Prakash Singh

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