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  • The electric field used in Thomson's experiment had a strength of <img alt="\mathrm{8.0 \times 10^3 N/C.}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B8.0%20%5Cti

The electric field used in Thomson's experiment had a strength of \mathrm{8.0 \times 10^3 N/C.} If a cathode ray particle with a mass of  \mathrm{ 9.11 \times 10^{-31} kg } and a charge of \mathrm{-1.6 \times 10^{-19} C } was deflected by an angle of 30 degrees in the electric field, what was the speed of the particle before entering the field?

Option: 1

\mathrm{1.4 \times 10^7 m/s }


Option: 2

\mathrm{ 2.0 \times 10^7 m/s }


Option: 3

\mathrm{2.5 \times 10^7 m/s }


Option: 4

\mathrm{3.0 \times 10^7 m/s }


Answers (1)

best_answer

The electric force experienced by the cathode ray particle in the electric field can be given as \mathrm{F = qE, }where q is the charge of the particle and E is the electric field strength. The force on the particle causes it to undergo circular motion, and the radius of the circular path can be given as \mathrm{ r = mv^2/Bq, } where m is the mass of the particle, v is its velocity, and B is the magnetic field strength. In this case, there is no magnetic field, so we can use the equation for the radius of the circular path in the electric field, which is \mathrm{ r = mv^2/2qE \sin(\theta)}, where \mathrm{ \theta} is the angle of deflection. Equating this to the radius of the circular path gives:

\mathrm{ \frac{mv^2}{2qE\sin\theta}= \frac{mv^2}{Bq}}

Simplifying and solving for v, we get:

\mathrm{v= \sqrt{\frac{2qE}{B\sin\theta}}}

 Plugging in the values given in the problem, we get:

\mathrm{v = \sqrt{\frac{2(-1.6\times10^{19} \mathrm{C})(8.0\times10^3 \mathrm{N/C})}{(9.11\times10^{31} \mathrm{kg})\sin(30^\circ)}}}

\mathrm{v \approx 2.5 \times 10^7 m/s}

Therefore, the correct answer is option 3)\mathrm{ 2.5 \times 10^7 m/s.}

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