A proton accelerated by a potential differenc | vedclass

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Question

(A) The kinetic energy of the proton is $K = qV = 1.6 	imes 10^{-19} 	imes 500 	imes 10^3 = 8 	imes 10^{-14} \ J$. The momentum $p$ of the proton

Vedclass · Accepted Answer

$0.01$

Vedclass · Answer

(A) The kinetic energy of the proton is $K = qV = 1.6 	imes 10^{-19} 	imes 500 	imes 10^3 = 8 	imes 10^{-14} \ J$. The momentum $p$ of the proton is given by $p = \sqrt{2mK} = \sqrt{2 	imes 1.6 	imes 10^{-27} 	imes 8 	imes 10^{-14}} = \sqrt{25.6 	imes 10^{-41}} = \sqrt{256 	imes 10^{-42}} = 1.6 	imes 10^{-20} \ kg \cdot m/s$. The radius of the circular path in the magnetic field is $R = \frac{p}{qB} = \frac{1.6 	imes 10^{-20}}{1.6 	imes 10^{-19} 	imes 0.1} = \frac{1.6 	imes 10^{-20}}{1.6 	imes 10^{-20}} = 1 \ m$. For a small thickness $d = 1.0 \ cm = 0.01 \ m$, the angle of deviation $	heta$ is given by $\sin 	heta = \frac{d}{R}$. Since $	heta$ is small, $\sin 	heta \approx 	heta = \frac{d}{R} = \frac{0.01}{1} = 0.01 \ rad$.

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