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(B) The radius of the circular path of a charged particle in a magnetic field is given by $r = \frac{mv}{qB}$.Since the proton is accelerated by a pot

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$30$

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(B) The radius of the circular path of a charged particle in a magnetic field is given by $r = \frac{mv}{qB}$.Since the proton is accelerated by a potential difference $V$,its kinetic energy is $K = qV = \frac{1}{2}mv^2$,which gives $v = \sqrt{\frac{2qV}{m}}$.Substituting $v$ in the radius formula,we get $r = \frac{m}{qB} \sqrt{\frac{2qV}{m}} = \frac{1}{B} \sqrt{\frac{2mV}{q}}$.Given $V = 500 	imes 10^3 \ V$,$B = 0.5 \ T$,$m = 1.6 	imes 10^{-27} \ kg$,$q = 1.6 	imes 10^{-19} \ C$,and $d = 0.1 \ m$.$r = \frac{1}{0.5} \sqrt{\frac{2 	imes 1.6 	imes 10^{-27} 	imes 500 	imes 10^3}{1.6 	imes 10^{-19}}} = 2 \sqrt{\frac{2 	imes 10^{-24} 	imes 500 	imes 10^3}{10^{-19}}} = 2 \sqrt{1000 	imes 10^{-2}} = 2 \sqrt{10} \approx 0.2 \ m$.From the geometry of the path,$\sin 	heta = \frac{d}{r} = \frac{0.1}{0.2} = 0.5$.Therefore,$	heta = \arcsin(0.5) = 30^{\circ}$.

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