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(B) From the geometry of the path of the proton in the magnetic field,we have $\sin 	heta = \frac{d}{r}$,where $d = 0.1 \; m$ is the width of the mag

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(B) From the geometry of the path of the proton in the magnetic field,we have $\sin 	heta = \frac{d}{r}$,where $d = 0.1 \; m$ is the width of the magnetic field region and $r$ is the radius of the circular path.The radius of the circular path is given by $r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB} = \frac{1}{B} \sqrt{\frac{2mV}{q}}$,where $V = 500 	imes 10^3 \; V$ is the accelerating potential.Substituting $r$ into the expression for $\sin 	heta$:$\sin 	heta = \frac{d}{r} = Bd \sqrt{\frac{q}{2mV}}$Given values: $B = 0.51 \; T$,$d = 0.1 \; m$,$q = 1.6 	imes 10^{-19} \; C$,$m = 1.67 	imes 10^{-27} \; kg$,$V = 500 	imes 10^3 \; V$.$\sin 	heta = 0.51 	imes 0.1 	imes \sqrt{\frac{1.6 	imes 10^{-19}}{2 	imes 1.67 	imes 10^{-27} 	imes 500 	imes 10^3}}$$\sin 	heta = 0.051 	imes \sqrt{\frac{1.6 	imes 10^{-19}}{1.67 	imes 10^{-21}}} \approx 0.051 	imes \sqrt{95.8} \approx 0.051 	imes 9.78 \approx 0.5$Since $\sin 	heta = 0.5$,we have $	heta = 30^o$.

$A$ proton accelerated by a potential difference of $500 \; kV$ moves through a transverse magnetic field of $0.51 \; T$ as shown in the figure. The angle $\theta$ through which the proton deviates from the initial direction of its motion is......$^o$

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