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(C) The radius of the circular path of the electron in the magnetic field is given by $R = \frac{mv}{qB} = \frac{\sqrt{2m(KE)}}{qB}$.Substituting the

Vedclass · Accepted Answer

$12.87$

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(C) The radius of the circular path of the electron in the magnetic field is given by $R = \frac{mv}{qB} = \frac{\sqrt{2m(KE)}}{qB}$.Substituting the given values: $R = \frac{\sqrt{2 	imes 9.1 	imes 10^{-31} 	imes 100 	imes 1.6 	imes 10^{-19}}}{1.6 	imes 10^{-19} 	imes 1.5 	imes 10^{-3}} \approx 0.02248 \, m = 2.248 \, cm$.From the geometry,$\sin 	heta = \frac{x}{R} = \frac{2}{2.248} \approx 0.8896$,so $	heta \approx 62.8^\circ$.The vertical displacement $y$ at the exit point $T$ is $y = R(1 - \cos 	heta) = 2.248(1 - \cos(62.8^\circ)) \approx 2.248(1 - 0.457) \approx 1.22 \, cm$.The electron travels in a straight line after leaving the magnetic field. The distance from the exit point $T$ to the screen is $8 \, cm - 2 \, cm = 6 \, cm$.The additional vertical displacement is $\Delta y = (6 \, cm) 	an 	heta = 6 	imes 	an(62.8^\circ) \approx 6 	imes 1.945 \approx 11.67 \, cm$.The total distance $d = y + \Delta y = 1.22 + 11.67 = 12.89 \, cm$. The closest option is $12.87 \, cm$.

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