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An ideal gas in a sealed container has an initial volume of 2-70 L- At.docx

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An electron in a cathode-ray tube is accelerated through a potential difference of 10 kV, then passes through the d 2.2-cm-wide region of uniform magnetic field in the figure(Figure 1). Part A What field strength will deflect the electron by 14Â° Express your answer to two significant figures and include the appropriate units BValue Units Figure 1 of1 Submit Request Answer Provide Feedback 0 V 10 kV Solution in a magnetic field magnetic force = centripetal force qvB = mv^2/r B = mv/qr if an electron is accelerated in a potential difference then 1/2mv^2 = q*V v = sqrt(2qV/m) v = sqrt(2*1.6*10^-19*10000/9.1*10^-31) v = 5.93*10^7 m/s from the figure sin(theta) = d/r r = 2.2*10^-2/sin14 r = 0.091 m B = 9.1*10^-31*5.93*10^7/1.6*10^-19*0.091 B = 3.71*10^-3 T .

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- An electron in a cathode-ray tube is accelerated through a potential difference of 10 kV, then passes through the d 2.2-cm-wide region of uniform magnetic field in the figure(Figure 1). Part A What field strength will deflect the electron by 14Â° Express your answer to two significant figures and include the appropriate units BValue Units Figure 1 of1 Submit Request Answer Provide Feedback 0 V 10 kV Solution in a magnetic field magnetic force = centripetal force qvB = mv^2/r B = mv/qr if an electron is accelerated in a potential difference then 1/2mv^2 = q*V v = sqrt(2qV/m) v = sqrt(2*1.6*10^-19*10000/9.1*10^-31) v = 5.93*10^7 m/s from the figure sin(theta) = d/r r = 2.2*10^-2/sin14 r = 0.091 m B = 9.1*10^-31*5.93*10^7/1.6*10^-19*0.091 B = 3.71*10^-3 T

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