A proton starts on a plate with potential of 5.0 x 10^3 V at y=0m. It then enter

Question

A proton starts on a plate with potential of 5.0 x 10^3 V at y=0m. It then enters a region in which there is magnetic field of 2.0 T pointing out of the page. In a field it travels a semicircular path. a) what is the radius of the semicircle, and where (what value of y) doed the proton leave the magnetic field? b) the proton now passes between a second set of plates with a potential difference of 5 x 10^3 V. What is its velocity after passing through these plates? c) The proton now enters another region in which there is a magnetic field of 2.0 T pointing out of the page, and it again describes a semicircular path. What is the radius of the semicircle, and where (what value of y) does the proton emerge from the field?

Explanation / Answer

(A) KE = q deltaV  

(1.67 x 10^-27) v^2 / 2 = (1.6 x 10^-19) (5 x 10^3)

v = 9.79 x 10^5 m/s

in magnetic field,

Fb= m a_c

q v B = m v^2 / r

r= m v / q B

r = (1.67 x 10^-27)(9.79 x 10^5) / (1.6 x 10^-19 x 2)

r = 5.11 x 10^-3 m or 5.11 mm

(B) KE = (1.67 x 10^-27) v^2 /2 = 1.6 x 10^-19 x 2 x 5 x 10^3

v = 1.384 x 10^6 m/s

(C) r = m v / q B = 7.22 x 10^-3 m Or 7.22 mm

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.