A proton is initially at rest and moves through three different regions as shown

Question

A proton is initially at rest and moves through three different regions as shown in the figure. In region 1, the proton accelerates across a potential difference of 3480 V. In region 2, there is a magnetic field of 1.30 T pointing out of the page and an electric field pointing perpendicular to the magnetic field and perpendicular to the proton's velocity. Finally, in region 3, there is no electric field, but just a 1.30 T magnetic field pointing out of the page.

(a) What is the speed of the proton as it leaves region 1 and enters region 2?
1........ m/s

(b) If the proton travels in a straight line through region 2, what is the magnitude and direction of the electric field?

(c) What will be the radius of the circular path the proton travels in region 3?
5 mm

magnitude  2..........V/m direction 3.......... A proton is initially at rest and moves through three different regions as shown in the figure. In region 1, the proton accelerates across a potential difference of 3480 V. In region 2, there is a magnetic field of 1.30 T pointing out of the page and an electric field pointing perpendicular to the magnetic field and perpendicular to the proton's velocity. Finally, in region 3, there is no electric field, but just a 1.30 T magnetic field pointing out of the page. (a) What is the speed of the proton as it leaves region 1 and enters region 2? ........ m/s (b) If the proton travels in a straight line through region 2, what is the magnitude and direction of the electric field? magnitude ..........V/m direction .......... (c) What will be the radius of the circular path the proton travels in region 3? mm

Explanation / Answer

(a)According to work energy theorem work done = V*q = change in kinetic energy    =  (1/2) mv^2 - (1/2) mu^2 iniitial velocity of the proton   u = 0.0 m/s   (1/2) mv^2   = V*q ==>   v = sqrt ( 2V*q / m).........(1) here q = charge of the proton + 1.6*10^-19 C V = potential = 3480V , m =mass of the proton = 1.67*10^-27 kg plug all values in (1) we get velovcity of the proton enter into the region (2) v = 8.16*10^5 m/s -------------------------------------------------------- (b) if a proton travell's straight line path electric foece =   magnetic force qE = Bvq Electric field E = Bv   = (1.30T) (8.16*10^5 m/s) = 1.06*10^6 N/C according to Fleming's left hand rule direction of electric field is toward's north (c) radius of the proton path in region (3) r = mv /Bq plug all the values we get r = 6.55*10^3 m = 6.55mm

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