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 1.35 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 1.35 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?

(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?

Explanation / Answer

a) decrease in potential energy = increase in kinetic energy now decrease in potential energy = qV = 1.6*10^(-19)*3480 = 5.568*10^(-16) J if speed is v after coming out of region 1, increase in kinetic energy = (1/2)mv^2 mass of proton = 1.67*10^(-27) kg so (1/2)*1.67*10^(-27)*v^2 = 5.568*10^(-16) ==> v = 8.16*10^5 m/s b) even though forces are acting as there is no change in direction, net force should be zero. force due to electric field will be qE and force due to magnetic field will be charge times cross product of velocity and magnetic field. let i be unit vector along east, j be unit vector along north and k be unit vector pointing out of the page. the velocity of proton is along i and B is along k. so v cross B is along -j . i.e. west. now as net force = 0 qE+q(v cross B) = 0 so E = -(v cross B) so direction of E is along j i.e. north. and magnitude = vB = 8.16*10^5 * 1.35 = 11.02*110^5 V/m c)radius of curvature of proton in only magnetic field is given by r = mv/qB =[ 1.67*10^(-27)*8.16*10^5 ] /[1.6*10^(-19) * 1.35 ] = 6.309*10^(-3) m so radius = 0.631 cm

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