In Figure, an electron with an initial kinetic energy of 4.0keV enters region 1

Question

In Figure, an electron with an initial kinetic energy of 4.0keV enters region 1 at time t=0. That region contains a uniform magnetic field directed into the page, with magnitude 0.01T. The electron goes through a half-circle and then exits region 1, headed toward region 2 across a gap of 25cm. There is an electric potential difference 2kV across the gap, with a polarity such that the electron's speed increases uniformly as it traverses the gap. Region 2 contains a uniform magnetic field directed out of the page, with magnitude 0.02T. The electron goes through a halfcircle and then leaves region 2. At what time t does it leave?

Explanation / Answer

Ans:- In region 1,

Charged particle circulating in a magnetic field. Applying Newton’s second law

Fnet = qvB = mv^2/r

qB = mv/r

v= qBr/m

T= 2r/v= 2r/(qBr/m) = 2m/qB

but since it only travels in a semi circle the time

t1=T/2 = m/qB = (3.14*9.1*10^-31)/(1.6*10^-19*0.01) = 1.78*10^-9 s

In region 2,

t2=T/2 = m/qB = (3.14*9.1*10^-31)/(1.6*10^-19*0.02)

= 8.92*10^-10 s

In the gap,

KE=1/2mv^2

4*10^3*1.6*10^-19 = ½*(9.1*10^-31)*v^2  

=> v= 3.75*10^7 m/s

Thus velocity of the electron before entering the gap = v= vi = 3.75*10^7 m/s

Acceleration in the region, a= F/m = qE/m = (V/d)*q/m = (2000*1.60*10^-19)/(0.25*9.11*10^-31) = 1.4*10^15m/s^2

so the time to cross the gap comes from x = vi*t + 1/2*a*t^2

0.25 = (3.75*10^7)*t +1/2*(1.4*10^15)*t^2    

=> t= 5.99*10^-9s

T = t1+t2+t3 = (1.78*10^-9)+( 8.92*10^-10)+( 5.99*10^-9) = 8.66*10^-9 s

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