Chapter 28, Problem 026 In the figure, a charged particle moves into a region of

Question

Chapter 28, Problem 026 In the figure, a charged particle moves into a region of uniform magnetic fieldgoes through half a circle, and then exits that region. The particle is either a proton or an electron (you must decide which). It spends 122 ns in the region. (a) What is the magnitude of (b) If the particle is sent back through the magnetic field (along the same initial path) but with 2.30 times its previous kinetic energy, how much time does it spend in the field during this trip? (a) Number UAni (b) Number Units SHOW HINT LINK TO TEXT LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM VIDEO MINI-LECTURE

Explanation / Answer

F = q (vXB)

mv^2/ r = q ( VXB)

mv/r = qB

m ( 2pir/ rT) = qB

m( 2pi / T) = qB

T = m ( 2pi ) / qB

but time spent is just ( T/2 , that is half circle)

2T = m ( 2pi ) / qB

2 (122 x 10^-9 ) = m ( 2 x 3.14 ) / ( 1.6 x10^-19 x B)

now, coming to the mass of particel, applying cross product, that is velocity is pointing downwards, magnetic field out of the page , and force towards left , implies a positive charge,

122x 10 ^-9 = ( 1.67 x 10^-27 x 3.14 ) / (1.6 x10^-19 x B)

B = ( 3.277x 10^-8 ) / ( 122 x 10^-9)= 0.268 T

b)doubling the speed will not have any change on time because, T is independent of velocity

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