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Chrome File Edit View History Bookmarks People Window Help Homework: HW 12 of Ch26 x c Chegg Study Guided So x C Secure www.flipitphysics.com 2551414&enrollmentID; 241092 100% until Monday, April 24 at 11:59 PM Tipler6 26.P,028. An electron of kinetic energy 3.8 kev moves in a circular orbit perpendicular to a magnetic field of 0.359 T. 1) (a) Find the radius of the orbit. 0,58 mm Submi You currently have 1 submissions for this question. Only 10 submission are allowed. You can make 9 more submissions for this question. Your submissions: 0,58 Computed value Submitted: Monday, April 24 at 4:29 PM eedback: Correct 2) (b) Find the frequency of the motion. GHz Submit You currently have 2 submissions for this question. Only 10 submission are allowed. You can make 8 more submissions for this question. 3) Find the period You currently have 3 submissions for this question, Only 10 submission are allowed, You can make 7 more submissions for this question. CAl E 5% D Mon 6:24 PM a E Print Assignment View nteractive Example 3r2c rd Exercise Two Loop RC Circuit Worked Example Magnetism optional Motion in B Field 28, nd Exercise Tipler 6 26. P.042 nd Exercise Tipler6 26 P.050 nd Exercise Tiplert 26 P.071 nd Exercise Rectangular Current Loop nd Exercise Right Triangular Current Loop

Explanation / Answer

given data:
KE = 3.8 keV =3.8*10^3*1.602*10^-19 Joules
magnetic field = 0.359 T
1. for radius, use formula:
E= (Bqr)^2 / 2m
q= 1.6 x 10^(-19) coulombs
m= 9.1 x 10^(-31) kg

E= (Bqr)^2 / 2m
3.8*10^3*1.602*10^-19 = ((0.359*1.6*10^(-19)*r)^2)/(2*9.1*10^(-31))
r = 0.000579 m
r = 0.579mm

2. for frequency,
f = Bq /2m
f = (0.359 * 1.6*10^(-19)) /(2 *3.142 * 9.1 8 10^(-31) )
= 1.255 * 10^(9) Hz or s-1
3. Period
T=1/f = 2m / Bq
= 1 / 1.255*10^-9
= 0.7968* 10 ^(-9) s

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