Please help me solve these!! THANKS 5.What is the diameter of the circular trajectory taken by a 1.0 C charge (m = 0.002 kg)moving at 3000 m/s in a 0.9 T uniform B-field?A. 13.3 mB. 13.9 mC. 14.2 mD. 12.9 m6How long does it take for the particle in question #5 to circulate one time?A. 0.007 sB. 0.014 SC. 0.021 SD. 0.028 s7.How much work is done by the magnetic field for one circulation of the particle inquestion # 5 ?A. 3000 JB. 1500 JC. 750 JD. 0J

Answered: What is the diameter of the circular…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

Please help me solve these!! THANKS

5.
What is the diameter of the circular trajectory taken by a 1.0 C charge (m = 0.002 kg)
moving at 3000 m/s in a 0.9 T uniform B-field?
A. 13.3 m
B. 13.9 m
C. 14.2 m
D. 12.9 m
6
How long does it take for the particle in question #5 to circulate one time?
A. 0.007 s
B. 0.014 S
C. 0.021 S
D. 0.028 s
7.
How much work is done by the magnetic field for one circulation of the particle in
question # 5 ?
A. 3000 J
B. 1500 J
C. 750 J
D. 0J

Expert Solution

Step 1

When a charge enters a magnetic field, it is acted upon by a magnetic force, given as

$F = q v B \sin θ q i s t h e m a g n i t u d e o f t h e c h a r g e v i s t h e s p e e d o f t h e c h a r g e B i s t h e m a g n e t i c f i e l d m a g n i t u d e θ i s t h e a n g l e b e t w e e n t h e v e l o c i t y d i r e c t i o n a n d t h e m a g n e t i c f i e l d$

Due to the influence of this magnetic field, the charge starts to follow a curved path and hence perform circular motion, during which a centripetal force acts on the charge, given as

$F = \frac{m v^{2}}{r} m i s t h e m a s s o f t h e c h a r g e r i s t h e r a d i u s o f t h e c u r v e d p a t h$

Since this centripetal force is provided by the magnetic force, thus

$q v B \sin θ = \frac{m v^{2}}{r} F o r a c h a r g e e n t e r i n g p e r p e n d i c u l a r t o t h e f i e l d, θ = 90 a n d \sin 90 = 1 q v B = \frac{m v^{2}}{r} q B = \frac{m v}{r} r = \frac{m v}{q B}$

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