DOC PREVIEW
UF PHY 2061 - Exam 2

This preview shows page 1-2-3-4 out of 13 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 13 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 13 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 13 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 13 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 13 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

Exam 2PHY206111-9-06Name:_______________________ ___Exam 2Closed book exam. A calculator is allowed, as is one 8.511” sheet of paper with your own written notes. Please show all work leading to your answer to receive full credit. Numerical answers should be calculated to at least 2 significant digits. Exam is worth 100 points, 25% of your total grade.UF Honor Code: “On my honor, I have neither given nor received unauthorized aid in doing this exam.”Page 1 of 13Sphere: 2 344 3.14159273S r V rp p p= = =e  16022 1019. C29.8 m/sg = 61 C 10 Cm-=61 F 10 Fm-=121 pF 10 F-=191 eV 1.6 10 J-= �9 2 2019 10 N m / C4Kpe= = �12 2 208.8542 10 C / N me-= � c  3 0 108. m/s70210 T m / A4Kkcmp-= = = �604 1.257 10 T m /Akm p-= = � � 0 021cme =1 2122ˆq qKr=F r0q=FEenc0ESqdeF = � =�Ε Α�0re�� =EV=- �E0UVq=CW U d=- D = ��F sCV dD =- ��E s qVCD =diV LdtD =-dqidt=eff 1 2C C C= +eff 1 21 1 1C C C= +LRAr=RCRCt =LRLRt =eff 1 2R R R= +eff 1 21 1 1R R R= +V iRD =22 VP Vi i RR= = =22QUC=212U Li=12 fLCw p= =112 2v c/t t 0LL0( )( )2/x x vtt t vx cgg�= ��= �y yz z21xxxu vuvuc��=�21yyxuuvucg�=� ��� �� �E mc2( )21KE mcg= -F E v B  q( ) i=F L× Bp umF p d dt/m c E p c2 4 2 2 2  03 4i ddrmp=s× rB0 enc Cd im� =�B s�0wire2iBrmp= 04arciBRmp= F p qBr==τ r× FNi=μ A=τ μ× BU =- �μ Bzz zdBFdzm=BSdF = ��BΑBdNdteF=-BL NiF=2 2002 2E Buem= +x x y y z za b a b a b�= + +a b( )( )( )y z y z x z x z x y x ya b b a a b b a a b b a� = - - - + -a b x y zPHY206111-9-06Name:_______________________ ___1. [8 points] In the circuit shown, the resistance112R = W. The battery voltages are identical:1 2 3 1 Ve e e= = =. What is the current (in amps) flowing through the middle branch from a to b? Page 2 of 131PHY206111-9-06Name:_______________________ ___2. [6 points] A capacitor of capacitance 102 10 FC-= � contains a charge q on one of its plates (and –q on the other). It is connected in to a resistor of resistance R = 200 such that it forms a closed circuit. How much time must elapse so that the charge on the capacitor is reduced to only 10% of its starting value?3. [6 points] A stationary flat conductor carries a constant current i >0 in the direction from top to bottom (ˆ- y) in the presence of a magnetic field B that points into plane the paper (ˆ- z). If the electrons that make up the current are not allowed to leave the conductor, and the magnitude of their drift velocity is dv, indicate the direction and determine the magnitude of any electric field created inside the conductor.Page 3 of 13xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxiBxyzPHY206111-9-06Name:_______________________ ___4. [6 points] An electron is accelerated from rest between 2 electric plates with a potential difference of 55 10 VVD = �. What is the velocity of the electron upon reaching the far plate? (Careful!) The mass of the electron is 319.11 10 kgem-= �.Page 4 of 13PHY206111-9-06Name:_______________________ ___5. [6 points] A long wire has an electric charge density of 2.5 / mCl m=+, where61 10C Cm-=, as measured in the rest frame of the wire. What is the magnitude of the electricfield in the rest frame of an electron traveling at a velocity of 82.8 10 m/sv = � parallel to the wire at a distance 1 cmr = from the wire? Page 5 of 13PHY206111-9-06Name:_______________________ ___6. [6 points] A long cylindrical wire of radius r = 3mm carries 100A of current. If the current density is uniform throughout the cross-section of the wire, what is the magnitude of the magnetic field at a radius of r = 1mm within the wire?Page 6 of 13PHY206111-9-06Name:_______________________ ___7. [8 points] An infinitely long insulated wire carrying a current I = 50 A is bent into a 270º arc (3 / 2pradians) of radius R=2 cm. The current comes in from infinitely far away from above, and exits to the left infinitely far away. Find the magnitude of the field B at the center of the arc. The wire begins and ends its turn at the locations indicated.Page 7 of 13RIRIIxyzPHY206111-9-06Name:_______________________ ___8. A stationary neutral atom resides at the center of a Cartesian coordinate system and has a magnetic dipole moment of 232.1 10 J/T-� aligned in the ˆ+y direction. (a) [6 points] What is the ratio of the magnitude of the magnetic field from the atomic dipole at y = 100 nm to that at y = 25 nm? (The radius of the atom is about 0.1 nm, where 1 nm = 10-9 m) (b) [6 points] If a magnetic field is present everywhere with the form0 0ˆ , where 0.5 T/mB B y B= =y, what is the magnitude of the acceleration of the dipole if the mass of the atom is 10-25 kg.Page 8 of 13PHY206111-9-06Name:_______________________ ___9. [6 points] The magnetic field of a large solenoid is used to keep a proton in a perfect circular orbit. The solenoid has 1000 windings per meter of length and has a radius of 1 m. If the proton has a velocity of 61.5 10 m/sv = �, what is the minimum current needed to keep the proton orbiting within the confines of the solenoid in a plane perpendicular to the solenoid axis? The proton mass is 271.67 10 kgpm-= � and its charge is 191.6 10 Cq-= �.Page 9 of 13prPHY206111-9-06Name:_______________________ ___10. [6 points] A rod of length L = 0.5 m and mass m = 0.5 kg carries a current I = 20 A in the direction shown. The rod is aligned parallel to the z axis, and a uniform magnetic field is present: 0 0ˆ, 0.5 TB B B=- =y. The rod is suspended by two massless wires which bring the current to and away from the rod. The acceleration due to gravity is 10 m/s2 in the ˆ- y direction. What is the angle  that the suspension wires make with respect to the magnetic field direction?Page 10 of 13IByxzPHY206111-9-06Name:_______________________ ___11. An oscillating LC circuit consists of a 0.002 H inductive coil and a 4 F capacitor. The capacitor has a voltage drop of 0.75 V when the current through the coil is 0.03A.(a) [6 points] Find the maximum charge on the capacitor.(b) [6 points] Find the maximum current through the coil.Page 11 of 13PHY206111-9-06Name:_______________________ ___12. [6 points] A circular conducting loop of wire increases in radius with time according tor vt= where v is a constant. It is immersed in a constant magnetic field B perpendicular to the plane of the loop. What is the induced EMF in the loop as a function of time?Page 12 of


View Full Document

UF PHY 2061 - Exam 2

Documents in this Course
Gauss Law

Gauss Law

14 pages

Exam 2

Exam 2

16 pages

Exam 3

Exam 3

11 pages

Exam2

Exam2

10 pages

Exam 1

Exam 1

12 pages

Exam 2

Exam 2

12 pages

Gauss Law

Gauss Law

14 pages

Exam 3

Exam 3

10 pages

Exam 1

Exam 1

10 pages

Exam 2

Exam 2

13 pages

Exam 2

Exam 2

12 pages

Exam 1

Exam 1

12 pages

Load more
Download Exam 2
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Exam 2 and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Exam 2 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?