Exam 2 Solutions. Applying the junction rule: i 1 Applying the loop rule to the left loop (LL), right loop (RL), and the full loop (FL) gives:

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1 PHY61 Eam Solutions 1. [8 points] In the circuit shown, the resistance R 1 = 1Ω. The batter voltages are identical: ε1 = ε = ε3 = 1 V. What is the current (in amps) flowing through the middle branch from a to b? i 1 i i 3 1 Appling the junction rule: i 1 + i = i 3 Appling the loop rule to the left loop (LL), right loop (RL), and the full loop (FL) gives: LL ε ir 1 ir 3 ε ir 1 = i1+ i3 = RL ε ir i3r ε ir = i + i3 = FL ε ir+ ir ε + ir ir= i = i Plugging the FL equation into the junction rule equation, and then into the RL equation gives: i + i = i = i = i = 1 3 Thus no current flows from top to bottom in the middle branch (or anwhere for that matter). Since all batteries have a potential difference of 1 V, one can see that the loop rule works out if all currents are zero. Page 1 of 9

2 PHY61 1. [6 points] A capacitor of capacitance C = 1 F contains a charge q on one of its plates (and q on the other). It is connected in to a resistor of resistance R = Ω such that it forms a closed circuit. How much time must elapse so that the charge on the capacitor is reduced to onl 1% of its starting value? Appling Kirchoff s loop rule, the equation to solve is: dq q R + = dt C t RC q t = q e () / Thus the elapsed time so that 1% of the charge remains is determined b:.1q ln1 = = q e t/ RC t RC 1 ( RC) ( )( )( ) t = ln1 = Ω 1 C.3 = 9ns 3. [6 points] A stationar flat conductor carries a constant current i > in the direction from top to bottom ( ˆ ) in the presence of a magnetic field B that points into plane the paper ( z ˆ ). If the electrons that make up the current are not allowed to leave the conductor, and the magnitude of their drift velocit is v d, indicate the direction and determine the magnitude of an electric field created inside the conductor. i e-, v d B z F B, E This is an eample of a Hall probe. If the current goes from top to bottom, the free electrons travel from bottom to top. The will feel a magnetic force according to: FB = ( e) v B which is in the + direction. In equilibrium, an electric field will be set up to balance this force, such that F= ( e) ( E+ v B ) =, so the electric field will point in the + direction with magnitude vb d Page of 9

3 PHY61 4. [6 points] An electron is accelerated from rest between electric plates with a potential 5 difference of Δ V = 5 1 V. What is the velocit of the electron upon reaching the far 31 plate? The mass of the electron is = kg. m e The potential energ of the electron when in starts from rest is translated into kinetic energ as it approaches the second plate: ( γ 1) U = eδ V = EK = mc We need to use special relativit here because the energ will be comparable to the rest mass energ. eδ V + 1 = γ = mc 1 v 1 c 1 v= c = c= eδv + 1 mc m/s 5. [6 points] A long wire has an electric charge densit of λ = +.5 μc / m, where 6 1μC = 1 C, as measured in the rest frame of the wire. What is the magnitude of the electric 8 field in the rest frame of an electron traveling at a velocit of v =.8 1 m/s parallel to the wire at a distance r = 1 cm from the wire? In the rest frame of the wire, and b using Gauss Law, we can solve for the electric field: λh E d A = π rhe = S ε λ E = πε r In the rest frame of the electron, however, the wire appears length contracted (since the electron s motion is parallel to the wire). Thus the charge per unit observed length has increased since charge is conserved but length is not: L L = γ q q λ = = γ = γλ L L So the field is: Page 3 of 9

4 PHY61 λ E = γ πε r 1 1 γ = = =.785 v c ( 9 1 Nm / C )(.5 1 C/m) 7 E = (.785) = N/C.1 m ( ) 6. [6 points] A long clindrical wire of radius r = 3mm carries 1A of current. If the current densit 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? i 1 A 6 The current densit is j = A/m A = π R = Use Ampere s Law to find the magnetic field b choosing a path C at a radius r = 1mm: B d s = πrb= μ i C i = jπ r enc ( i r ) ( πr)( πr ) enc μ π μi 3 B= = r =. 1 T π R Page 4 of 9

5 PHY61 7. [8 points] An infinitel long insulated wire carring a current I = 5 A is bent into a 7º arc (3 π /radians) of radius R= cm. The current comes in from infinitel far awa from above, and eits to the left infinitel far awa. Find the magnitude of the field B at the center of the arc. The wire begins and ends its turn at the locations indicated. I I R R I z There are 3 components to the field: semi-infinite segments of wire, and the ¾ circle. The magnitude of the field from a semi-infinite wire is half of that of an infinite wire (see homework, or look at direct integration of the Biot-Savart law): 1 μi 4 Bsemi = =.5 1 T in z direction. πr From the circular wire, the field is: μi 3π Barc = Φ where Φ = 4π R 3μi 3 B T in ˆ arc = = z 8R The magnetic field from each section contributes in the same direction (z). Thus the total magnitude is: B= B + B = arc 3 semi T Page 5 of 9

6 PHY61 8. A stationar neutral atom resides at the center of a Cartesian coordinate sstem and has a 3 magnetic dipole moment of.1 1 J/T aligned in the + ˆ direction. (a) [6 points] What is the ratio of the magnitude of the magnetic field from the atomic dipole at = 1 nm to that at = 5 nm? (The radius of the atom is about.1 nm, where 1 nm = 1-9 m) The field of a dipole at distances large compared to its size falls off with the cube power of distance: μ B 3 So the ratio is: ( = ) ( = ) 3 B B 1nm 1 1 = = 5nm 4 64 (b) [6 points] If a magnetic field is present everwhere with the form B= B ˆ, where B =.5 T/m, what is the magnitude of the acceleration of the dipole if the mass of the atom is 1-5 kg. F db = μ = μ B = ( )( ) = d N 5 15 m/s F a = = = m 1 kg J/T.5 T/m N Page 6 of 9

7 PHY61 9. [6 points] The magnetic field of a large solenoid is used to keep a proton in a perfect circular orbit. The solenoid has 1 windings per meter of length and has a radius of 1 m. If 6 the proton has a velocit of v = m/s, what is the minimum current needed to keep the proton orbiting within the confines of the solenoid in a plane perpendicular to the 7 19 solenoid ais? The proton mass is = kg and its charge is q = C. m p r p B= μ ni p = qbr = qrμ ni 7 6 ( kg)( m/s) ( C)( 1m )( 4π1 )( 1) mv i = = = 1.5 A qrμ n 1. [6 points] A rod of length L =.5 m and mass m =.5 kg carries a current I = A in the direction shown. The rod is aligned parallel to the z ais, and a uniform magnetic field is present: B= B ˆ, B =.5 T. The rod is suspended b two massless wires which bring the current to and awa from the rod. The acceleration due to gravit is 1 m/s in the ˆ direction. What is the angle θ that the suspension wires make with respect to the magnetic field direction? B z θ I F M F g F= il B ( )(.5)(.5) (.5)( 1) FB ilb tanθ = = = = 1 θ = 45 F mg g Page 7 of 9

8 PHY An oscillating LC circuit consists of a. H inductive coil and a 4 μf capacitor. The capacitor has a voltage drop of.75 V when the current through the coil is.3a. (a) [6 points] Find the maimum charge on the capacitor. (b) [6 points] Find the maimum current through the coil. The total energ in the circuit is: q U = + Li = CV + Li C U = ( 4 1 )(.75) + (.)(.3) =.5 1 J This must equal the energ when all charge is on the capacitor, or all current in the inductor: q U = + Li = CV + Li C 6 qma U =.5 1 J = C qma = 4.μC 6 1 U =.5 1 J = Lima i =.45 A ma Page 8 of 9

9 PHY61 1. [6 points] A circular conducting loop of wire increases in radius with time according to r = 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? dφb d da d dr ε = = ( BA) = B = B ( πr ) = πbr dt dt dt dt dt ε = π Bv t 13. A solenoid 1 cm long has a circular cross-section of 17 cm. There are 1 turns of wire carring a current of.5a. (a) [6 points] Find the magnitude of the magnetic field inside the solenoid assuming that it is essentiall infinite in length (neglect end effects). The magnetic field of a solenoid, obtained via Ampere s Law for eample, is: 1 B= μni= 4π1.5= π (b) [6 points] Find the total energ stored in the magnetic field inside the volume of the solenoid. Neglect end effects. U ( π ) ( ) B 1 1 m = = = μ cm 3 3 V T 7 4 π Page 9 of 9

Louisiana State University Physics 2102, Exam 3 April 2nd, 2009.

PRINT Your Name: Instructor: Louisiana State University Physics 2102, Exam 3 April 2nd, 2009. Please be sure to PRINT your name and class instructor above. The test consists of 4 questions (multiple choice),