PHY 2049 SPRING 2001 EXAM 3

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Kenneth Godwin Sanders
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1 PHY 2049 SPING 2001 EXAM 3 1. A negatively charged particle moves with velocity v in a uniform magnetic field as shown in figure (both vector and vector v are in the plane of the paper). What is the direction of the force on the particle? Answer: out of the paper v - y the right hand rule for the cross product,the direction of the force would be into the page if the charge was positive. However,the charge is negative. The force is out of the paper. 2. A charged particle rotates with velocity v in a uniform magnetic field.howdoesthe period of rotations depend on particle s velocity? Answer: independent of v For circular motion, F c = ma c = mv 2 /r. Since the magnetic force causes the circular motion, F = qv. Thisgivesr = mv/q. TheperiodT is the time needed to complete the orbit. The total distance traveled by the orbit is the speed times the total time,2πr = vt. This gives 2π mv q = vt T = 2πm q. Since v is absent from the above equation,the period is independent of the speed. 3. A wire 1.8 m long carries a current of 13 A and makes and angle of 35 with a uniform magnetic field =1.5 T. Calculate the magnetic force on the wire. Answer: 20 N The magnetic force on a current carrying wire is The magnitude of the force is F = I L. F = IL sin = (13 A)(1.8 m)(1.5 T)sin35 =20N. 4. There are a number of wires in space carrying different currents (see figure). What is the result of integrating the -field along a circular path of radius r =1masshownin figure. Answer: Tm

2 2 A 4 A A 3A Ampere s law states d s = µ0 i enc. The sign of the currents is found by curling the fingers of the right hand in the direction of integration. If the current passes through the area in the same direction as the outstretched thumb,the current is positive. If the current is in the opposite direction,the current is negative. This sign convention for the currents makes the 1 A current negative and the 2 A current positive. The 3 A current pierces the area in both directions and cancels itself. The 4 A current is eternal to the line integral and contributes nothing. Therefore, d s = µ0 i enc =( Tm/A)( 1 A+2A)= Tm. 5. Two infinitely long and parallel wires running along the z-direction cross the (, y) plane at the points (0,d/2) and (0, d/2) as shown in the figure. The currents in each of the two wires are identical. At which point along the positive -ais ( >0) is the magnitude of magnetic field maimum? Answer: = d/2 d/2 Wire 1 r -d/2 Wire 2 r 1 2 The magnetic field due to each wire will the same for all points on the ais since all

3 points are equidistant from the currents. The field due to one wire is 1 = 2 = µ 0i 2πr = µ 0 i 2π 2 +(d/2). 2 The net magnetic field is = = ( 1 cos î 1 sin ĵ)+( 2 cos î 2 sin ĵ) = 2 1 sin ĵ = µ 2 0 i 2π 2 +(d/2) 2 2 +(d/2) 2 µ 0 i = π( 2 +(d/2) 2 ). To find the maimum,use the first derivative test, d d = µ 0i π ( 1( 2 +(d/2) 2 ) ) (2)() 0 = 2 + d = d 2. ( 2 +(d/2) 2 ) 2 6. An infinitely long wire carrying a current of 1 A is bent 90 with a radius of curvature of the bend equal to 1 cm. What is the magnitude of the magnetic field at the center point of the bend O. Answer: 36 µt O The figure consists of two half-infinite wires and one quarter of a circular loop. The fields due to each section is in the same direction at O. The total field is the sum of the fields due to each section. = 2 1 ( ) µ0 i + 1 ( ) µ0 i 2 2π 4 2 = µ ( 0i 1 2 π 4) + 1 = (4π ( 10 7 Tm/A)(1 A) 1 2(0.01 m) π + 1 4) = T.

4 7. One of the future high energy physics eperiments (Compact Muon Solenoid) will have the largest solenoid in the world. It will be 8 m in diameter,10 m long,and will have 4 T field inside. Find the amount of energy stored in this solenoid. Answer: J The energy density is The total energy is u = 2 2µ 0 = (4 T) 2 2(4π 10 7 Tm/A) = J/m 3. U = u (Volume) = u (πr 2 L)=( J/m 3 )π(4 m) 2 (10 m) = J. 8. A uniform magnetic field is perpendicular to the plane of a circular wire loop of radius r. The magnitude of the field varies with time according to = 0 e t/τ,where 0 and τ are constants. Find the emf in the loop at t =0. Answer: πr 2 0 /τ The emf created by a changing magnetic flu is found from Faraday s law,. The magnetic flu through a circular loop is Φ = A =( 0 e t/τ )(πr 2 ). Differentiating, Evaluating the emf at t =0, ( ) 1 = πr 2 0 e t/τ τ = πr2 0 e t/τ. τ E = πr2 0. τ 9. A stiff wire bent into a circle of radius r is rotated with frequency f in a uniform magnetic field. The loop includes a light bulb of resistance. What is the maimum current going through the light bulb? Answer: 2π 2 r 2 0 f/ The magnetic flu is Φ = A = 0 A cos. In this situation both 0 and A are constant. It is that changes in time. Therefore,

5 = d ( 0A cos ) = 0 A( sin ) d = 0 πr 2 ω sin where ω = d/ is the angular speed of the rotating loop. The angular speed is related to the linear frequency, ω =2πf. Thisgives E =2 0 π 2 r 2 f sin. The current in the loop is found from Ohm s law to be I = E/. The maimum current will occur during the maimum emf. That occurs when sin = 1. The maimum current is I = E = 2 0π 2 r 2 f. 10. A solenoid having and inductance of 6.3 µh is connected in series with a 1.2 kohm resistor. If a 14 V battery is switched across the pair,how long will it take for the current through the resistor to reach 80% of its final value? Answer: 8.5 ns The epression for the current in a charging inductor is i = E ( ) r/τ L. The final current is E/. Substituting, When the current is 80% of the final value, i =0.80(E/). i = E ( ) r/τ L 0.80 E = E ( ) r/τ L 0.80 = t/τ 0.20 = e t/τ ln 0.20 = t τ L t = τ ln The time constant is τ L = L/ =( H)/( Ω) = s. This gives t = ( s)(ln 0.20) = s.

Exam 2 Solutions. ε 3. ε 1. Problem 1

Exam 2 Solutions. ε 3. ε 1. Problem 1 Exam 2 Solutions Problem 1 In the circuit shown, R1=100 Ω, R2=25 Ω, and the ideal batteries have EMFs of ε1 = 6.0 V, ε2 = 3.0 V, and ε3 = 1.5 V. What is the magnitude of the current flowing through resistor