202 Exam 3
The figure shows a cross section of a long cylindrical conductor of radius a = 2 cm, carrying a uniformly distributed current i = 40 A into the page. Evaluate B(x) at the values asked below. B(x = 0 cm)? B(x = 0.5 cm)? B(x = 1. cm)? B(x = 1.5 cm)? B(x = 2 cm)? B(x = 3 cm)? B(x = 4 cm)? B(x = 5 cm)? B(x = 6 cm)?
-if r=0, then B=0 -B for x is less than or equal to 2 =((a*10^-7) x current x (xE10^-2))/radius^2 -B for x greater than 2 =((a*10^-7) x current)/x
When a 43.0-V emf device is placed across two resistors in series, a current of 11.0 A is flowing in each of the resistors. When the same emf device is placed across the same two resistors in parallel, the current through the emf device is 55.0 A. What is the magnitude of the larger of the two resistances?
1) I=V/(R1+R2) solve for R1+R2 2) I2=V(R1+R2)/R1R2 solve for R1R2 3) R1^2-(R1+R2)R1+R1R2=0 4) quadratic formula
A Nichrome heater dissipates 500 W when the applied potential difference is 110 V and the wire temperature is 825°C. What would be the dissipation rate if the wire temperature were held at 200°C by immersing the wire in a bath of cooking oil? The applied potential difference remains the same, and α for Nichrome at 825°C is 4.00×10-4/K.
1) P=E^2/R0 solve for R0 2)Rt=R0(1 +(α x change in t)) 3)P=E^2/Rt solve for P
An unknown resistor is connected between the terminals of a 2.0 V battery. Energy is dissipated in the resistor at the rate of 0.540 W. The same resistor is then connected between the terminals of a 0.70 V battery. At what rate is energy now dissipated?
1) R=V1^2/P 2) P=V2^2/R
You are given a number of 10 Ω resistors, each capable of dissipating only 1.0 W without being destroyed. What is the minimum number of such resistors that you need to combine in series or parallel to make a 10 Ω resistance that is capable of dissipating at least 5.0 W?
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Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 1.2 mm. Conductor B is a hollow tube of outside diameter 6.3 mm and inside diameter 5.3 mm. What is the resistance ratio RA/RB , measured between their ends?
Aa=pi x r^2 Ab=(pi x r^2outer)-(pi x r^2inner) r=d/2 x 10^-3
A long circular pipe with outside radius R carries a (uniformly distributed) current i = 50 A into the page as shown in the figure. A wire runs parallel to the pipe at a distance 3R from center to center. Find the magnitude and direction of the current in the wire such that the resultant magnetic field at point P between them and a distance R from the wire has the same magnitude as the resultant field at the center of the pipe but is in the opposite direction. Take positive current as into the page and negative as out of the page.
B1=(muknot * current)/(2pir) where r=distance from the pipe B2=((muknotx current)/4piR)-((muknot x current)/2piR) B1=B2 solve for i
Suppose in the figure that the four identical currents i = 20 A, into or out of the page as shown, form a square of side 90 cm. What is the force per unit length (magnitude and direction) on the wire in the bottom left hand corner? (N/m) Take the positive y direction as up and the positive x direction as to the right. a)What is the magnitude of the force per unit length? b)What is the x component of the force per unit length? c)What is the y component of the force per unit length?
F1=(4pix10^-7 times current^2)/(2pi times a) F2=(4pix10^-7 times current^2)/(2pi times a(sqrt2)) F3=(4pix10^-7 times current^2)/(2pi times a) Fx=-(F3+F2Cos(45)) Fy=F1-F2sin(45) Ftotal=sqrt(Fx^2+Fy^2)
A battery of ε = 2.90 V and internal resistance R = 0.600 Ω is driving a motor. The motor is lifting a 2.0 N mass at constant speed v = 0.50 m/s. Assuming no energy losses, find the current i in the circuit. a) Enter the lower current. b) Enter the higher current c) Find the potential difference V across the terminals of the motor for the lower current. d)Find the potential difference V across the terminals of the motor for the higher current.
Power used by motor=Force * Velocity Rnet=R+internal resistance Current In circuit=E/Rnet Voltage across R=(R*E)/Rnet a/b)E^2 * R=(R + Rinternal)^2 and solve for higher and lower R c/d) V=Power used by moter/current for higher and lower
What is the equivalent resistance of the five resistors in the circuit in the figure, if R1 = 32 Ω?
R1 and R2 are in series so R1 + R2 R12 is parallel to R4 so R124=R12 x R4/(R12 + R4) R124 is parallel to R5 so R1245=R124 x R5/(R124 + R5) R1245 is in series with R3 so R12453=R1245 + R3
A common flashlight bulb is rated at 0.300 A and 2.70 V (the values of the current and voltage under operating conditions). If the resistance of the bulb filament at room temperature (20°C) is 2.40 Ω, what is the temperature of the filament, in degrees Celcius (degC), when the bulb is on? The filament is made of tungsten.
Resistance when operating=V/I R=R0[1+a (T-T0)] where a is 0.004403 for tungsten solve for T
An electron has an initial velocity of (13.0j + 15.0k) km/s and a constant acceleration of (1.50×1012 m/s2)i in a region in which uniform electric and magnetic fields are present. If B =(400μT)i, find the electric field E. a)Ex? b)Ey? c)Ez?
VB=(i + j + k):of initial velocity*(i + j + k):of b field a) mass of electron * constant acceleration of x/(-e) then subtract x component of VB b) mass of electron * constant acceleration of y/(-e) then subtract y component of VB c)mass of electron * constant acceleration of z/(-e) then subtract z component of VB where mass of electron=9.1x10^-31 -e=-1.6x10^-19
Eight wires cut the page perpendicularly at the points shown in the figure. A wire labeled with the integer k (k = 1,2,...,8) carries the current ki where i = 1 A. For those with odd k, the current is out of the page; for those with even k, it is into the page. Evaluate the closed integral B dot dl along the closed path in the direction shown.
choose the points that the current is passing through -add the first two together and the second two together and then subtract the second from the first -4pix10^-7 x ans step 1=final answer
A certain substance has a dielectric constant of 2.8 and a dielectric strength of 18.0 MV/m. If it is used as the dielectric material in a parallel-plate capacitor, what minimum area should the plates of the capacitor have to obtain a capacitance of 5.00×10-8 F and to ensure that the capacitor will be able to withstand a potential difference of 5.7 kV?
d=potential difference/dielectric strength A=Capacitance*d/constant*E0)
Consider the circuit in the figure (b). The curved segments are arcs of circles of angle 15o and radii a = 28 cm and b = 14 cm. The straight segments are along radii. Find the magnetic field B at point P ( the center of curvature), assuming a current of 17 A clockwise in the circuit. (Positive out of page and negative into page)
magnetic field at P=Kx I x theta(1/rb-1/ra) theta in radian
A long solenoid has 110 turns/cm and carries current i. An electron moves within the solenoid in a circle of radius 2.10 cm perpendicular to the solenoid axis. The speed of the electron is 0.042 c (c = speed of light). Find the current i in the solenoid.
mv/r=q(4pix10^-7 times n times current) m=mass of electron v=speed of electron x10^8 * 3 r=radius of circle q=charge of electron n=turns per meters solve for current
A metal wire of mass m = 0.250 kg slides without friction on two horizontal rails spaced a distance d = 0.26 m apart, as in the figure. The track lies in a vertical uniform magnetic field B = 1.20 T. There is a constant current i = 0.35 A through generator G, along one rail, across the wire, and back down the other rail. Find the speed and direction of the wire's motion as a function of time, assuming it to be stationary at t = 0. Evaluate for t = 0.7 s. Take positive to the right and negative to the left.
Step 1) F=ILB where I is current given L is the distance given B is the magnetic field given Step 2) F=ma where F=ans from step 1 m=mass given -solve for a Step 3) Vf=Vi +at where Vi=0 a=answer for step 2 t=time given Step 4) moves to the left
The figure shows a cross section of a long conductor of a type called a coaxial cable and gives its radii (a,b,c) = (2.0, 1.8, 0.4 cm). Equal but opposite currents i = 110 A are uniformly distributed in the two conductors. The current in the inner cylinder is into the page, while the current in the outer is out of the page. Evaluate the magnitude of B(x) for the values asked below. B(0.0 cm)? B(0.2 cm)? B(0.4 cm)? B(1.0 cm)? B(1.9 cm)? B(2.0 cm)? B(4.0 cm)?
a) 0 b) (4pix10^-7)(current)(cm x 10^-2)/(2pi x smallest radii^2) c) (4pix10^-7)(current)/(2pi times cm^2) d)
Each of two long straight parallel wires 12 cm apart carries a current of 125 A. The figure shows a cross section, with the wires running perpendicular to the page and point P lying on the perpendicular bisector of the line between the wires. Find the magnitude and direction of the magnetic field at P when the current in the left-hand wire is out of the page and the current in the right-hand wire is also out of the page. a)What is the magnitude of B? b)What is the value of Bx? c)What is the value of Bz? d)Find the magnitude and direction of the magnetic field at P when the current in the left-hand wire is out of the page and the current in the right-hand wire is into the page. What is the magnitude of B? e)What is the value of Bx? f)What is the value of By? g)What is the value of Bz?
a) B=(4x10^-7)(current)/2r then multiply times 2 b) 0 c) opposite of a d) 0 e) same as a f) 0 g) 0 h) same direction as e
A long hairpin is formed by bending a very long wire, as shown. a)If the wire carries a current of 5 A, what is the magnitude and direction of the magnetic field at the point a? Take R = 8 mm. Take the sign of B to be positive if the direction is out of the paper and negative if B is into the paper. You need to enter the number with its sign and a space and the unit. b)What is the magnitude and direction of the magnetic field at the midpoint b? The distance between a and b are much larger than R (each straight section is "infinite"). Take the sign of B to be positive if the direction is out of the paper and negative if B is into the paper
a) Ba=((4pix10^-7 * current)/(2 times RE10^-3))(1/pi + 1/2) b) Bb=((4pix10^-7 * current)/(pi times RE10^-3)
For the vectors in the figure, a = 4.7, b = 2.4 and c = 5.28. a)Calculate a x b. A positive value is for the z-component coming out of the page. The units are m. b) Calculate a x c c) Calculate b x c.
a) ai x bj = answer b) ai x -bj=answer where c is on hypotenuse c) ai x bj=answer
A solar cell generates a potential difference of 0.1731 V when a 500 Ω resistor is connected across it, and a potential difference of 0.2500 V when a 1000 Ω resistor is substituted. a)What is the internal resistance of the solar cell? b)What is the emf of the solar cell? c)The area of the cell is 5.0 cm2, and the rate per unit area at which it receives energy from light is 2.0m W/ cm2. What is the efficiency of the cell for converting the light energy to thermal energy in the 1000 Ω external resistor?
a) i1=V1/R1 and i2=V2/R2 E1=i1(R1 + r) E2=i2(R2 + r) set E1=E2 and solve for r b) E=i2(R2 + r) and plug in r from a c) Efficiency=V2^2/(R2 x A of cell x unit area x 10^-3)
An electron in a TV camera tube is moving at 5.80×106 m/s in a magnetic field of strength 89 mT. Without knowing the direction of the field, what can you say about the greatest and least magnitude of the force acting on the electron due to the field? a)Maximum force? b)minimum force? c)At one point the acceleration of the electron is 3.101×1016 m/s2. What is the angle between the electron velocity and the magnetic field? (deg)
a) max=qVB where q is charge of an electron V is velocity given B is magnetic field given x10^-2 b)minimum foce=0 c) sin^-1((9.11x10^-31)(acceleration given)/answer to a)
An electron is accelerated from rest by a potential difference of 430 V. It then enters a uniform magnetic field of magnitude 230 mT with its velocity perpendicular to the field. a)Calculate the speed of the electron. b)Calculate the radius of its path in the magnetic field.
a) qv=1/2 MeV^2 electron has q of 1.6x10^-19 and Me of 9.1x10^-31 solve for V b) R=mV/qB m=9.1x10^-31 V=answer to a q=1.6x10^-19 B=magnitude x10^-3
A 100 W lightbulb is plugged into a standard 120 V outlet. a)How much does it cost per month (30 days) to leave the light turned on continuosly? Assume electrical energy costs 4.5 cents /kWh. b)What is the resistance of the bulb? c)What is the current in the bulb?
a)24 x P x t=answ convert to Kw/h answ x 4.5 and convert to dollars b) R=V^2/P c) I=V/R
A cylindrical resistor of radius 4.2 mm and length 2.4 cm is made of material that has a resistivity of 3.50×10-5 Ω m. a) What is the current density when the energy dissipation rate in the resistor is 1.0 W? b)What is the potential difference?
a)J=sqrt(P/(rho x L x pir^2)) b)V=P/pir^2J
In a nuclear experiment a proton with kinetic energy 6.0 MeV moves in a circular path in a uniform magnetic field. a)If the magnetic field is B = 2.8 T what is the radius of the orbit? b)What energy must an alpha particle (q = +2e, m = 4.0 u) and a deuteron (q = +e, m = 2.0 u) have if they are to circulate in the same orbit? Energy of the alpha? (MeV) c)Energy of the deuteron?
a)r=mv/qB where m is mass of proton=1.67x10^-27 q is charge of a proton=1.6x10^-19 B= magnetic field v=kinetic energy of a proton=3.39x10^7 b)V=2qrB/m where q=charge of proton=1.6x10^-19 r=answer to a B=magnetic field m=4 times mass of proton Ek=1/2 mv^2 =1/2 (4 times mass of proton)*v^2=ans =ans/1.6E-19 =alpha c)use same velocity from b E=1/2 mv^2 1/2 (2 times mass of proton)*v^2=ans ans/1.6E-19=deuteron
