(a) What is the direction of the magnetic field at point P (i.e., into or out of the page), and why?

Transcription

1 Physics 9 Fall 2010 Midterm 2 s For the midterm, you may use one sheet of notes with whatever you want to put on it, front and back Please sit every other seat, and please don t cheat! If something isn t clear, please ask You may use calculators All problems are weighted equally PLEASE BOX YOUR FINAL ANSWERS! You have the full length of the class If you attach any additional scratch work, then make sure that your name is on every sheet of your work Good luck! 1 A closed circuit consists of two semicircles of radius R and R/2 that are connected by straight segments, as seen in the figure to the right A current I exists in this circuit and has a clockwise direction a) What is the direction of the magnetic field at point P ie, into or out of the page), and why? b) Calculate the magnitude of the magnetic field at point P c) Suppose that r = 40 cm, and I = 3 amps What s the magnetic field? a) By the right-hand rule for currents, the magnetic field from the circuit points into the page b) The total magnetic field is just the sum of the two semicircles the straight-line segments contribute nothing, since they are inline with the point P) We can use the Biot-Savart law to determine the magnetic field from each semicircle From the top, since the distance from the wire to the point is constant, r = R, and the current is always tangent to the point, the Biot-Savart law says B top = µ 0I 4π = µ 0I = µ 0I d s ˆr r 2 4πR ds 2, 4R where we have recalled that the distance around the semicircle is s = πr The magnetic field from the bottom wire is done in exactly the same way, but now the radius is only r = R/2 So, B bottom = µ 0I Combining the two fields gives the 2R total, B = 3µ 0I 4R c) Plugging in the numbers gives B = 3µ 0I 4R = 3 4π = 7 µt

2 2 An inductor and resistor is connected to a 60 Hz generator with a peak voltage of 100 V At this frequency, the circuit has an impedance of 10 Ω and an inductive reactance of 8 Ω a) What is the resistance, R, of the resistor? b) What is the peak current in the coil? c) What is the phase angle in degrees) between the current and the applied voltage? d) A capacitor is put in series with the inductor and the generator What capacitance is required so that the current is in phase with the generator voltage? e) What is the peak voltage measured across the capacitor for this in-phase condition? a) For a circuit with only a resistor and inductor, the impedance Z = R 2 + X 2 L, where R is the resistance, and X L = ωl is the inductive reactance Since Z = 10 and X L = 8, then R = 6 Ω b) The peak current in the circuit is I 0 = E 0 /Z = 100/10 = 10 amps c) The phase angle is given by tan φ = X L /R = 8/6 φ = 53 d) When the current and voltage are in phase, then φ = 0 When we include a capacitor, then tan φ = X L X C, and so the in-phase condition sets X R L = X C = 8 Ω Since X C = 1/ωC, then C = 1/ωX C = 1/120π 8) = 330 µf e) When the current and voltage are in-phase, then X C = X L, and so the impedance Z = R 2 + X L X C ) 2 = R Thus, the peak current is I = E 0 /Z = E 0 /R = 100/6 = 167 amps, and the peak voltage through the capacitor is V C = I 0 X C = = 133 V 2

3 3 A circular loop is constructed from a flexible, conducting wire of resistance R The wire is being cooled such that it s radius is shrinking at a rate rt) = r 0 1 t ), τ where r 0 is the initial, uncooled radius, τ is a constant, and t is the time The loop is perpendicular to a uniform, constant magnetic field B a) Find an expression for the induced voltage in the loop as a function of time b) What is the induced current in the loop? a) The induced voltage, E, is given by E = d dt Φ m, where Φ m is the magnetic flux In this case, since the magnetic field is constant, and perpendicular to the loop, then the flux is just Φ m = BA = Bπr 2 Plugging in for the radius we have Φ m t) = πr0b 2 1 t ) 2 τ To get the induced current we just need to take the derivative Doing so gives E = 2πr2 0B 1 t ) τ τ b) The induced current is just the voltage, divided by the resistance, I = E/R, or It) = 2πr2 0B 1 t ) Rτ τ 3

4 4 The maximum electric field strength in air is 30 MV/m Stronger electric fields ionize the air and create a spark What is the maximum power that can be delivered by a 10 cm diameter laser beam propagating through the air? The power of a laser is related to its intensity by P = IA, where A is the area of the beam Furthermore, the intensity is the average of the Poynting vector, I = S = 1 2µ 0 E 0 B 0, where E 0 and B 0 are the amplitude of the electric and magnetic fields, respectively But these amplitudes are related by E 0 = cb 0, and so I = 1 2µ 0 c E2 0 Thus, the power is So, plugging in the numbers gives P = IA = E2 0 2µ 0 c A P = IA = E2 0 2µ 0 c A = ) 2 24π 10 7 ) ) π ) 2 = W 4

5 Extra Credit Question!! The following is worth 10 extra credit points! Theorists have speculated about the existence of magnetic monopoles, and several experimental searches for such monopoles have occurred Suppose magnetic monopoles were found and that the magnetic field at a distance r from a monopole of strength q m is given by B = µ 0 q m 4π r 2 Modify the Gauss s law for magnetism equation to be consistent with such a discovery Gauss s law for magnetism says that B d A = 0, which says that magnetic monopoles don t exist; we want to adjust this expression to include monopoles We ll do it by analogy with Gauss s law for the electric field, which says that E da = Q encl ɛ 0 For a single electric point charge of strength q, then a spherical Gaussian surface of radius r gives the electric field as E = 1 4πɛ 0 q r 2, as we ve seen many times before In analogy with Gauss s law for the electric field, if we had individual magnetic charges, q m, then we expect that we could write Gauss s law for these charges as B da = µ 0 Q m encl We can easily check that this gives the expected magnetic field of a monopole Take a spherical Gaussian surface of radius r, enclosing a magnetic point charge q m Then, the magnetic field is constant over the surface, and points along the direction of the normal to the surface, and so the left-hand-side of the modified Gauss s law gives B da = B da = B 4πr 2 ), just as in the electric case The right-hand-side reads µ 0 Q m encl = µ 0 q m, and so B = µ 0 q m 4π r, 2 which is the expected field Thus we have the correct generalization 5

6 Some Useful Constants Some Possibly Useful Information Coulomb s Law constant k 1 4πɛ 0 = Nm2 C 2 The magnetic permeability constant µ 0 = 4π 10 7 N A 2 Speed of Light c = m/s 11 Nm2 Newton s Gravitational Constant G = kg 2 The charge on the proton e = C The mass of the electron, m e = kg The mass of the proton, m p = kg Boltzmann s constant, k B = J/K 1 ev = Joules 1 MeV = 10 6 ev 1 Å = meters Planck s constant, h = J s = ev s The reduced Planck s constant, h 2π = J s = ev s Some Useful Mathematical Ideas { x n+1 x n n 1, n+1 dx = ln x) n = 1 dx a = ln x + a 2 + x 2) 2 +x 2 x dx a = a 2 + x 2 2 +x 2 Other Useful Stuff The force on an object moving in a circle is F = mv2 r Kinetmatic equations xt) = x 0 + v 0x t a xt 2, yt) = y 0 + v 0y t a yt 2 The binomial expansion, 1 + x) n 1 + nx, if x 1 6