EXAM 3: SOLUTIONS. B = B. A 2 = BA 2 cos 0 o = BA 2. =Φ(2) B A 2 = A 1 cos 60 o = A 1 2 =0.5m2

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1 EXAM : S Q.The normal to a certain m area makes an angle of 6 o with a uniform magnetic field. The magnetic flux through this area is the same as the flux through a second area that is perpendicular to the field if the second area is: Φ () B = B. A = BA cos 6 o Φ () B = B. A = BA cos o = BA Φ () B A =m A =? =Φ() B A = A cos 6 o = A =.5m Q.A rectangular loop of wire is placed midway between two long straight parallel conductors as shown. The conductors carry currents i and i as indicated. If i is increasing and i is constant, then the induced current in the loop is : The currents i produces a magnetic field which is directed into the page and the current i produces a magnetic field which is directed out of the page. The total flux in the loop has contributions from both the fields but only the former is incresing with time while the latter is a constant. The total flux depends on the relative strengths of i and i. There are two cases, i (t) >i and i (t) <i. For both of them Lenz s law gives the direction of induced current as counter-clockwise. ˆΦ net ˆΦind î ind i (t) >i -increasing CC i (t) <i -decreasing CC (NOTE: CC - Counter-clockwise.)

2 Q.The circuit shown is in a uniform magnetic field that is into the page. The current in the circuit is. A. At what rate is the magnitude of the magnetic field changing? Is it increasing or decreasing? V net = V app + ξ ind = ir. ξ ind = ir V app. Now, the direction of current was not mentioned. This ambiguity is resolved by choosing a sign convention for the current i. Ifwechoosei to be positive for clockwise direction and negative otherwise, then { (. ) = V if î is clockwise ξ = ir V app = (. ) = 6V if î is counter-clockwise The opposing emf induces a flux which is directed into the page indicating that the primary flux is decreasing. To find the rate of change of the primary field, use the Faraday s law db dt = ξ A = ξ = dφ B = A db dt dt { (/( ) ) = T/s if î is clockwise (6/( ) ) = T/s if î is counter-clockwise If we relax the sign condition on current then we have two correct choices and both are considered right answers. QWe desire to make an LC circuit that oscillates at Hz using an inductance of.5h We also need a capacitance of : f = Hz; L =.5H; C =? ω =πf = LC C = π f L =. 6 F µ F.

3 Q5.An LC circuit has a capacitance of µ F and an inductance of 5 m H. At time t =the charge on the capacitor is µ C and the current is ma. The maximum charge on the capacitor is: The charge and current through the circuit at time t is q(t) = Q cos(ωt + φ) i(t) = Qω sin(ωt + φ) Defining q q(t =)andi i(t =), q = Q cos φ cos φ = q i Q = Qω sin φ i = Q ω sin φ = Q ω ( cos φ) i ( ) = Q q = Q q ω Q Q = q + i ω = q + i LC q =µ C; i =ma; L =5mH; C =µ F. Q =6.7 6 C 7µ C. Q6.The primary of a :step-up transformer is connected to a source and the secondary is connected to a resistor R. The power dissipated by R in this situation is P.IfR is connected directly to the source it will dissipate a power of : : For a step-up transformer, N s >N p N s N p == V s V p ; R eq p ( ) = R Np = R/9 N s With the resistor connected in the secondary circuit, power is P = V p Rp eq = 9V p R If the same resistor is connected in the primary circuit, power is P = V p R = P 9

4 Q7.A bar magnet is placed vertically with the S pole up and the N pole down. The B field at its center is : The field lines emerge out of the N pole and enter into the S pole. Inside the magnet they go S-N. SowiththeS pole facing up, the direction of the B field will be down. Q8.The impedence of the circuit shown is : (Ref. to the picture) : ω =πf = π rad s ; L =.5H; C = µ F; R = Ω. X L = Lω =5πΩ X C = Cω = 5 π Ω. Impedence, Z = R +(X L X C ) = = 7Ω. Q9.The diagram shows one plate of a parallel-plate capacitor from within the capacitor. The plate is circular and has a radius R. The dashed circles are four integration paths that have radii of r = R/, r = R/, r =R/ and r =R. Rank the paths according to the magnitude of B.d s around the paths during the discharging of the capacitor, least to greatest. Between the plates of a discharging capacitor, the electric flux is changing with time giving rise to a fictitious displacement current i d. This current produces a circular magnetic field around the plate s central axis. By the Maxwell-Ampere s law, B.d s = µ i d (r). If i d is the net displacement current between the plates, then the displacement current density is J d = i d /(πr ). The displacement current enclosed by a circular path of radius r is i d (r), where i d (r) = B.d s = ( J d πr = i r d R) (interior) J d πr = i d (exterior) ( µ i r d R) (interior; r,r ) µ o i d (exterior; r,r )

5 ( ) r µ i d B.d s = µ i d = R 6 ( ) r µ i d B.d s = µ i d = R B.d s = B.d s = µ i d =6 = B.d s B.d s B.d s < B.d s < B.d s = B.d s. 5