cancel each other out. Thus, we only need to consider magnetic field produced by wire carrying current 2.

Transcription

1 PC /2012 Exam Solutions Question 1 a) Assumption: shells are conductors. Notes: the system given is a capacitor. Make use of spherical symmetry. Energy density, =. in this case means electric field strength. To find, we use gauss law. = 4 = = 1 4 = = = where V represents volume. = = = = Remember to include upper and lower limit when doing integration! b) = +. represents dielectric constant. c) is smaller than, since there are induced charges on the dielectric surface and work is done by system to induce these charges, leading to a decrease in stored energy. Question 2 a) = = = = T b) i) Use =. Using the right hand grip rule, the magnetic fields due to the wires carrying current cancel each other out. Thus, we only need to consider magnetic field produced by wire carrying current 2. = = 2

2 = 2 2 = 2 2 = 2 2 b) ii) Arrows defining positive x and positive y. = 2 2 = + = 2 2 = 2 2 = sin 45 = 2 2 = 2 = cos 45 = 2 = 2 = 2 = 2 2 = 0 = 2 2 = 0 So, = 0. Question 3 a) At steady state, before the switch is flipped: By loop rule, + E = 0 At steady state, = 0 E = = 0 = E After the switch is flipped from a to b: By loop rule, = 0 = = = 0 = E

3 Must memorise solution to above differential equation! b) Power dissipated in resistor = = E Total energy dissipated = = E = E Integrate from = 0 to =, total energy dissipated, E = E = = 0 = Energy stored in inductor at = 0. (shown) Question 4 a) For RLC is parallel, is the same phase as the source voltage. = = =. = sin = sin lag by radian (Note: must know this for exam!) = cos lead by radian (Note: must know this for exam!) = = cos = cos b) Power dissipated in circuit = Power dissipated in resistor = = sin = sin c) Energy stored in circuit = Energy in capacitor + Energy in inductor = = 1 2 sin cos = 1 2 sin cos Question 5 a) Concept: Direction of propagation is same as direction of. Direction of propagation is direction (given) and is in direction, since it is given that is perpendicular to direction and it must also be perpendicular to the direction. Note:, and in solution to this question refer to unit vectors. =, = cos + ῳ = cos + ῳ, = cos + ῳ Find ῳ using ῳ = 2, is given in the question to be 100MHz. Find using = ῳ. b) =

4 = 1/μ cos + ῳ cos + ῳ = 1 μ cos + ῳ c) = = = Given = 1000W/m, and given μ and, we can solve for. Then, find with formula =. Question 6 a) i) In this solution, and refer to unit vectors. = 1 4 = = + Unit vector, = = = 2 = = = = = Because charge distribution is symmetrical about axis, net = 0. a) ii) When line of charge is infinitely long,, = 4 + = where =. 4 = 4 b) i) =. Consider Gauss Law, =, 2 =, where is the charge on one cylinder. = 1 2 Since =, =,

5 = 2 = 1 2 Take integration limits from = to =, = ln 2 = = ln = 2 ln 2 b) ii) Given. Consider the very thin cylindrical strip, by Gauss Law, =, 2 = 2 = = 2 = 2 = = = 2 2 But I don t know how to comment on my answer. =( Question 7 a) i) Making use of Biot-Savart s Law: = 1 4 Let = cos + sin = sin + cos = = = cos sin + = + Unit vector or, = cos sin + = 4 + sin + cos cos sin + + = 4 + sin + sin + cos + cos = sin + cos = = 4 + = = 1 2 +

6 a) ii) = 1 2 = 2 + = + For approaching : + = = b) = = = c) i) and represent unit vectors. At point, = = cos + sin = sin + cos Using Biot-Savart s Law, = 1 4 = = cos + sin For point far away from loop, = cos + = cos + =, since point on axis is very far away from loop. = 1 cos + = 4 sin + cos 1 cos + = 4 sin + cos = 4 sin + cos = 4 cos = cos = 4 For point very far away from loop, is very large. = 0. c) ii) Ampere s Law: = Consider a closed loop from = to, from = 0 to point.

7 Consider over this closed loop, = This is because the system is symmetric about axis, so on line above axis cancels out with on line below axis, and on line cutting through is 0 because we approximate = 0 at any point along this line. Question 8 a) i) By Faraday's Law, = Φ In this case, Φ = 1 Note: consider in direction only, because in direction doesn t contribute to flux. = 2 1 = 2 = 2 = + = = 2 + a) ii) Draw a graph, same shape as = graph. Must say that asymptote is at = 0 and intercept is. a) iii) = = = = 2 + a) iv) The emf induced as a result of changing magnetic flux through wire loop is non conservative and depends on path taken. Even if =, with respect to, current flows from to, has higher potential. With respect to, current flows from to, has higher potential. So the readings on both voltmeters will be different (have different sign). b) Use the definition: flux through triangle loop = mutual inductance current through straight wire Consider similar triangles: From trigonometry ratios, tan 60 = 3 =

8 = 3 Consider many horizontal rectangular strips, = 2 = 2 3 cutting through the rectangular strip, = Flux through the triangle, = = 3 = 3 = = 3 + ln + = ln 3. Using the definition: flux through triangle loop = mutual inductance current through straight wire ln = 3 = ln 3 Answers provided by Formatted and edited by : Woo Jia QIan : John Soo NUS Physics Society (2012)