Physics 8.02 Exam Two Mashup Spring 2003

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Edmund Hart
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1 Physics 8.0 Exam Two Mashup Spring 003 Some (possibly useful) Relations: closedsurface da Q κ d = ε E A inside points from inside to outside b V = V V = E d s moving from a to b b a E d s = 0 V many point charges C = C Q V parallelplate = o q N i i= 1 ε o r ri Q V = C ε A o = d U = 1 C V = Q C 1 1 = + 1 C C C C series 1 = C +C parallel 1 E= ρj where ρ is the resistivity V = i R R = ρl A 1 R parallel = 1 R R R series = R 1 + R V Pohmic heating = i V = i R = R µ ˆ o q v r B= v < < c r I d µ ˆ o s r db = r rˆ pointsfromsource to observer a B ds =µ oi contour through where I through is the current flowing through any open surface bounded by the contour: Ithrough = J da open surface ds right-handed with respect to da d dφ Eds = ε = N dt BdA dt F= q v B Fcent. ext = mv r µ = IAn n perpendicular to loop, sgl loop df= Ids B right-handed with respect to I τ = µ B dbz Fz = µ z dz Cross-products of unit vectors: ˆi ˆi = ˆj ˆj= kˆ k ˆ = 0 ˆi ˆj = k ˆ ˆj kˆ = ˆi k ˆ ˆ i = ˆ j Useful integrals: dθ = 1/ ( c+ fθ ) ( c+ fθ ) f dθ 1 ln( c fθ ) ( c+ fθ ) = f + dθ 1 1 = + θ + θ 1/ ( c f ) f ( c f ) Useful Small Argument Approximations (1 + ε) n 1 + nε for ε<<1 ln(1 + ε ) ε for ε<<1 ext

2 Problem 1 (5 Points) Circle your choice for the correct answer to the five questions below. A: A current I is uniformly distributed over the cross-section of a conducting wire of radius a. For distances r from the center of the wire with r < a, 1. The magnitude of the magnetic field does not depend on r.. The magnitude of the magnetic field is proportional to 1/r 3. The magnitude of the magnetic field is proportional to r 4. The magnitude of the magnetic field is proportional to 1/r 5. The magnitude of the magnetic field is proportional to r B: A segment of a wire of length dl carries a current I (see sketch below). The magnitude of the magnetic field due to this wire segment at the location P shown in the sketch is µ I o a dl b µ I o a dl b µ oi b dl a + b µ I o b dl a + b µ I o dl a + b [ a + ] 3 / [ a + ] [ ] 3 / [ ] [ ] Page

C: Two wires run parallel to the z-azis, which is out of the

page. The wire on the left carries a current of I = I / also

representation of the magnetic field of these two wires shown

In an iron filings representation, the magnetic fields are

3 C: Two wires run parallel to the z-azis, which is out of the page. The wire on the right carries a current I > 0 out of the page. The wire on the left carries a current of I = I / also out of the page (see sketch) Which of the four iron filings representation of the magnetic field of these two wires shown below is correct? In an iron filings representation, the magnetic fields are parallel (or anti-parallel) to the streaks, and an x structure denotes a zero in the field strength. (a) (b) (c) (d) Page 3

4 D: Two charged particles of identical mass and charge move in circular orbits in the same constant magnetic field B = zˆ. The two particles have different speeds. B o The orbit of the particle with the larger speed will: E: 1. have a larger radius and a longer period as compared to that of the slower particle. have a smaller radius and a shorter period as compared to that of the slower particle 3. have a larger radius and the same period as compared to that of the slower particle 4. have a smaller radius and the same period as compared to that of the slower particle A circuit in the form of a rectangular loop of wire is pushed toward a long wire carrying current I in the direction shown in the sketch. 1. The induced current in the loop is clockwise and the net force on the loop is to the left.. The induced current in the loop is clockwise and the net force on the loop is to the right. 3. The induced current in the loop is counterclockwise and the net force on the loop is to the left. 4. The induced current in the loop is counterclockwise and the net force on the loop is to the right. Page 4

F. A coil of wire defines an open surface whose normal da points downward, as shown in the sketch. The coil is below a magnet whose North pole points upward.

5 F. A coil of wire defines an open surface whose normal da points downward, as shown in the sketch. The coil is below a magnet whose North pole points upward. As the coil moves from well below the magnet to well above that magnet, the magnetic flux through the coil due to the field produced by the magnet only looks like: (1) () (3) (4) Page 5

6 Problem (5 points): A current I flows around a continuous path that consists of portions of two concentric circles of radii a and a/, respectively, and two straight radial segments. The point P is at the common center of the two circle segments. (a) Use the Biot-Savart Law to calculate db at P due only to that segment of the path dl shown in the sketch. Indicate on the sketch the vector r ˆ you use and the direction of db. Give the magnitude of db in terms of I, a, dl, and µ o. (b) Derive an expression for the magnetic field at P due to the larger circle segment only. Give its magnitude and direction. Answers without work will receive no credit. (c) What is the total field B at P? Give its magnitude and direction. Page 6

7 Problem 3 (5 points): An infinitely long solenoid with radius R carries current I and has n turns per unit length. The current in the solenoid circulates counterclockwise when viewed from the top. You may assume that the magnetic field inside the solenoid is constant and that the magnetic field outside the solenoid is zero. (a) On the figure above, draw the Amperean loop that will allow you to find the magnetic field inside the solenoid at the point P. Indicate the dimensions of the loop. (b) What is the total current flowing through your Amperean Loop in terms of its dimensions and the quantities given? (c) Using Ampere s Law, derive an expression for the magnetic field inside the solenoid. Give its magnitude and direction in terms of the quantities given and µ o. Answers without work will receive no credit. Page 7

8 Problem 3 continued (d) We now insert a second solenoid of radius R/ into the first solenoid. The second solenoid carries the same current I as the first, except that the current in the second solenoid circulates clockwise as seen from the top. Also, the number of turns per unit length of the second solenoid is twice that of the first solenoid (see sketch). What is the magnetic field along the common axis of the two solenoids? Give its magnitude and direction in terms of the quantities given and µ o. You must justify your answer using either the Biot-Savart Law, Ampere s Law, or the Principle of Superposition. Page 8

9 Problem 4 (5 points) A pie-shaped circuit is made from a straight vertical conducting rod of length a welded to a conducting rod bent into the shape of a semi-circle with radius a (see sketch). The circuit is completed by a conducting rod of length a pivoted at the center of the semicircle, Point P, and free to rotate about that point. This moving rod makes electrical contact with the vertical rod at one end and the semi-circular rod at the other end. The angle θ is the angle between the vertical rod and the moving rod, as shown. The circuit sits in a constant magnetic field B ext pointing out of the page. (a) If the angle θ is increasing with time, what is the direction of the resultant current flow around the pieshaped circuit. Draw the direction of the current flow on the diagram to the left. To get credit for the right answer, you must justify your answer, using a Lenz s Law argument or some other argument. For the next two parts, assume that the angle θ is increasing at a constant rate, θ () t = ω t, where ω is a constant. (b) What is the magnitude of the rate of change of the magnetic flux through the pieshaped circuit due to B ext only (do not include the magnetic field associated with any induced current in the circuit)? Page 9

10 (c) If the pie-shaped circuit has a constant resistance R, what is the magnitude and direction of the magnetic force due to the external field on the moving rod in terms of the quantities given. Show the direction of the force on the sketch below. Page 10

Physics 8.02 Exam Two Equation Sheet Spring 2004

Physics 8.02 Exam Two Equation Sheet Spring 2004 Physics 8.0 Exam Two Equation Sheet Spring 004 closed surface EdA Q inside da points from inside o to outside I dsrˆ db 4o r rˆ points from source to observer V moving from a to b E ds 0 V b V a b E ds