hw18
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hw18

Course Number: PHY 303K, Fall 2007

College/University: A.T. Still University

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homework 18 LEE, BENJAMIN Due: Oct 16 2007, 4:00 am Question 1, chap 30, sect 1. part 1 of 3 10 points A device ("source") emits a bunch of charged ions (particles) with a range of velocities (see figure). Some of these ions pass through the left slit and enter "Region I" in which there is a vertical uniform electric field (in the -^ direction) and a 0.2 T uniform mag ^ netic field (aligned...

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18 homework LEE, BENJAMIN Due: Oct 16 2007, 4:00 am Question 1, chap 30, sect 1. part 1 of 3 10 points A device ("source") emits a bunch of charged ions (particles) with a range of velocities (see figure). Some of these ions pass through the left slit and enter "Region I" in which there is a vertical uniform electric field (in the -^ direction) and a 0.2 T uniform mag ^ netic field (aligned with the k-direction) as shown in the figure by the shaded area. +2800 V q m Region of Magnetic Field 0.2 T FE = -FB . 1 Therefore, the forces are equal and opposite and the magnitude of forces are equal; i.e., FE = FB . The force due to the magnetic field provides the centripetal force that causes the positive ions to move in the semicircle. As the negatively charged ion exits the region of the electric field, FB = q v B, so by the right-hand rule the magnetic field must point out of the page or in the -z-direction ^ -k , since the force F is in the direction down the page; i.e., "-^ " q =- |q| v = +^ i v B B FB FB the vector product ^ k = -^ , and since i ^ F = q v B = F (-^) F F = q |q| B v v B =? = -^ , and 39 cm 1 cm y x z Region I Region II Figure: ^ is in the direction +x i (to the right), ^ is in the direction ^ +y (up the page), and k is in the direction +z (out of the page). In which direction (relative to the coordinate system shown above) should the magnetic field point in order for negatively charged ions to move along the path shown by the dotted line in the diagram above? 1. B ^ = -k correct ^ = - (+^) -k i B B = -^ , ^ = -k B 2. B = 0 ; direction undetermined 3. B ^ = +k consequently is correct. B Explanation: To obtain a straight orbit, the upward and downward forces need to cancel. The force on a charged particle is F = FE + FB = q (E + v B) . For the force to be zero, we need FE + FB = 0 , or Question 2, chap 30, sect 1. part 2 of 3 10 points In "Region I", the electric potential between the plates is 2800 V, the distance between the plates is 1 cm, and the magnetic field in both "Regions I and II" is 0.2 T . What is the speed of a singly charged ion that passes through both slits and makes it into "Region II"? homework 18 LEE, BENJAMIN Due: Oct 16 2007, 4:00 am Correct answer: 1.4 106 m/s (tolerance 1 %). Explanation: Let : B = 0.2 T , and V (2800 V) E = d (1 cm) = 280000 N/C . Since the electric and magnetic forces on the ion are equal, qE = qvB 280000 N/C E = = 1.4 106 m/s . v= B 0.2 T Question 3, chap 30, sect 1. part 3 of 3 10 points The ions that make it into "Region II" are observed to be deflected downward and then follow a circular path with a radius of r = 0.39 m. The charge on each ion is 4.4 10-18 C. What is the mass of the ions? Correct answer: 2.4514310-25 kg (tolerance 1 %). Explanation: 2 An infinite wire is bent as shown in the figure. The current is I. It consists of three segments, AB, BCD, and DE. AB and DE are parallel and both infinite in length. BCD is a semi-circular arc, with radius OB = OC = OD = r. The following questions are concerned with the magnetic field vector, B, at O. E D y C O II III z I x IV A B What is the direction of B at O due to the current through the entire wire, ABCDE? 1. into the xy plane correct 2. out of the xy plane 3. in quadrant II 4. in quadrant IV 5. in the positive y direction Let : r = 0.39 m = 0.39 m q = 4.4 10-18 C . and 6. in quadrant I 7. in the positive x direction 8. in the negative y direction 9. in the negative x direction 10. in quadrant III The radius of a circular path taken by a charged particle in a magnetic field is given by mv r= . qB Br m=q v = (4.4 10-18 C) (0.2 T)(0.39 m) 1.4 106 m/s = 2.45143 10-25 kg . Question 4, chap 29, sect 5. part 1 of 2 10 points Explanation: From the right hand rule, we can see that the direction of B from all three segments, AB, BCD, and DE is directed into the page. Question 5, chap 29, sect 5. part 2 of 2 10 points homework 18 LEE, BENJAMIN Due: Oct 16 2007, 4:00 am The permeability of free space is 1.25664 10-6 T m/A. The radius of curvature is 0.599 m . Find the magnitude of B at O due to a 4 A current through the entire wire ABCDE. Correct answer: 3.43345 10-6 T (tolerance 1 %). Explanation: 3 I b P a Let r = 0.599 m I = 4 A and 0 = 1.25664 10-6 T m/A . O r s I A What is the magnitude of the component of the magnetic field ( Ba Ba ) at P due to the current in the inner semicircle (r = a)? 1. Ba = 2. Ba = 3. Ba = 4. Ba = 5. Ba = 6. Ba = 0 I a 8 4 0 I a 0 I 4a 0 I correct 4a 4 0 I a 0 I 2a C B x x Note: The magnetic field due to DE is equal in magnitude and direction to the magnetic field due to AB. The semicircular arc BCD subtends radian. Thus the total magnetic field is Btotal = BAB + BDE + BBCD 0 I 0 I + =2 4r 4r 0 I 2 +1 = 4r (1.25664 10-6 T m/A) (4 A) = 4 (0.599 m) 2 +1 = 3.43345 10-6 T . Question 6, chap 29, sect 5. part 1 of 4 10 points A current I flows in the directions indicated by the arrows in the figure along a wire bent to form concentric semicircles. 7. Ba = 0 0 I a 0 I 9. Ba = a 0 I 10. Ba = 8a Explanation: Solution: If we are not concerned with the field's direction, the Biot-Savart law gives 8. Ba = 0 I ds 4 a2 since to distance point is fixed; i.e. r = a. Integration over the semicircle yields dB = B= Thus Ba = 0 I . 4a 0 I 4 a2 ds = 0 I ( a) . 4 a2 homework 18 LEE, BENJAMIN Due: Oct 16 2007, 4:00 am Question 7, chap 29, sect 5. part 2 of 4 10 points What is the magnitude of the component of the magnetic field ( Bb Bb ) at P due to the current in the outer semicircle (r = b)? 1. Bb = 0 I b 0 I 2. Bb = 8b 0 I 3. Bb = 2b 0 I 4. Bb = b 5. Bb = 0 4 0 I b 0 I 7. Bb = 4b 4 0 I 8. Bb = b 0 I 9. Bb = correct 4b 0 I b 10. Bb = 4 Explanation: The integral here is similar, only the radius is b, not a 0 I 0 I Bb = ds = ( b) . 2 4b 4 b2 Thus 0 I Bb = . 4b 6. Bb = Question 8, chap 29, sect 5. part 3 of 4 10 points What is the magnitude of the magnetic field ( B B) at P due to the current in the entire loop? 0 I 1. B = 4 0 I 2. B = 2 1 1 + a b 1 1 - a b 0 I 2 0 I 4. B = 2 0 I 5. B = 4 0 I 6. B = 2 3. B = 7. B = 0 0 I 4 0 I 9. B = 4 8. B = 1 1 - a b 1 1 + a b 1 a 1 a 1 a 1 a 1 b 1 + b 1 - b 1 - b + 4 correct Explanation: The integral along the x-axis is zero since the cross product is zero. The magnitude of the magnetic field due to the inner semicircle and outer semi-circle subtract since the currents are in opposite directions. Thus B= 0 I 0 I 0 I - = 4a 4b 4 1 1 - a b . Question 9, chap 29, sect 5. part 4 of 4 10 points What is the direction of the magnetic field at P due to the current in the entire loop? 1. Into the page correct 2. Out of the page 3. The magnitude is 0, thus the direction is undetermined. 4. The direction cannot be determined using simple techniques. Explanation: The magnetic fields produced from the inner semi-circle and outer semi-circle subtract since the currents are in opposite directions. The inner circle contributes more strongly to homework 18 LEE, BENJAMIN Due: Oct 16 2007, 4:00 am the direction of the magnitude field. Grab it by your right hand with your thumb in the direction of the current, and your fingers, when projected at point P, point in the same direction as the magnetic field. Question 10, chap 29, sect 5. part 1 of 2 10 points A conductor in the shape of a square of edge length carries a clockwise current I as shown in the figure below. By the Biot-Savart law, 5 0 I ds ^ r . 2 4 r Consider a thin, straight wire carring a constant current I along the x-axis with the yaxis pointing towards the center of the square, as in the following figure. y P dB = r ^ r ds a P I What is the magnitude of the magnetic field at point P (at the center of the square loop) due to the current in only one of the sides of the wire? 0 I 1. B = 2 0 I 2. B = 2 0 I 3. B = 2 2 0 I 4. B = 2 2 0 I 2 5. B = 0 I 2 6. B = 3 2 0 I 7. B = 0 I 8. B = 2 3 0 I 9. B = 2 0 I 10. B = 2 correct 2 Explanation: x O I Let us calculate the total magnetic field at the point P located at a distance a from the wire. An element ds is at a distance r from P . The direction of the field at P due to this element is out of the paper, since ds ^ is r out of the paper. In fact, all elements give a contribution directly out of the paper at P . Therefore, we have only to determine the magnitude of the field at P . In fact, taking the origin at O and letting P be along the ^ positive y axis, with k being a unit vector pointing out of the paper, we see that x ^ ds ^ = k |ds ^| = k (dx sin ) . r ^ r Substituting into Biot-Savart law gives ^ dB = k dB, with 0 I dx sin . (1) 4 r2 In order to integrate this expression, we must relate the variables , x, and r. One approach is to express x and r in terms of . From the geometry in the figure and some simple differentiation, we obtain the following relationship a = a csc . (2) r= sin a Since tan = - from the right triangle in x the figure, dB = x = -a cot and dx = a csc2 d . (3) homework 18 LEE, BENJAMIN Due: Oct 16 2007, 4:00 am Substituting Eqs. 2 and 3 into Eq. 1 gives 0 I a csc2 sin d dB = 4 a2 csc2 0 I sin d . = 4a 2. B is zero. 3. B is into the page. correct (4) 4. B is to the right. 5. B is out of the page. 6. B is up the page. 7. B is down the page. 6 Thus, we have reduced the expression to one involving only the variable . We can now obtain the total field at P by integrating Eq. 4 over all elements subtending angles ranging from 1 to 2 as defined in the figure. This gives 0 I 2 B= sin d 4 a 1 0 I = (cos 1 - cos 2 ) . 4a We can apply this result to this problem. For a segment of wire, as set up above, the magnetic field at a point P is B= 0 I (cos 1 - cos 2 ) . 4a Explanation: The direction of the magnetic field due to a current element is determined by the cross product in the definition of the magnetic field B= 0 I 4 ds ^ r . 2 r For the present case the right hand rule gives the direction of the magnetic field as into the page or in for short. For a square wire loop consider the bottom segment. Using the above general formula for this case gives Bone = 0 I 4 2 0 I cos 3 - cos 4 4 2 , 2 B = 4 Bone 2 0 I = 2, = where since all four sides contribute and the direction and magnitude of the field is the same for each side. Question 11, chap 29, sect 5. part 2 of 2 10 points What is the direction of the magnetic field BP at point P due to the upward current in the left-hand side of the square wire? 1. B is to the left.

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A.T. Still University - PHY - 303K
homework 19 LEE, BENJAMIN Due: Oct 18 2007, 4:00 am of the solenoid. Question 1, chap 29, sect 5. part 1 of 1 10 points A cross section of a long solenoid that carries current I is shown. I (into the page) rI1Question 2, chap 29, sect 5. part
A.T. Still University - PHY - 303K
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A.T. Still University - PHY - 303K
oldhomewk 10 LEE, BENJAMIN Due: Sep 26 2007, 4:00 am Question 1, chap 26, sect 1. part 1 of 1 10 points A small electrically charged object is suspended by a thread between the vertical plates of a parallel-plate capacitor. The acceleration of grav
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oldhomewk 11 LEE, BENJAMIN Due: Sep 28 2007, 4:00 am Question 1, chap -1, sect -1. part 1 of 4 10 points An infinite chain of capacitors is pictured below with C1 = 8 F, C2 = 4.37 F, and C3 = 11.3 F.a C1 C1 C11C1 C2 C3 2 = Ceq + C2 Ceq C1 + C3
A.T. Still University - PHY - 303K
oldhomewk 12 LEE, BENJAMIN Due: Sep 30 2007, 4:00 am Question 1, chap -1, sect -1. part 1 of 1 10 points Why are thick wires rather than thin wires usually used to carry large current? 1. Thick wires are difficult to break. 2. Thick wires have a la
A.T. Still University - PHY - 303K
oldhomewk 13 LEE, BENJAMIN Due: Oct 2 2007, 4:00 am and the total charge is Question 1, chap 27, sect 1. part 1 of 1 10 points How long does it take electrons to get from the car battery to the starting motor? Assume the current is 113 A and the el
A.T. Still University - PHY - 303K
oldhomewk 14 LEE, BENJAMIN Due: Oct 4 2007, 4:00 am Rb = R3 + R4 = 1 + 15 = 16 Question 1, chap 28, sect 7. part 1 of 2 10 points The circuit has been connected as shown in the figure for a "long" time. Ib = It = E 16 V = =1A Rt 16 16 V E = =1A
A.T. Still University - PHY - 303K
oldhomewk 15 LEE, BENJAMIN Due: Oct 7 2007, 4:00 am E2 r1 = -i1 r1 r2 + I r1 (R + r2 ) Question 1, chap 28, sect 4. part 1 of 1 10 points 30 V 6 Adding, E1 r2 + E2 r1 = I [r2 R + r1 (R + r2 )]115 V5 I= 22 E 1 r2 + E 2 r1 r2 R + r1 (R + r2 )
A.T. Still University - PHY - 303K
oldhomewk 16 LEE, BENJAMIN Due: Oct 9 2007, 4:00 am Question 1, chap 28, sect 8. part 1 of 4 10 points A 154 lamp, a 21 electric heater, and a 45 fan are connected in parallel across a 135 V line. What total current is supplied to the circuit? C
A.T. Still University - PHY - 303K
oldhomewk 17 LEE, BENJAMIN Due: Oct 11 2007, 4:00 pm Question 1, chap 29, sect 1. part 1 of 1 10 points A duck flying due north at 65 m/s passes over Atlanta, where the Earth's magnetic field is 7.6 10-5 T in a direction 19 below the horizontal l
A.T. Still University - PHY - 303K
oldhomewk 18 LEE, BENJAMIN Due: Oct 14 2007, 4:00 am Question 1, chap 30, sect 3. part 1 of 1 10 points What is the maximum torque on a 200-turn circular coil of radius 0.95 cm that carries a current of 1.3 mA and resides in a uniform magnetic fiel
A.T. Still University - PHY - 303K
oldhomewk 19 LEE, BENJAMIN Due: Oct 16 2007, 4:00 am Question 1, chap 29, sect 5. part 1 of 2 10 points The wire is carrying a current I. y 1801I OI Rin from , points towards O; i.e., in the xdirection. We need to find d s ^ to use the r B
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