Exercise Exercise Correct. Correct. Part A

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1 Heimadæmi 7 Due: :00pm on Thursday, March 3, 206 You will receive no credit for items you complete after the assignment is due. Grading Policy Exercise 28. z A +z direction mm A long, straight wire lies along the axis and carries a 3.80 current in the. Find the magnetic field (magnitude and direction) produced at the following points by a segment of the wire centered at the origin. Part A x = 2.00 m, y = 0, z = 0 Enter your answers numerically separated by commas.,, B = 0, x B y B z,0 T All attempts used; correct answer displayed Part B x = 0, y = 2.00 m, z = 0 Enter your answers numerically separated by commas.,, B x B y B z = ,0,0 T Part C x = 2.00 m, y = 2.00 m, z = 0 Enter your answers numerically separated by commas.,, B x B y B z = ,2.02 0,0 T Part D x = 0, y = 0, z = 2.00 m Enter your answers numerically separated by commas.

2 ,, B x B y B z = 0,0,0 T Exercise 28.8 Certain fish, such as the Nile fish (Gnathonemus), concentrate charges in their head and tail, thereby producing an electric field in the water around them. This field creates a potential difference of a few volts between the head and tail, which in turn causes current to flow in the conducting seawater. As the fish swims, it passes near objects that have resistivities different from that of seawater, which in turn causes the current to vary. Cells in the skin of the fish are sensitive to this current and can detect changes in it. The changes in the current allow the fish to navigate.since the electric field is weak far from the fish, we shall consider only the field running directly from the head to the tail. We can model the seawater through which that field passes as a conducting tube of area and having a potential difference across its ends. These fish navigate by responding to changes in the current in seawater. This current is due to a potential difference of around 3.00 V generated by the fish and is about 2.0 ma within a centimeter or so from the fish. Receptor cells in the fish are sensitive to the current. Since the current is at some distance from the fish, their sensitivity suggests that these cells might be responding to the magnetic field created by the current. To get some estimate of how sensitive the cells are, we can model the current as that of a long, straight wire with the receptor cells 2.00 away. cm Part A What is the strength of the magnetic field at the receptor cells? B = 0.2 μt Exercise cm Two long, straight, parallel wires, 0.0 apart carry equal 4.00 currents in the same direction, as shown in the figure. A

3 Part A Find the magnitude of the magnetic field at point P, midway between the wires. B = 0 T Part B What is its direction? to the left to the right upward downward no field Part C Find the magnitude of the magnetic field at point P2, 25.0 cm to the right of P. B = T

4 Part D What is its direction? to the left to the right upward downward no field Part E Find the magnitude of the magnetic field at point P3, 20.0 cm directly above P. B = T Part F What is its direction? to the left to the right upward downward no field Problem The figure is a sectional view of two circular coils with radius a, each wound with N turns of wire carrying a current I, circulating in the same direction in both coils. The coils are separated by a distance a equal to their radii. In this configuration the coils are called Helmholtz coils; they produce a very uniform magnetic field in the region between them.

5 Part A Derive the expression for the magnitude B of the magnetic field at a point on the axis a distance x to the right of point P, which is midway between the coils. N I B= ( + ) μ0 a2 2 ((x+a/2) 2 a2) 3/2 ((x a/2) 2 a2) 3/2 N I B= ( ) μ0 a2 2 ((x+a/2) 2 a2) 3/2 ((x a/2) 2 + a2) 3/2 N I B= ( + ) μ0 a2 2 ((x+a/2) 2 + a2) 3/2 ((x a/2) 2 + a2) 3/2 N I B= ( ) μ0 a2 2 ((x+a/2) 2 + a2) 3/2 ((x a/2) 2 + a2) 3/2 Part B From part (a), obtain an expression for the magnitude of the magnetic field at point P. Express your answer in terms of the variables N, I, a, and appropriate constant ( μ 0 ). B = μ0ni a Part C P I A a cm Calculate the magnitude of the magnetic field at if 30 turns, = 7.00, and = 7.00.

6 B = T Part D Calculate at. db/dx P(x = 0) db dx x=0 = 0 Part E Calculate B/d at. d 2 x 2 P(x = 0) d 2 B dx2 x=0 = 0 Part F Discuss how your results in parts d) and e) show that the field is very uniform in the vicinity of P Character(s) remaining Breytingin er engin við agnarsmáa breytingu dx Graded, see 'My Answers' for details Force between an Infinitely Long Wire and a Square Loop A square loop of wire with side length a carries a current I. The center of the loop is located a distance d from an infinite wire carrying a current I2. The infinite wire and loop are in the same plane; two sides of the square loop are parallel to the wire and two are perpendicular as shown.

7 Part A What is the magnitude, F, of the net force on the loop? Express the force in terms of I, I2, a, d, and μ 0. Hint. How to approach the problem You need to find the total force as the sum of the forces on each straight segment of the wire loop. You'll save some work if you think ahead of time about which forces might cancel. Hint 2. Determine the direction of force Which of the following diagrams correctly indicates the direction of the force on each individual line segment? Hint. Direction of the magnetic field In the region of the loop, the magnetic field points into the plane of the paper (by the right hand rule). Hint 2. Formula for the force on a current carrying conductor The magnetic force on a straight wire segment of length l, carrying a current I with a uniform magnetic field B along its length, is F = I l B where l is a vector along the wire in the direction of the current.,

8 a b c d Hint 3. Determine the magnitude of force Which of the following diagrams correctly indicates the relative magnitudes of the forces on the parallel wire segments? Hint. Find the magnetic field due to the wire What is the magnitude, B, of the wire's magnetic field as a function of perpendicular distance from the wire, r. Express the magnetic field magnitude in terms of I2, r, and μ 0. Hint. Ampère's law Use Ampère's law to obtain the magnetic field. Ampère's law states that = B dl μ0 Iencl, where the line integral can be done around any closed loop. B = μ 0 I2 2πr a b c d

9 Hint 4. Find the force on the section of the loop closest to the wire What is the magnitude of the force from it? Express your answer in terms of μ 0, I, I2, a, and d. F on the section of the loop closest to the wire, that is, a distance d a/2 Hint. Formula for the force on a current carrying conductor The magnetic force on a straight wire segment of length l, carrying a current I with a uniform magnetic field B along its length, is F = I l B where l is a vector along the wire in the direction of the current. Hint 2. Find the magnetic field due to the wire What is the magnitude, B, of the wire's magnetic field as a function of perpendicular distance from the wire, r. Express the magnetic field magnitude in terms of I2, r, and μ 0., Hint. Ampère's law Use Ampère's law to obtain the magnetic field. Ampère's law states that = B dl μ0 Iencl, where the line integral can be done around any closed loop. B = μ0i2 2πr F = a I 2 μ0i 2π(d ) a 2 Hint 5. Find the magnetic field due to the wire What is the magnitude, B, of the wire's magnetic field as a function of perpendicular distance from the wire, r. Express the magnetic field magnitude in terms of I2, r, and μ 0. Hint. Ampère's law Use Ampère's law to obtain the magnetic field. Ampère's law states that = B dl μ0 Iencl, where the line integral can be done around any closed loop.

10 B = μ 0 I2 2πr ( F = ) μ 0 II2a 2π d a 2 d+ a 2 Part B The magnetic moment of a current loop is defined as the vector whose magnitude equals the area of the loop times the magnitude of the current flowing in it ( m = IA), and whose direction is perpendicular to the plane in which the current flows. Find the magnitude, F, of the force on the loop from Part A in terms of the magnitude of its magnetic moment. Express F in terms of m, I2, a, d, and μ 0. m μ0i2m F = ( ) 2πa d a 2 d+ a 2 The direction of the net force would be reversed if the direction of the current in either the wire or the loop were reversed. The general result is that "like currents" (i.e., currents in the same direction) attract each other (or, more correctly, cause the wires to attract each other), whereas oppositely directed currents repel. Here, since the like currents were closer to each other than the unlike ones, the net force was attractive. The corresponding situation for an electric dipole is shown in the figure below. Magnetic Field due to a Wire Conceptual Question The same amount of current I is flowing through two wires, labeled and 2 in the figure, in the directions indicated by the arrows. In this problem you will determine the direction of the net magnetic field at each of the indicated points (A C).

11 Part A What is the direction of the magnetic field at point A? Recall that the currents in the two wires have equal magnitudes. Hint. The magnitude of the magnetic field due to a long, straight current carrying wire The magnitude B of a magnetic field B is directly proportional to the amount of current I flowing in the wire and inversely proportional to the distance r from the wire: B= μ 0 I. 2πr Hint 2. The direction of the magnetic field due to a long, straight current carrying wire The magnetic field surrounding a long, straight wire encircles the wire, as shown in the figure: The direction of the field is determined by a right hand rule: Grasp the wire with the thumb of your right hand in the direction of the current flow. The direction in which your fingers encircle the wire is the direction in which the magnetic field encircles the wire. Hint 3. How to approach the problem To determine the direction of the magnetic field at point A, you must determine the contribution to the field from both of the wires. The field at point A is the vector sum of these two contributions. Because point A is in the same plane as the wires, the contribution to the net magnetic field at point A from wire or wire 2 will either point into or out of the screen. Keep in mind that if the magnetic fields are in opposite

12 directions, the larger field will decide the direction of the net magnetic field. If they are the same size, the net magnetic field will be zero. Hint 4. Find the direction of the magnetic field at point A due to wire Is the magnetic field from wire directed into or out of the screen at point A? Be sure to point your thumb in the direction of the current, in this case to the right. in out Hint 5. Find the direction of the magnetic field at point A due to wire 2 Is the magnetic field from wire 2 directed into or out of the screen at point A? Be sure to point your thumb in the direction of the current, in this case downward. in out points out of the screen at A. points into the screen at A. points neither out of nor into the screen and B at A. net 0 = 0 at A. Part B What is the direction of the magnetic field at point B? Hint. Find the direction of the magnetic field at point B due to wire Is the magnetic field from wire directed into or out of the screen at point B? in out Hint 2. Find the direction of the magnetic field at point B due to wire 2 Is the magnetic field from wire 2 directed into or out of the screen at point B?

13 in out points out of the screen at B. points into the screen at B. B points neither out of nor into the screen and at B net 0 = 0 at B. Part C What is the direction of the magnetic field at point C? Hint. Find the direction of the magnetic field at point C due to wire Is the magnetic field from wire directed into or out of the screen at point C? in out Hint 2. Find the direction of the magnetic field at point C due to wire 2 Is the magnetic field from wire 2 directed into or out of the screen at point C? in out points out of the screen at C. points into the screen at C. points neither out of nor into the screen and B at C. net 0 = 0 at C. Score Summary:

14 Your score on this assignment is 94.9%. You received 6.65 out of a possible total of 7 points.