The right-hand rule is a way to determine the direction of the magnetic field produced by the moving charges. Imagine wrapping your right hand around the path of the charges so that the positive charges flow from the little finger side of your fist to the thumb side. Then your finger circles the path in the same direction as the magnetic field lines. Use this rule to verify the direction of the magnetic field around moving negative charges.

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Sagar YadavLv10

22 Nov 2020

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Learning Goal: To work through astraightforward application of Faraday's law to find the EMF andthe electric field surrounding a region of increasing flux

Faraday's law describes how electric fields and electromotiveforces are generated from changing magnetic fields. This problem isa prototypical example in which an increasing magnetic fluxgenerates a finite line integral of the electric field around aclosed loop that surrounds the changing magnetic flux through asurface bounded by that loop. A cylindrical iron rod withcross-sectional area is oriented with its symmetry axis coincidentwith the z axis of a cylindrical coordinate system as shown. It hasa uniform magnetic field inside that varies according to . In otherwords, the magentic field is always in the positive z direction,and it has no other components.

For your convenience, we restate Faraday's law here: , where is theline integral of the electric field, and the magnetic flux is givenby , where is the angle between the magnetic field and the localnormal to the surface bounded by the closed loop.
Direction: The line integral and surface integral reverse theirsigns if the reference direction of or is reversed. The right-handrule applies here: If the thumb of your right hand is taken along ,then the fingers point along . You are free to take the loopanywhere you choose, although usually it makes sense to choose itto lie along the path of the circuit you are considering.

Find , the electromotive force (EMF) around a loop that is atdistance from the z axis, where is restricted to the region outsidethe iron rod as shown. Take the direction shown in the figure aspositive.
Express in terms of , , , , and any needed constants such as , ,and .
Due to the cylindrical symmetry of this problem, the inducedelectric field can depend only on the distance from the z axis,where is restricted to the region outside the iron rod. Find thisfield.
Express in terms of quantities given in the introduction (andconstants), using the unit vectors in the cylindrical coordinatesystem,r^ ,theta^ , and z^ .

<DIV>Suppose in the figure that the four identical currents i = 6 A, into or out of the page as shown, form a square of side 118 cm. What is the force per unit length (magnitude and direction) on the wire in the bottom left hand corner? (N/m) Take the positive y direction as up and the positive x direction as to the right. I do know all of the answers are supposed to be in N/m http://loncapa2.physics.sc.edu/res/sc/gblanpied/courses/usclib/hrw8/hrwpictures/30-42.jpg (a)What is the magnitude of the force per unit length? (b)What is the x component of the force per unit length? (c)What is the y component of the force per unit length? ******** This method was previously submitted and the answers were found incorrect so im reposting. Thanx guys. first find magnetic field. then use F=BiL B=uI/2piR u is 4piX10^-7 I is current, 6am R is distance from between teh wire. u get 1.02X10^-6 from top left or bot right wire. And 7.19X10^-7 for the top right. then plug into equation L is 1 cuz it says unit length. So u get x-component=1.02X10^-6X6+7.19X10^-7X6XCos45=9.17X10^-6 N y-component=-1.02X10^-6X6+7.19X10^-7X6Xsin45=-3.07X10^-6 N Then u use pythragram theorem square root of Fx^2+Fy^2=ur magnitude. which is 1.02X10^-5N. the answer to b, and C is already above. b is Fx or ur x component. c is Fy or ur y component. </DIV>

A Simple Atwood Machine consists of two masses m1 and m1 that are connected by a string wound over a pulley, as seen in the figure below. m2 is larger than m1. On a piece of paper, draw two free body diagrams; one for each of the masses, showing all forces acting on each mass. Then answer the following questions. (a) Based on your free body diagram, identify the forces acting on the mass m1. Select all that apply. (m1+m2)a m1g FN m1a (m1+m2)g m2a f m2g T (b) Using the direction rosette indicate the direction for each of the forces acting on m1 that you identified in part (a). If any of the forces listed below is not applicable, enter NA. m1a= m2a= m1g= m2g= (m1+m2)a= (m1+m2)g= f= T= FN= (c) Based on your free body diagram, identify the forces acting on the mass m2. Select all that apply. T (m1+m2)g m2g m1g (m1+m2)a m1a m2a f FN (d) Using the direction rosette indicate the direction for each of the forces acting on m2 that you identified in part (c). If any of the forces listed below is not applicable, enter NA. m1a= m2a= m1g= m2g= (m1+m2)a= (m1+m2)g= f= T= FN= (e) Write Newton's Second Law equation (in the form Fnet = ma) for m1 in the y-direction. Use the first answer box for the left hand side (LHS) of the equation and the second answer box for the right hand side (RHS) of the equation. It is important to use the correct signs for each of the forces. Motion in the upward direction is positive and motion in the downward direction is negative. (Use m1, g, a, and T as appropriate.) LHS = RHS = (f) Write Newton's Second Law equation (in the form Fnet = ma) for m2 in the y-direction. Use the first answer box for the left hand side (LHS) of the equation and the second answer box for the right hand side (RHS) of the equation. It is important to use the correct signs for each of the forces. Motion in the upward direction is positive and motion in the downward direction is negative.(Use m2, g, a, and T as appropriate.) LHS = RHS = (g) Use your answers from parts (e) and (f) to get an expression for the magnitude of the tension T. (Use m1, m2, and g as appropriate.) magnitude of T = (h) Use your answers from parts (e), (f), and (g) to find the magnitude of a, the acceleration of the system. (Use m1, m2, and g as appropriate.) magnitude of a = (i) In what direction will each mass accelerate. For mass m1: For mass m2: (j) Suppose m1 = 7 kg and m2 = 11 kg. What is the tension in the string? T = (k) Suppose m1 = 7 kg and m2 = 11 kg. What is the value of the acceleration? a = (l) Suppose that m2 starts from rest at a height of 5 m. Use the Kinematic Equations to determine how long it takes for m2 to hit the ground.

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