# Chapter 28: Mastering Physics, Test 4

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Torque on a Current Loop in a Magnetic Field
Learning Goal To understand the origin of the torque on a current loop due to the magnetic forces on the current-carrying wires. This problem will show you how to calculate the torque on a magnetic dipole in a uniform magnetic field. We start with a rectangular current loop, the shape of which allows us to calculate the electromagnetic forces explicitly. Then we generalize our result. Even if you already know the general formula to solve this problem, you might find it instructive to discover where it comes from.
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A current I flows in a plane rectangular current loop with height w and horizontal sides b. The loop is placed into a uniform magnetic field B? in such a way that the sides of length w are perpendicular to B? (Figure 1) , and there is an angle ? between the sides of length b and B? (Figure 2) . Calculate ?, the magnitude of the torque about the vertical axis of the current loop due to the interaction of the current through the loop with the magnetic field.
The magnetic field due to the length of the wire: F=BIw So torque=bFcos(?) Which is: ? = BIwbcos(?)
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Give a more general expression for the magnitude of the torque ?. Rewrite the answer found in Part A in terms of the magnitude of the magnetic dipole moment of the current loop ?. Define the angle between the vector perpendicular to the plane of the coil and the magnetic field to be ?, noting that this angle is the complement of angle ? in Part A.
The torque on the loop is: torque=?Bsin? The more general vector form of this expression is ? =? Ã—B
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A current I flows around a plane circular loop of radius r, giving the loop a magnetic dipole moment of magnitude ?. The loop is placed in a uniform magnetic field B? , with an angle ? between the direction of the field lines and the magnetic dipole moment as shown in the figure. (Figure 3) Find an expression for the magnitude of the torque ? on the current loop.
Torque on the loop: IABsin?=I(pi*r^2)Bsin?
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AmpÃ¨re's Law Explained
Learning Goal To understand AmpÃ¨re's law and its application. AmpÃ¨re's law is often written ?B (r )?dl =?0Iencl.
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The integral on the left is
the line integral along the (magnetic field along a) closed loop
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What physical property does the symbol Iencl represent?
The net current through the loop The positive direction of the line integral and the positive direction for the current are related by the right-hand rule: Wrap your right-hand fingers around the closed path, then the direction of your fingers is the positive direction for dl? and the direction of your thumb is the positive direction for the net current. Note also that the angle the current-carrying wire makes with the surface enclosed by the loop doesn't matter. (If the wire is at an angle, the normal component of the current is decreased, but the area of intersection of the wire and the surface is correspondingly increased.)
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The circle on the integral means that B (r ) must be integrated
along any closed path that you choose.
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Which of the following choices of path allow you to use AmpÃ¨re's law to find B (r )? a. The path must pass through the point r . b. The path must have enough symmetry so that B (r )?dl is constant along large parts of it. c. The path must be a circle.
a and b
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AmpÃ¨re's law can be used to find the magnetic field around a straight current-carrying wire.
True
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AmpÃ¨re's law can be used to find the magnetic field at the center of a square loop carrying a constant current.
False The key point is that to be able to use AmpÃ¨re's law, the path along which you take the line integral of B must have sufficient symmetry to allow you to pull the magnitude of B outside the integral. Whether the current distribution has symmetry is incidental.
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The key point is that to be able to use AmpÃ¨re's law, the path along which you take the line integral of B? must have sufficient symmetry to allow you to pull the magnitude of B outside the integral. Whether the current distribution has symmetry is incidental.
False
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AmpÃ¨re's law can be used to find the magnetic field inside a toroid. (A toroid is a doughnut shape wound uniformly with many turns of wire.)
True
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Current Sheet
Consider an infinite sheet of parallel wires. The sheet lies in the xy plane. A current I runs in the -y direction through each wire. There are N/a wires per unit length in the x direction. (Figure 1)
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Write an expression for B(d), the magnetic field a distance d above the xy plane of the sheet. Use ?0 for the permeability of free space.
Current through one turn=I Total current through n turns I' = I(n/a) L Using Ampere's law B 2L =?o I (n/a) L So... B_vec(d) = -[(- ?oIn/a)/2] i
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Magnetic Field of a Current-Carrying Wire
Find the magnetic field a distance r from the center of a long wire that has radius a and carries a uniform current per unit area j in the positive z direction.
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First find the magnetic field, B out(r ), outside the wire (i.e., when the distance r is greater than a). (Figure 1)
The magnetic field outside current carrying wires: Ampere's circuital law gives: integral of Bdl=mu(knot)I(enclosed) I=current enclosed mu=permiability B=strength of magnetic field. The current density: J=I/A---> I/(pi*a^2) I=J((pi/)a^2) Bout2(pi)r=mu(knot)I Bout=mu(knot)I/(2(pi)r) Insert I Bout=(mu(knot)Ja^2/2r)theta hat
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Now find the magnetic field B? in(r? ) inside the wire (i.e., when the distance r is less than a). (Figure 2)
Current enclosed of wire of radius a: pi*a^2-->I I'=(Ir^2)/a^2 Circuital Law: Bin(2pi*r)=mu(knot)I' after some algebra: Bin=(mu(knot)Jr/2)theta hat
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Magnetic Field inside a Very Long Solenoid
Learning Goal: To apply AmpÃ¨re's law to find the magnetic field inside an infinite solenoid. In this problem we will apply AmpÃ¨re's law, written ?B? (r? )?dl? =?0Iencl, to calculate the magnetic field inside a very long solenoid (only a relatively short segment of the solenoid is shown in the pictures). The solenoid has length L, diameter D, and n turns per unit length with each carrying current I. (Figure 1) It is usual to assume that the component of the current along the z axis is negligible. (This may be assured by winding two layers of closely spaced wires that spiral in opposite directions.) From symmetry considerations it is possible to show that far from the ends of the solenoid, the magnetic field is axial.
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Which figure shows the loop that the must be used as the AmpÃ¨rian path for finding Bin(r) for r inside the solenoid?
path must pass through r and must have symmetry B.
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Assume that loop B (in the Part A figure) has length L along k^ (the z direction). What is the loop integral in AmpÃ¨re's law? Assume that the top end of the loop is very far from the solenoid (even though it may not look like it in the figure), so that the field there is assumed to be small and can be ignored.
integral B*dl=BinL
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What physical property does the symbol Iencl represent?
The net current through the AmpÃ¨rian path The positive direction of the line integral and the positive direction for the current are related by the right-hand rule: Wrap your right-hand fingers around the closed path, then the direction of your fingers is the positive direction for dl? and the direction of your thumb is the positive direction for the net current. Note also that the angle the current-carrying wire makes with the surface enclosed by the loop doesn't matter. (If the wire is at an angle, the normal component of the current is decreased, but the area of intersection of the wire and the surface is correspondingly increased.)
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What is Iencl, the current passing through the chosen loop?
Calculate magnetic field using ampere's law: Iencl = InL
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Find Bin, the z component of the magnetic field inside the solenoid where AmpÃ¨re's law applies.
Bin = ?0nI
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Which of the following conditions must hold to allow you to use AmpÃ¨re's law to find a good approximation?
Consider only locations where the distance from the ends is many times D.
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Direction of the Magnetic Field due to a Wire Conceptual Question
Find the direction of the magnetic field at each of the indicated points.
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What is the direction of the magnetic field BA at Point A?
BA is out of the page.
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What is the direction of the magnetic field BB at Point B?
BB is into the page.
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Note that the bottom wire carries a current of magnitude 2I. What is the direction of the magnetic field B? C at Point C?
BC is out of the page.
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What is the direction of the magnetic field BD at Point D?
BD is out of the page.
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What is the direction of the magnetic field B? E at Point E?
B? E is into the page.
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Force between Two Infinite, Parallel Wires
You are given two infinite, parallel wires each carrying current I. The wires are separated by a distance d, and the current in the two wires is flowing in the same direction. This problem concerns the force per unit length between the wires.
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Is the force between the wires attractive or repulsive?
Attractive: current flows in the same direction The second right hand rule assists us in solving this problem. Since we have two parallel wires with equal current going the same direction, that vector represents each of the wires velocity vectors. By pointing our right thumb in this direction and using our pointer finger to represent the magnetic field (pointing back toward ourselves), at any point, the Force of the wire will be represented by our middle finger. Thinking of this for both wires, the force vectors are pointing towards each other, resulting in an attractive relationship.
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What is the force per unit length F/L between the two wires?
B=mu(knot)I/2(pi)d F=ILB--->F/L=ID= I(mu(knot)I/2(pi)d)---> mu(knot)I^2/2(pi)d First some relevant formulas: FB=qvBsin(?) = qv?Ã—B? on a long straight wire, F=ILBsin(?) = I?LÃ—B? where L is the length of the wire So the magnetic field at wire 2 from the current in wire 1 will be B=?0I1 ? 1/[2?d] The force on a length ?L of wire 2 will be F=?LI2Ã—B The force per unit length in terms of the currents will be, F/?L = ?0I1I2 ? 1/[2?d] since I1 = I2, F/L = ?0I2 / [2?d]
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In the SI system, the unit of current, the ampere, is defined by this relationship using an apparatus called an AmpÃ¨re balance. What would be the force per unit length of two infinitely long wires, separated by a distance 1m, if 1A of current were flowing through each of them?
If I = 1 A, & d = 1 m, plugging these values into the formula from the previous section, F/L = ?0(1 A)2 / [2?(1m)] = ?0/ [2?] ? A2/m ?0 = 4?Ã—10-7 ? T?m/A ? 4?Ã—10-7 / [2?] ?A2/m ?T?m/A ? 2Ã—10-7 ? T?A 1 T = 1 N / [A?m] ?2Ã—10e-7 ? N/m
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Magnetic Field due to Semicircular Wires
A loop of wire is in the shape of two concentric semicircles as shown. (Figure 1) The inner circle has radius a; the outer circle has radius b. A current I flows clockwise through the outer wire and counterclockwise through the inner wire.
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What is the magnitude, B, of the magnetic field at the center of the semicircles?
For thesemicirle, Themagnetic field at the point P due to the inner wire of radius 'a'is Ba = (1/2)(?0I / 2a) Themagnetic field at the point P due to the inner wire of radius 'b'is Bb = (1/2)(?0I / 2b) Now thedirections of these two magnetic field's are in the oppositedirections, as a result the resulatnat magnetic field at the pointP is B = Ba - Bb =(1/2)(?0I / 2a) - (1/2)(?0I /2b)
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What is the direction of the magnetic field at the center of the semicircles?
out of the screen
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Magnetic Field inside a Toroid
A toroid is a solenoid bent into the shape of a doughnut. It looks similar to a toy SlinkyÂ® with ends joined to make a circle. Consider a toroid consisting of N turns of a single wire with current I flowing through it. (Figure 1) Consider the toroid to be lying in the r ? plane of a cylindrical coordinate system, with the z axis along the axis of the toroid (pointing out of the screen). Let ? represent the angular position around the toroid, and let r be the distance from the axis of the toroid. For now, treat the toroid as ideal; that is, ignore the component of the current in the ?^ direction.
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The magnetic field inside the toroid varies as a function of which parameters?
Evaluate the line integral and find Iencl B(r) = ?0IN2?r
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Note that whether the field points upward or downward depends on the direction of the current, that is, on whether the coil is wound clockwise or counterclockwise.
simplify the problem B =?0I2R This is the same expression that you would derive for the magnetic field at the center of a circular loop of current-carrying wire. To see why this makes sense, imagine that the local diameter d of the coils gets so small that it is negligible in comparison to the radius of the toroid. The wire makes one complete turn around the axis of the toroid. So, to a point in the center, the toroid looks like a simple current loop.
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Force between Moving Charges