Chapter 28: Mastering Physics, Test 4

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.

The magnetic field due to the length of the wire: F=BIw So torque=bFcos(?) Which is: ? = BIwbcos(?)

The torque on the loop is: torque=?Bsin? The more general vector form of this expression is ? =? ×B

Torque on the loop: IABsin?=I(pi*r^2)Bsin?

Learning Goal To understand Ampère's law and its application. Ampère's law is often written ?B (r )?dl =?0Iencl.

the line integral along the (magnetic field along a) closed loop

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.)

along any closed path that you choose.

a and b

True

False The key point is that to be able to use Ampère'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

True

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)

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

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.

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

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

Learning Goal: To apply Ampère's law to find the magnetic field inside an infinite solenoid. In this problem we will apply Ampère'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.

path must pass through r and must have symmetry B.

integral B*dl=BinL

The net current through the Ampèrian 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.)

Calculate magnetic field using ampere's law: Iencl = InL

Bin = ?0nI

Consider only locations where the distance from the ends is many times D.

Find the direction of the magnetic field at each of the indicated points.

BA is out of the page.

BB is into the page.

BC is out of the page.

BD is out of the page.

B? E is into the page.

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.

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.

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?×B? on a long straight wire, F=ILBsin(?) = I?L×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=?LI2×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]

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?×10-7 ? T?m/A ? 4?×10-7 / [2?] ?A2/m ?T?m/A ? 2×10-7 ? T?A 1 T = 1 N / [A?m] ?2×10e-7 ? N/m

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.

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)

out of the screen

A toroid is a solenoid bent into the shape of a doughnut. It looks similar to a toy Slinky® 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.

Evaluate the line integral and find Iencl B(r) = ?0IN2?r

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.

(Figure 1) Two point charges, with charges q1 and q2, are each moving with speed v toward the origin. At the instant shown q1 is at position (0, d) and q2 is at ( d, 0). (Note that the signs of the charges are not given because they are not needed to determine the magnitude of the forces between the charges.)

Apply coulombs law find the value R^2 F = q1q2/4??02d2

Find the magnetic field of q1 on q2. Magnitude of the magnetic field. The Biot-Savart law, which gives the magnetic field produced by a moving charge, can be written , where is the permeability of free space and is the vector from the charge to the point where the magnetic field is produced. Note we have in the numerator, not , necessitating an extra power of in the denominator. Determine the cross product Find the direction of the magnetic field. Computing the force. F = q1q2v^2?0/4?2?2d^2

Determine which force has a greater magnitude by finding the ratio of the electric force to the magnetic force and then applying the approximation. The magnitude of the electric force is greater than the magnitude of the magnetic force. This result holds quite generally: Magnetic forces between moving charges are much smaller than electric forces as long as the speeds of the charges are nonrelativistic.