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Figure 29-62 shows, in cross section, two long straight wires held against a plastic cylinder of radius20.0cm. Wire 1 carries currentrole="math" localid="1663151041166" i1=60.0mAout of the page and is fixed in place at the left side of the cylinder. Wire 2 carries currenti2=40.0mAout of the page and can be moved around the cylinder. At what (positive) angleθ2should wire 2 be positioned such that, at the origin, the net magnetic field due to the two currents has magnitude80.0nT?

Short Answer

Expert verified

The angle θ2is104°

Step by step solution

01

Identification of given data

  1. The radius of the plastic cylinder R=20.0cmor0.200m
  2. Wire 1 carries current i1=60.0mAor60.0×10-3A
  3. Wire 2 carries current i2=40.0mAor40.0×10-3A
  4. The net magnetic field due toi1andi2isB=80.0nTor80×10-9T
02

Understanding the concept

When two parallel current-carrying conductors are placed near each other, they exert a force on each other, either attractive or repulsive, depending on the direction of the current. The fields are expressed by Biot-Savart’s law. It is stated as,

dB=μ04πids×r^r2

Here, dBis the field produced by the current element idsat a point located at rfrom the current element.

Using Biot-Savart’s law and Lorentz force equation, we can calculate the force on the current carrying conductor by another current carrying conductor.

By using, Equation29-4we can find the magnitude of themagnetic fields.Also, we have to find the net componentsB2xandB2yofB2. Now, by usingPythagoras theorem and substituting the given quantities, we can find the angleθ2.

Formulas:

From , 29-4the magnitude of the magnetic field is given by

B=μ0i2πR

Here, Bis magnetic field, μ0is permeability constant, iis current, Ris the distance between the point and the current conducting wire.

03

Determining the angle θ2

From,equation29-4the magnitude of the magnetic field is given by

B=μ0i2πR

By the right-hand rule, the field caused by wirecurrent, evaluated at the coordinate origin, is along+yaxis.

The field caused by the current of wire2's will generally have both an xand ay component which are related to its magnitudeB2and sines and cosines of some angle. When wire 2 is at an angleθ, its net components are given by

B2x=B2sinθ2andB2y=-B2cosθ2

By Pythagoras theorem, the magnitude-squared of the net field is equal to the sum of the square of their net x-component and square of their net y-component.

Therefore,

B2=B2sinθ22+B1-B2cosθ22=B12+B22sin2θ+B22cos2θ-2B1B2cosθ2

B2=B12+B22sin2θ+B22cos2θ-2B1B2cosθ2

Since,sin2θ+cos2θ=1

B2=B12+B22-2B1B2cosθ2

Therefore, angleθ2is

θ2=cos-1B12+B22-B22B1B2

Now, we have

B1=μ0i12πR=4π×10-7T·m/A×60.0×10-3A2×π×0.20m=60×10-9T

Similarly,

B2=μ0i22πR=4π×10-7T·m/A×40.0×10-3A2×π×0.20m=40×10-9T

With the requirement that the net field has magnitudeB=80nT, we find

θ2=cos-1B12+B22-B22B1B2=cos-160×10-9T2+40×10-9T2-80×10-9T22×60×10-9T40×10-9T=cos-1-0.25=104°

Therefore, the angle θ2is104°.

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Most popular questions from this chapter

Three long wires all lie in an xyplane parallel to the xaxis. They are spaced equally,10 cm apart. The two outer wires each carry a current of 5.0 Ain the positive xdirection. What is the magnitude of the force on a3.0 m section of either of the outer wires if the current in the center wire is 3.2 A(a) in the positive xdirection and (b) in the negative xdirection?

Figure 29-80 shows a cross-section of a long cylindrical conductor of radius a=4cmcontaining a long cylindrical hole of radiusb=1.50cm. The central axes of the cylinder and hole are parallel and are distanced=2cmapart; currentis uniformly distributed over the tinted area. (a) What is the magnitude of the magnetic field at the center of the hole? (b) Discuss the two special casesb=0andd=0.

Figure 29-29 gives, as a function of radial distance r, the magnitude Bof the magnetic field inside and outside four wires (a, b, c, and d), each of which carries a current that is uniformly distributed across the wire’s cross section. Overlapping portions of the plots (drawn slightly separated) are indicated by double labels. Rank the wires according to (a) radius, (b) the magnitude of the magnetic field on the surface, and (c) the value of the current, greatest first. (d) Is the magnitude of the current density in wire agreater than, less than, or equal to that in wire c?

In unit-vector notation, what is the magnetic field at pointPin Fig. 29-86 ifi=10Aanda=8.0cm? (Note that the wires are notlong.)

A long wire is known to have a radius greater than4.00mmand to carry a current that is uniformly distributed over its cross section. The magnitude of the magnetic field due to that current is0.28mTat a point4.0mmfrom the axis of the wire, and0.20mTat a point 10 mm from the axis of the wire. What is the radius of the wire?

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