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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

Experiments show that a steady current I in a long wire produces a magnetic field B that is tangent to any circle that lies in the plane perpendicular to the wire and whose center is the axis of the wire (as in the figure). Ampère’s Law relates the electric current to its magnetic effects and states that

C B · dr = μ0I

where I is the net current that passes through any surface bounded by a closed curve C, and μ0 is a constant called the permeability of free space. By taking C to be a circle with radius r, show that the magnitude B = |B| of the magnetic field at a distance r from the center of the wire is

B = μ 0 I 2 π r

To determine

To show: The magnitude of the magnetic field at a distance r from the center of the wire as B=μ0I2πr .

Explanation

Given data:

Refer to the Figure at the bottom the question 52 in the textbook.

Write the equation of Ampere’s law as follows.

CBdr=μ0I (1)

Here,

B is the magnetic field,

μ0 is the permeability of the free space,

I is the current passes through any surface bounded by a closed curve, and

C is the circle with radius r .

Formula used:

Write the expression to find the unit tangent vector T for a curve with the vector function r .

T=r|r| (2)

Here,

r is the tangent vector, which is the derivative of vector r .

Consider the parametric equations of the circle C as follows.

x=rcosθ,y=rsinθ,0θ2π

Write the vector equation from the parametric equations as follows.

r=rcosθ,rsinθ

As the magnetic field B is tangent to any circle, the magnetic field lies in the plane perpendicular to the wire B=|B|T .

B=|B|T (3)

Here,

T is the unit tangent vector to the circle C .

Calculation of tangent vector r :

To find the derivative of the vector function, differentiate each component of the vector function.

Differentiate each component of the vector function r=rcosθ,rsinθ as follows.

ddt[r]=ddtrcosθ,rsinθr=ddt(rcosθ),ddt(rsinθ)

Rewrite and compute the expression as follows.

r=rsinθ,rcosθ

Calculation of unit tangent vector T :

Substitute rsinθ,rcosθ for r in equation (2),

T=rsinθ,rcosθ|rsinθ,rcosθ|=rsinθ,rcosθ(rsinθ)2+(rcosθ)2=rsinθ,rcosθr2(sin2θ+cos2θ)=rsinθ,rcosθr2(1)

Simplify the expression as follows.

T=rsinθ,rcosθr=sinθ,cosθ

Substitute sinθ,cosθ for T in equation (3),

B=|B|sinθ,cosθ

Calculation of dr :

Differentiate the vector r with respect to θ and obtain dr as follows

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Chapter 16 Solutions

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