Consider a planeloop of wire that carries a steady current I;we

want to calculate the magnetic field at a point in the plane. We might as well take

that point to be the origin (it could be inside or outside the loop). The shape of the

wire is given, in polar coordinates, by a specified function r(θ)(Fig. 5.62).

(a) Show that the magnitude of the field is

role="math" localid="1658927560350" B=μ0I4π(5.92)

(b) Test this formula by calculating the field at the center of a circular loop.

(c) The "lituus spiral" is defined by a

r(θ)=aθ     0<θ2π

(for some constant a).Sketch this figure, and complete the loop with a straight

segment along the xaxis. What is the magnetic field at the origin?

(d) For a conic section with focus at the origin,

r(θ)=p1+ecosθ

where pisthe semi-latus rectum (the y intercept) and eis the eccentricity (e= 0

for a circle, 0 < e< 1 for an ellipse, e= 1 for a parabola). Show that the field is

B=μ0I2pregardless of the eccentricity.

Short Answer

Expert verified

(a) The magnitude of the magnetic field of a closed plane loop carrying current I is B=μ0I4πdθr.

(b) If the loop is a circle, the field at its center isμ0I2R.

(c) If the loop has formrole="math" localid="1658927823158" r(θ)=aθ , the magnitude of the field isμ0I2π3a. The sketch for the loop is obtained.

(d) If the loop has form r(θ)=p1+ecosθ, the magnitude of the field is μ0I2p.

Step by step solution

01

Given data

(a) There is a closed loop carrying currentI.

(c) The loop has trajectory r(θ)=aθ     0<θ2π.

(d) The loop has trajectoryr(θ)=p1+ecosθ.

02

Determine magnetic field of a current carrying wire

The magnetic field of a current carrying wireIis:

B=μ0I4πdl×r^r2 …… (1)

Here, μ0 is the permeability of free space and dl is an infinitesimal length element on the wire.

03

Determine magnetic field of a closed plane current carrying loop

(a)

For the given configuration,

|dl×r^|=rdθ

Thus, from equation (1) write as:

B=μ0I4πdθr …… (2)

04

Determine the magnetic field at the center of a current carrying circular loop

(b)

For a circular loop,r=R. Here,Ris the radius of the loop.

Thus, from equation (2),

B=μ0I4πRdθ=μ0I2R

This agrees with the field at the center of a current carrying loop.

05

Determine the magnetic field for a loop defined by   r(θ)=aθ

(c)

The sketch of r(θ)=aθ     0<θ2π is given in the following figure.

Thus, from equation (2) is as follows:

B=μ0I4πaθ=μ0I4πa02πθ=μ0I4πa23[θ3/2]02π=μ0I2π3a

Thus, the field is μ0I2π3a.

06

Determine the magnetic field of a conic section

(d)

The field forr(θ)=p1+ecosθis obtained from equation (2) as:

B=μ0I4π1+ecosθp=μ0I4πp02π(1+ecosθ)=μ0I4πp[θesinθ]02π=μ0I2p

Thus, the field is obtained as: μ0I2p.

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Most popular questions from this chapter

A thin glass rod of radius Rand length Lcarries a uniform surface charge σ. It is set spinning about its axis, at an angular velocityω. Find the magnetic field at a distances sRfrom the axis, in the xyplane (Fig. 5.66). [Hint: treat it as a stack of magnetic dipoles.]

Question: (a) Find the force on a square loop placed as shown in Fig. 5.24(a), near an infinite straight wire. Both the loop and the wire carry a steady current I.

(b) Find the force on the triangular loop in Fig. 5.24(b).

In 1897, J. J. Thomson "discovered" the electron by measuring the

charge-to-mass ratio of "cathode rays" (actually, streams of electrons, with charge qand mass m)as follows:

(a) First he passed the beam through uniform crossed electric and magnetic fields Eand B(mutually perpendicular, and both of them perpendicular to the beam), and adjusted the electric field until he got zero deflection. What, then, was the speed of the particles in terms of Eand B)?

(b) Then he turned off the electric field, and measured the radius of curvature, R,

of the beam, as deflected by the magnetic field alone. In terms of E, B,and R,

what is the charge-to-mass ratio (qlm)of the particles?

(a) Complete the proof of Theorem 2, Sect. 1.6.2. That is, show that any divergenceless vector field F can be written as the curl of a vector potential . What you have to do is find Ax,Ayand Azsuch that (i) Az/y-Ay/z=Fx; (ii) Ax/z-Az/x=Fy; and (iii) Ay/x-Ax/y=Fz. Here's one way to do it: Pick Ax=0, and solve (ii) and (iii) for Ayand Az. Note that the "constants of integration" are themselves functions of y and z -they're constant only with respect to x. Now plug these expressions into (i), and use the fact that F=0to obtain

Ay=0xFz(x',y,z)dx';Az=0yFx(0,y',z)dy'-0yFy(x',y,z)dx'

(b) By direct differentiation, check that the you obtained in part (a) satisfies ×A=F. Is divergenceless? [This was a very asymmetrical construction, and it would be surprising if it were-although we know that there exists a vector whose curl is F and whose divergence is zero.]

(c) As an example, let F=yx^+zy^+xz^. Calculate , and confirm that ×A=F. (For further discussion, see Prob. 5.53.)

Magnetostatics treats the "source current" (the one that sets up the field) and the "recipient current" (the one that experiences the force) so asymmetrically that it is by no means obvious that the magnetic force between two current loops is consistent with Newton's third law. Show, starting with the Biot-Savart law (Eq. 5.34) and the Lorentz force law (Eq. 5.16), that the force on loop 2 due to loop 1 (Fig. 5.61) can be written as

F2=μ04πl1l2r^r2dl1dl2

Figure 5.60

Figure 5.61

In this form, it is clear that F2=-F1, since role="math" localid="1657622030111" r^changes direction when the roles of 1 and 2 are interchanged. (If you seem to be getting an "extra" term, it will help to note thatdl2r^=dr.)

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