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A straight conductor carrying current i=5.0Asplits into identical semicircular arcs as shown in Figure. What is the magnetic field at the center C of the resulting circular loop?

Short Answer

Expert verified

The magnetic field at the center isB=0.

Step by step solution

01

Given Information

Angle made by arc=Ï€

02

Determining the formula for the magnetic field

Formula:

Magnetic field at center of circular arc is given by:

BC=μ04π×iϕR

Here, ϕis the angle of arc and R is the radius of the arc.

03

Calculating the magnetic field at the center C of the resulting circular loop 

Magnetic field at center due to arc is given by:

B=μ04π×iϕR

Fields due to both arcs are opposite to each other.

Angle due to semicircular arcs =Ï€,

Net field due to both semicircular arcs is

B=μ04π×iπR+μ04π×i-πR

B=0

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Most popular questions from this chapter

A current is set up in a wire loop consisting of a semicircle of radius4.00cm,a smaller concentric semicircle, and tworadial straight lengths, all in the same plane. Figure 29-47ashows the arrangement but is not drawn to scale. The magnitude of the magnetic field produced at the center of curvature is 47.25μ°Õ. The smaller semicircle is then flipped over (rotated) until the loop is again entirely in the same plane (Figure29-47 b).The magnetic field produced at the (same) center of curvature now has magnitude 15.75μ°Õ, and its direction is reversed. What is the radius of the smaller semicircle.

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Question: Figure 29-72 shows an arrangement known as a Helmholtz coil. It consists of two circular coaxial coils, each of200turnsand radiusR=25.0cm, separated by a distances=R. The two coils carry equal currentsi=12.2mAin the same direction. Find the magnitude of the net magnetic field at P, midway between the coils.

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