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In Fig. 29-83, two infinitely long wires carry equal currents i. Each follows a 90°arc on the circumference of the same circle of radius R. Show that the magnetic field Bat the center of the circle is the same as the field Ba distance R below an infinite straight wire carrying a current Ito the left.

Short Answer

Expert verified

Proved.

Step by step solution

01

Given

Figure 29-83.

02

Understanding the concept

In this case, we take the vector sum of the magnetic fields and then use the relation for magnetic field in terms of current, distance, and angle subtended to get the required result. We will use the usual convention as which will represent the direction of the field going into the page and will represent the direction of the field coming out of page

Formula:

Magnetic field due to the circular arc at the center of arc with current (i).

B=μ0i2πRϕ

direction of B is determined by right hand thumb rule

03

 Step 3: Show that the magnetic field B→  at the center of the circle is the same as the field B→ a distance R below an infinite straight wire carrying a current I to the left

Let’s find the magnetic field due to the first and second wire separately. Adding them, we will get the net field due to both wires at the center of the arc using equation 29-9

B1=μ0i4πR+μ0i2πR·π2+μ0i4πR

This field is pointing out of the page.

For wire 2

B2=0+μ0i2πR·π2+0

This field is pointing into the page.

So the net field at the center is

Bc=B1+B2=μ0i4πR+μ0i2πR·π2+μ0i4πR-μ0i2πR·π2Bc=μ0i2πR

We can see that this expression is exactly the same as the equation for the field by a single infinite straight wire (equation 29-4).

Hence, it is proved.

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Most popular questions from this chapter

Figure 29-89 is an idealized schematic drawing of a rail gun. Projectile Psits between two wide rails of circular cross section; a source of current sends current through the rails and through the(conducting) projectile (a fuse is not used). (a) Let wbe the distance between the rails, Rthe radius of each rail, and i the current. Show that the force on the projectile is directed to the right along the rails and is given approximately byF=i2μ02π·ln(w+RR)

(b) If the projectile starts from the left end of the rails at rest, find the speed vat which it is expelled at the right. Assume that I = 450 kA, w = 12 mm, R = 6.7 cm, L = 4.0 m, and the projectile mass is 10 g.

In a particular region there is a uniform current density of 15A/m2in the positive z direction. What is the value of B.dswhen that line integral is calculated along the three straight-line segments from (x, y, z) coordinates (4d, 0, 0) to (4d, 3d, 0) to (0, 0, 0) to (4d, 0, 0), where d=20cm?

Figure 29-81 shows a wire segment of length Δs=3cm, centered at the origin, carrying current i=2A in the positive ydirection (as part of some complete circuit). To calculate the magnitude of the magnetic field produced by the segment at a point several meters from the origin, we can use B=μ04πiΔs×r^r2 as the Biot–Savart law. This is because r and u are essentially constant over the segment. Calculate (in unit-vector notation) at the(x,y,z)coordinates (a)localid="1663057128028" (0,0,5m)(b)localid="1663057196663" (0,6m,0)(c) localid="1663057223833" (7m,7m,0)and (d)(-3m,-4m,0)

Shows four identical currents iand five Amperian paths (athrough e) encircling them. Rank the paths according to the value of B.dstaken in the directions shown, most positive first.

One long wire lies along an xaxis and carries a current of30A in the positive xdirection. A second long wire is perpendicular to the xyplane, passes through the point 0,4.0m,0, and carries a current of 40A in the positive zdirection. What is the magnitude of the resulting magnetic field at the point0,2.0m,0?

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