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In Fig 29-59, length a is4.7cm(short) and current iis13A. (a) What is the magnitude (into or out of the page) of the magnetic field at point P? (b) What is the direction (into or out of the page) of the magnetic field at point P?

Short Answer

Expert verified
  1. The magnitude of the magnetic field at the point P is,2.0×10-5T.
  2. The direction of the magnetic field at the point P is into the page.

Step by step solution

01

Given

  1. The length a is,4.7cm=4.7×10-2m.
  2. The current is, i=13A.
02

Understanding the concept

The magnetic field at a point due to a current-carrying wire depends upon the distance of the point from the wire and the magnitude of the current in the wire. It is determined using the Biot-Savart law. The direction of the magnetic field is decided by the right hand rule. The net magnetic field at the point is the vector sum of the magnetic fields due to all the wires.

Formula:

B=μ0i4πR

03

(a) Calculate The magnitude of the magnetic field at the point P

We can write the magnetic field using the Biot Savart law as below:

dB=μ04π.idl×r^r2

dlrepresents the displacement vector for the length of the wire. Let’s assume theleft bottom point of the loop as origin, upward direction as positive y direction, and horizontal right direction as positive x direction. In +y direction,dlwill vary from 0 to2a.Therefore, we can write

dl=dyj^

Now to find ther^,we need to findr and its magnitude. r is the position vector between dl and the point under consideration. The position vector can be written as

r=2ai^+2a-yj^

We can writethemagnitude of this as

r=2a2+2a-y2

Using this to write the unit vector r^,we get

r^=2ai^+2a-yj^2a2+2a-y2

We can use this in the Biot Savart law equation. So we get

dB=μ04π.ir2dl×r^

B=μ04π.i2a2+2a-y2dyj^×2ai^+2a-yj^2a2+2a-y2dB=μ04π.-i2adyk^[2a2+2a-y]3/2

To find the magnetic field due to the long wire, we can integrate the above equation between the limits0 and2a .

B1=02adBB1=02aμ04π·-i2adyk^2a2+2a-y3/2B1=-2aiμ04π·02adyk^2a2+2a-y3/2

Integrating this, we get

B1=-2aiμ04π.2a-y4a24a2+2a-y202ak^

Simplifying this, we have

B1=μ0i82.π.ak^

Similarly, for shorter wire, we can write

B2=-μ0i42.π.ak^

The negative sign here is because of the opposite direction of the current in the shorter wire.

So the total field is

B=B1+B2B=2μ0i82.π.ak^+2-μ0i42.π.ak^

We have multiplied it by 2 because there are 2 long and 2 short wires in the loop.

Simplifying this, we get

B=-μ0i42.π.ak^

Now substituting the given values,

B=-4π×10-7T·m/A×13A42×π×4.7×10-2mk^T

Calculating this, we get

B=-1.96×10-5Tk^-2.0×10-5Tk^

Therefore, the magnitude of the net field at the given point is 2.0×10-5T.

04

(b) Calculate the direction of the magnetic field at the point P

From the above value of B, we can see that the direction of the field is along -k^. It implies that it is into the plane of the page.

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Most popular questions from this chapter

In Fig. 29-48 part of a long insulated wire carrying currenti=5.78mAis bentinto a circular section of radius R=1.89cm. In unit-vector notation, what is the magnetic field at the center of curvature Cif the circular section (a) lies in the plane of the page as shown and (b) is perpendicular to the plane of the page after being rotated 90°counterclockwise as indicated?

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Figure 29-58ashows, in cross section, the cylinder and wire 1 but not wire 2. With wire 2 fixed in place, wire 1 is moved around the cylinder, from angle localid="1663154367897" θ1=0°to angle localid="1663154390159" θ1=180°, through the first and second quadrants of the xycoordinate system. The net magnetic field Bat the center of the cylinder is measured as a function of θ1. Figure 29-58b gives the x component Bxof that field as a function of θ1(the vertical scale is set by Bxs=6.0μT), and Fig. 29-58c gives the y component(the vertical scale is set by Bys=4.0μT). (a) At what angle θ2 is wire 2 located? What are the (b) size and (c) direction (into or out of the page) of the current in wire 1 and the (d) size and (e) direction of the current in wire 2?

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