Figure a shows an element of length ds=1.00μmin a very long straight wire carrying current. The current in that element sets up a differential magnetic field at points in the surrounding space. Figure b gives the magnitudedBof the field for points2.5cmfrom the element, as a function of angle u between the wire and a straight line to the point. The vertical scale is set bydBs=60.0pT. What is the magnitude of the magnetic field set up by the entire wire at perpendicular distance2.5cmfrom the wire?


Short Answer

Expert verified

The value of the magnetic field is3.0μT.

Step by step solution

01

Understanding the concept

Use the Biot- savart law to calculate the current through the wire. Then, using this current value and the equation for magnetic field due to an infinitely long wire carrying current, find the magnetic field.

dB=μ04πidssinθr2

02

Calculating the magnitude of the magnetic field set

Let us find the current through the wire first

Forθ=900

dB=μ04πidsr2

Hence,i=0.375A

Now, to find magnetic field (B):

Magnetic field due to infinite wire carrying current(i):B=μ0i2πR

Substitute the values and solve as:

B=(4π×10-7)×0.3752π×0.025B=3×106B=3.0μT

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Most popular questions from this chapter

Figure 29-24 shows three circuits, each consisting of two radial lengths and two concentric circular arcs, one of radius rand the other of radius R>r. The circuits have the same current through them and the same angle between the two radial lengths. Rank the circuits according to the magnitude of the net magnetic field at the center, greatest first

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