In Fig. 29-44 point P1is at distance R=13.1cmon the perpendicular bisector of a straight wire of length L=18.0cm. carrying current. (Note that the wire is notlong.) What is the magnitude of the magnetic field at P1due to i?

Short Answer

Expert verified

The magnitude of the magnetic field at P1due to i is5.03×10-8T

Step by step solution

01

Given

  1. Current i=58.2mA
  2. Distance of point P on the perpendicular bisector of wireR=13.1cm=0.131m
  3. Length of wireL=18.0cm=0.18m
02

Determine the formulas:

Use the formula of Biot-Savarts law to find the magnitude of the field at P1due to i

Formula:

dB=μ0idl4πsinθr2

03

Calculate the magnitude of the magnetic field at P1 due to i

According to Bio-Savarts law

dB=μ0idl4πsinθr2

If r makes an angle θwithl then

r=l2+R2

And

sinθ=RR=Rl2+R2l=L2=0.18m2=0.09m

Integrating an equation of Bio-Savarts law

B=00.09dB=00.09μ0idl4πsinθr2B=μ0i4π00.09sinθdlr2B=μ0i4π00.09Rl2+R2dll2+R22B=μ0iR4π00.09dll2+R23/2

Solve further as:

B=μ0iR4π1R2ll2+R21/200.09B=μ0i2πRll2+R21/2B=4π×10-7TmA0.0582A2π0.131m0.09m0.09m2+0.131m21/2B=5.03×10-8T

Therefore, the magnitude of the magnetic field at P1due to i is5.03×10-8T

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