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Fundamentals Of Physics
Found in: Page 857

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Short Answer

In Fig. 29-44 point P1 is at distance R= 13.1 cm on the perpendicular bisector of a straight wire of length L= 18.0 cm. carrying current. (Note that the wire is not long.) What is the magnitude of the magnetic field at P1 due to i?

The magnitude of the magnetic field at P1 due to i is 5.03×10-8 T

See the step by step solution

Step by Step Solution

Step 1: Given

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

Step 2: Determine the formulas:

Use the formula of Biot-Savarts law to find the magnitude of the field at P1 due to i

Formula:

dB=μ0i dl4πsinθr2

Step 3: Calculate the magnitude of the magnetic field at P1 due to i

According to Bio-Savarts law

dB=μ0i dl4πsinθr2

If r makes an angle θ with l then

r=l2+R2

And

sinθ=RR=Rl2+R2 l=L2 =0.18 m2 =0.09 m

Integrating an equation of Bio-Savarts law

B=00.09dB =00.09μ0i dl4πsinθr2B=μ0i 4π00.09sinθdlr2B=μ0i 4π00.09Rl2+R2dll2+R22B=μ0iR 4π00.09dll2+R23/2

Solve further as:

B=μ0i R4π1R2ll2+R21/200.09 B=μ0i 2πRll2+R21/2B=4π×10-7 TmA0.0582 A 2π0.131 m0.09 m0.09 m2+0.131 m21/2B=5.03×10-8 T

Therefore, the magnitude of the magnetic field at P1 due to i is 5.03×10-8 T

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