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Figure 29-45 shows two current segments. The lower segment carries a current of i1=0.40Aand includes a semicircular arc with radius 5.0cm, angle 180°, and center point P. The upper segment carries current i2=2i1and includes a circular arc with radius 4.0cm, angle 120°, and the same center point P. What are the(a) magnitude and (b) direction of the net magnetic field at Pfor the indicated current directions? What are the (c)magnitude of if i1 is reversed and (d) direction B of if i1 is reversed?

Short Answer

Expert verified
  1. The magnitude of the net magnetic fieldB at Pfor the indicated current directions is 1.7×10-6T.
  2. The direction of the net magnetic fieldB at Pfor the indicated current directions is-k^.
  3. The magnitude of the magnetic field Bifi1 is reversed is 6.7×10-6T.
  4. The direction of the magnetic fieldB ifi1 is reversed is -k^.

Step by step solution

01

Given

  1. Current in the lower segmenti1=0.40A
  2. Radius of the lower segmentr1=5.0cm=0.05m
  3. Angle of the lower arcθ1=1800=πrad
  4. Current in the upper segmenti2=2i1
  5. Radius of the lower segmentr1=4.0cm=0.04m
  6. Angle of the lower arcθ2=1200=2π3rad
02

Determine the formulas

Write the formula for the magnetic field:

B=μ0iθ4πR

03

(a) Calculate the magnitude of the net magnetic field B→ at P for the indicated current directions

The magnitude of the magnetic field is given by

B=μ0i1θ14πr1k^-μ0i2θ24πr2

Substitute the values and solve as:

B=μ00.40Aπrad4π0.05mk^-μ00.80A2π3rad4π0.04mk^B=μ04π8π-403πk^B=μ04π-5.33πk^B=-1.7×10-6k^

Therefore, the magnitude of the net magnetic fieldB at P for the indicated current directions is 1.7×10-6T.

04

(b) Calculate the direction of the net magnetic field B→ at P for the indicated current directions

The result of part a shows that the direction of the magnitude of the net magnetic fieldB at P for the indicated current directions is into the page, therefore, the direction is -k^

05

(c) Calculate The magnitude of the magnetic field B→ if i1 is reversed

For the reverse direction of the currenti1,the magnitude of the magnetic field is given by

B=-μ0i1θ14πr1k^-μ0i2θ24πr2

Substitute the values and solve as:

B=-μ00.40A2πrad4π0.05mk^-μ00.80A2π3rad4π0.04mk^B=-μ04π8π-403πk^B=μ04π-21.33πk^B=-6.7×10-6k^

Therefore, the magnitude of the magnetic fieldB ifi1 is reversed is 6.7×10-6T.

06

(d) Calculate the direction of the magnetic field B→ if i1 is reversed

The result of part c shows that the direction of the magnitude of the net magnetic fieldB at P for the indicated current directions is into the page, therefore, the direction is -k^.

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Most popular questions from this chapter

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?

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