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?

Short Answer

Expert verified
  1. Magnetic Field at the center of curvature C when circular section lies in the plane will be2.53×10-7Tk^.
  2. Magnetic Field at the center of curvature C when circular section is perpendicular to the plane will be1.92×10-7Ti^+6.12×10-8Tk^.

Step by step solution

01

Determine the concept and the formulas:

As the loop and the wire lies in the same plane, the magnitude of the magnetic field due to both wire and loop will get added to the resultant at the center of curvature. The loop is perpendicular, and the wire lies in the plane, the components will be added to find the resultant between them.

Formula:

Bwire=μ0I2πR

Bloop=μ0I2R

02

(a) Calculate magnetic Field at the center of curvature C when circular section lies in the plane

Consider the required equation is obtained as:

BC=Bwire+Bloop

BC=μ0I2πR+μ0I2R

BC=μ0I2R1π+1

BC=μ0I1+π2Rπ

Substitute the values and solve as:

BC=4π×10-7×5.78×10-31+π2×π×0.0189

BC=2.53×10-7T

Using the right hand rule direction for theBCwill be in positivek^. So,

BC=2.53×10-7Tk^

03

(b) Magnetic Field at the center of curvature C when circular section is perpendicular to the plane

Now the loop is perpendicular to the plane

So the magnetic field at point C will be

BC=Bloopi^+Bwirek^

BC=μ0I2Ri^+μ0I2πRk^

Substitute the values and solve as:

BC=4π×10-7×5.78×10-32×0.0189i^+4π×10-7×5.78×10-32π×0.0189k^

BC=1.92×10-7Ti^+6.12×10-8Tk^

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