Question

In Fig. 29-48 part of a long insulated wire carrying current$\mathbf{i}\mathbf{=}\mathbf{}\mathbf{5}\mathbf{.}\mathbf{78}\mathbf{}\mathbf{mA}\mathbf{}$is bentinto a circular section of radius $\mathbf{}\mathbf{R}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{89}\mathbf{}\mathbf{cm}$. 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 $\mathbf{90}\mathbf{°}$counterclockwise as indicated?

Short answer

1. Magnetic Field at the center of curvature C when circular section lies in the plane will be$\left(2.53\times {10}^{-7}\text{T}\right)\hat{k}$.
2. Magnetic Field at the center of curvature C when circular section is perpendicular to the plane will be$\left(1.92\times {10}^{-7}\text{T}\right)\hat{i}+\left(6.12\times {10}^{-8}\text{T}\right)\hat{k}$.

Step-by-step solution

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:

$B_{wire}=\frac{{\mu }_{0}I}{2\pi R}$

$B_{loop}=\frac{{\mu }_{0}I}{2R}$

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

Consider the required equation is obtained as:

$B_C=B_{wire}+B_{loop}$

$B_C=\frac{{\mu }_{0}I}{2\pi R}+\frac{{\mu }_{0}I}{2R}$

$B_C=\frac{{\mu }_{0}I}{2R}\left(\frac{1}{\pi }+1\right)$

$B_C=\frac{{\mu }_{0}I\left(1+\pi \right)}{2R\pi }$

Substitute the values and solve as:

$B_C=\frac{4\pi \times {10}^{-7}\times 5.78\times {10}^{-3}\left(1+\pi \right)}{2\times \pi \times 0.0189}$

$B_C=2.53\times {10}^{-7}\text{T}$

Using the right hand rule direction for the$B_C$will be in positive$\hat{k}$. So,

$B_C=\left(2.53\times {10}^{-7}T\right)\hat{k}$

(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

$B_C=\left(B_{loop}\right)\hat{i}+\left(B_{wire}\right)\hat{k}$

$B_C=\left(\frac{{\mu }_{0}I}{2R}\right)\hat{i}+\left(\frac{{\mu }_{0}I}{2\pi R}\right)\hat{k}$

Substitute the values and solve as:

$B_C=\left(\frac{4\pi \times {10}^{-7}\times 5.78\times {10}^{-3}}{2\times 0.0189}\right)\hat{i}+\left(\frac{4\pi \times {10}^{-7}\times 5.78\times {10}^{-3}}{2\pi \times 0.0189}\right)\hat{k}$

$B_C=\left(1.92\times {10}^{-7}T\right)\hat{i}+\left(6.12\times {10}^{-8}T\right)\hat{k}$