Chapter 22, Problem 41P

(a)

To determine

The magnetic field at the center of the square.

Answer to Problem 41P

The magnetic field at the center of the square is $28.3\text{μT}$ into the page.

Explanation of Solution

Write the expression for the magnetic field at the point $A$.

  $B=\frac{{\mu }_{o}i}{4\pi \left(\frac{l}{2}\right)}\left(\cos \theta -\cos \left(180°-\varphi \right)\right)$        (I)

Here, $B_a$ is the magnetic field at the center due to the segment a, $i$ is the current in the coil, $\frac{l}{2}$ is the perpendicular distance between the point $A$ and the side $BC$ , ${\mu }_0$ is the permeability of the free space, $\theta$ is the angle between the side of the triangle $AB$ and $BC$ .

Conclusion:

Substitute $4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}$ for ${\mu }_0$ , $45°$ for $\theta$ ,  $45°$ for $\varphi$, $0.400\text{m}$ for $l$ and $10.0\text{A}$ for $i$ in the expression (I)  due to segments $a$

    $\begin{array}{l}{B}_a=\frac{{\mu }_{o}i}{4\pi \left(\frac{l}{2}\right)}\left(\cos \theta -\cos \left(180°-\varphi \right)\right)\\ =\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}\right)\left(10.0\text{A}\right)}{4\pi \left(\frac{0.400\text{m}}{2}\right)}\left(\cos 45°-\cos \left(180°-45°\right)\right)\\ =\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}\right)\left(10.0\text{A}\right)}{4\pi \left(\frac{0.400\text{m}}{2}\right)}\left(\cos 45°-\cos \left(135°\right)\right)\end{array}$

Similarly from the expression (I) write the expression for segment $b$  of the square.

    $\begin{array}{c}{B}_b=\frac{{\mu }_{o}i}{4\pi \left(\frac{l}{2}\right)}\left(\cos \theta -\cos \left(180°-\varphi \right)\right)\\ =\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}\right)\left(10.0\text{A}\right)}{4\pi \left(\frac{0.400\text{m}}{2}\right)}\left(\cos 45°-\cos \left(180°-45°\right)\right)\\ =\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}\right)\left(10.0\text{A}\right)}{4\pi \left(\frac{0.400\text{m}}{2}\right)}\left(\cos 45°-\cos \left(135°\right)\right)\end{array}$

Similarly from the expression (I) write the expression for segment $c$ of the square.

    $\begin{array}{c}{B}_c=\frac{{\mu }_{o}i}{4\pi \left(\frac{l}{2}\right)}\left(\cos \theta -\cos \left(180°-\varphi \right)\right)\\ =\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}\right)\left(10.0\text{A}\right)}{4\pi \left(\frac{0.400\text{m}}{2}\right)}\left(\cos 45°-\cos \left(180°-45°\right)\right)\\ =\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}\right)\left(10.0\text{A}\right)}{4\pi \left(\frac{0.400\text{m}}{2}\right)}\left(\cos 45°-\cos \left(135°\right)\right)\end{array}$

Similarly from the expression (I) write the expression for segment $d$ of the square.

    $\begin{array}{c}{B}_d=\frac{{\mu }_{o}i}{4\pi \left(\frac{l}{2}\right)}\left(\cos \theta -\cos \left(180°-\varphi \right)\right)\\ =\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}\right)\left(10.0\text{A}\right)}{4\pi \left(\frac{0.400\text{m}}{2}\right)}\left(\cos 45°-\cos \left(180°-45°\right)\right)\\ =\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}\right)\left(10.0\text{A}\right)}{4\pi \left(\frac{0.400\text{m}}{2}\right)}\left(\cos 45°-\cos \left(135°\right)\right)\end{array}$

Write the expression for the magnetic field at the center of the square due to the current in the wire $i$.

    $B=B_a+B_b+B_c+B_d$        (II)

Substitute $\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}\right)\left(10.0\text{A}\right)}{4\pi \left(\frac{0.400\text{m}}{2}\right)}\left(\cos 45°-\cos \left(135°\right)\right)$ for $B_a,B_b,B_c,B_d$ in the expression (II),

    $\begin{array}{c}B=4\times \left(\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}\right)\left(10.0\text{A}\right)}{4\pi \left(\frac{0.400\text{m}}{2}\right)}\left(\cos 45°-\cos \left(135°\right)\right)\right)\\ =4\times \left(\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}\right)\left(10.0\text{A}\right)}{4\pi \left(\frac{0.400\text{m}}{2}\right)}\left(0.7071+0.7071\right)\right)\\ =28.3\times {10}^{-6}\text{T}\times \frac{{10}^6\text{μT}}{1\text{T}}\\ =28.3\text{μT}\end{array}$

The magnetic field at the center of the square is $28.3\text{μT}$ into the page.

(b)

To determine

The magnetic field at the center of the circle.

Answer to Problem 41P

The magnetic field at the center of the circle is $28.7\text{μT}$ and into the page.

Explanation of Solution

Write the expression for the perimeter of the circle

  $\begin{array}{c}4l=2\pi r\\ 2r=\frac{4l}{\pi }\end{array}$        (I)

Here, $4l$ is the perimeter of the square.

Write the expression for the magnetic field at the center of the circle.

  $B_c=\frac{{\mu }_{0}i}{2r}$        (II)

Here, ${\mu }_0$ is the permeability of the free space, $i$ is the current in the wire, $r$ is the radius of the circle.                      

Conclusion:

Substitute $\frac{4l}{\pi }$ for $2r$ in expression (II)

    $\begin{array}{c}{B}_c=\frac{{\mu }_{0}i}{\frac{4l}{\pi }}\\ =\frac{{\mu }_{0}i\pi }{4l}\end{array}$        (III)

Substitute $4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}$ for ${\mu }_0$,, $0.400\text{m}$ for $l$ and $10.0\text{A}$ for $i$ in the expression (III)  for the magnetic field at the center of the circle.

    $\begin{array}{c}{B}_c=\frac{{\mu }_{0}i\pi }{4l}\\ =\frac{\pi \left(4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}\right)\left(10.0\text{A}\right)}{4\left(0.400\text{m}\right)}\\ =24.7\times {10}^{-6}\text{T}\times \left(\frac{{10}^6\text{μT}}{1\text{T}}\right)\\ =24.7\text{μT}\end{array}$

The magnetic field at the center of the circle is $28.7\text{μT}$ and into the page.