Chapter 19, Problem 71AP

a)

To determine

The magnitude and direction of magnetic field at points $A$.

Answer to Problem 71AP

The magnetic field at point $A$ will have a magnitude of $53\text{μΤ}$ and will point downward.

Explanation of Solution

Given info: The three conductors are parallel and carry equal currents of magnitude $I=2.0\text{A}$ which is directed out of the page.

Explanation:

The arrangement of the wires is given by the following figure.

The magnetic field due to a straight current carrying wire is given by,

$B=\frac{{\mu }_{0}I}{2\pi r}$      (1)

• ${\mu }_0$ is the permeability of free space
• $I$ is the current in the wire
• $r$ is the distance of the point from the wire

Let $B_1$ be the magnitude of the magnetic field due to the conductor at point $1$, Let $B_2$ be the magnitude of the magnetic field due to the conductor at point $2$, Let $B_3$ be the magnitude of the magnetic field due to the conductor at point $3$.

At point $A$ the $B_1$ will be,

$B_1=\frac{{\mu }_{0}I}{2\pi \left(a\sqrt{2}\right)}$

At point $A$ the $B_2$ will be,

$B_2=\frac{{\mu }_{0}I}{2\pi \left(a\sqrt{2}\right)}$

At point $A$ the $B_2$ will be,

$B_3=\frac{{\mu }_{0}I}{2\pi \left(3a\right)}$

The horizontal component of $B_1$ and $B_2$ will cancel out each other, where as the vertical component of the magnetic field get added up. The magnetic field at $A$ will have only vertical component and will be directed downward.

The vertical component of the field will be,

$\begin{array}{c}{B}_{ver}=B_1\cos 45°+B_2\cos 45°+B_3\\ =\frac{{\mu }_{0}I}{2\pi \left(a\sqrt{2}\right)}\cos 45°+\frac{{\mu }_{0}I}{2\pi \left(a\sqrt{2}\right)}\cos 45°+\frac{{\mu }_{0}I}{2\pi \left(3a\right)}\\ =\frac{{\mu }_{0}I}{2\pi a}\left(\sqrt{2}\cos 45°+\frac{1}{3}\right)\end{array}$

Substitute $2.0\text{A}$ for $I$, $4\pi \times {10}^{-7}\text{T}\cdot \text{m}\cdot {\text{A}}^{\text{-1}}$ for ${\mu }_0$, $1.0\text{cm}$ for $a$ to determine the vertical component of the net magnetic field at $A$,

$\begin{array}{c}{B}_{ver}=\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}\cdot {\text{A}}^{\text{-1}}\right)\left(2.0\text{A}\right)}{2\pi \left(1.0\text{cm}\right)}\left(\sqrt{2}\cos 45°+\frac{1}{3}\right)\\ =\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}\cdot {\text{A}}^{\text{-1}}\right)\left(2.0\text{A}\right)}{2\pi \left(1.0\times {10}^{-2}\text{m}\right)}\left(\sqrt{2}\cos 45°+\frac{1}{3}\right)\\ =53\text{μΤ}\end{array}$

Conclusion: The magnetic field at point $A$ will have a magnitude of $53\text{μΤ}$ and will point downward.

b)

To determine

The magnitude and direction of magnetic field at points $B$.

Answer to Problem 71AP

The magnetic field at point $B$ will have a magnitude of $20\text{μΤ}$ and will point downward on the plane of the page.

Explanation of Solution

Given info: The three conductors are parallel and carry equal currents of magnitude $I=2.0\text{A}$ which is directed out of the page.

Explanation:

The arrangement of the wires is given by the following figure.

The magnetic field due to a straight current carrying wire is given by,

$B=\frac{{\mu }_{0}I}{2\pi r}$      (1)

• ${\mu }_0$ is the permeability of free space
• $I$ is the current in the wire
• $r$ is the distance of the point from the wire

Let $B_1$ be the magnitude of the magnetic field due to the conductor at point $1$, Let $B_2$ be the magnitude of the magnetic field due to the conductor at point $2$, Let $B_3$ be the magnitude of the magnetic field due to the conductor at point $3$.

At point $B$ the $B_1$ will be,

$B_1=\frac{{\mu }_{0}I}{2\pi \left(a\right)}$

At point $B$ the $B_2$ will be,

$B_2=\frac{{\mu }_{0}I}{2\pi \left(a\right)}$

At point $B$ the $B_3$ will be,

$B_3=\frac{{\mu }_{0}I}{2\pi \left(2a\right)}$

At the point $B$, $B_1$ and $B_2$ will cancel out each other as they have equal magnitude and opposite direction. Only $B_3$ will contribute to the magnetic field at point $B$ and will be pointing downward on the plane of the paper.

The magnetic field at point $B$ will be

$\begin{array}{c}{B}_{}=B_3\\ =\frac{{\mu }_{0}I}{2\pi \left(2a\right)}\end{array}$

Substitute $2.0\text{A}$ for $I$, $4\pi \times {10}^{-7}\text{T}\cdot \text{m}\cdot {\text{A}}^{\text{-1}}$ for ${\mu }_0$, $1.0\text{cm}$ for $a$ to determine the vertical component of the net magnetic field at $A$,

$\begin{array}{c}B=\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}\cdot {\text{A}}^{\text{-1}}\right)\left(2.0\text{A}\right)}{2\pi \left(2\left(1.0\text{cm}\right)\right)}\\ =\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}\cdot {\text{A}}^{\text{-1}}\right)\left(2.0\text{A}\right)}{2\pi \left(2\left(1.0\times {10}^{-2}\text{m}\right)\right)}\\ =20\text{μΤ}\end{array}$

Conclusion: The magnetic field at point $B$ will have a magnitude of $20\text{μΤ}$ and will point downward on the plane of the page.

c)

To determine

The magnitude and direction of magnetic field at point $C$.

Answer to Problem 71AP

The magnetic field at point $C$ will be zero.

Explanation of Solution

Given info: The three conductors are parallel and carry equal currents of magnitude $I=2.0\text{A}$ which is directed out of the page.

Explanation:

The arrangement of the wires is given by the following figure.

The magnetic field due to a straight current carrying wire is given by,

$B=\frac{{\mu }_{0}I}{2\pi r}$      (1)

• ${\mu }_0$ is the permeability of free space
• $I$ is the current in the wire
• $r$ is the distance of the point from the wire

Let $B_1$ be the magnitude of the magnetic field due to the conductor at point $1$, Let $B_2$ be the magnitude of the magnetic field due to the conductor at point $2$, Let $B_3$ be the magnitude of the magnetic field due to the conductor at point $3$.

At point $C$ the $B_1$ will be,

$B_1=\frac{{\mu }_{0}I}{2\pi \left(a\sqrt{2}\right)}$

At point $C$ the $B_2$ will be,

$B_2=\frac{{\mu }_{0}I}{2\pi \left(a\sqrt{2}\right)}$

At point $C$ the $B_3$ will be,

$B_3=\frac{{\mu }_{0}I}{2\pi \left(a\right)}$

The horizontal component of $B_1$ and $B_2$ will cancel out each other, where as the vertical component of the magnetic field get added up. Considering the upward direction to be positive the net magnetic field will be

The vertical component of the field will be,

$\begin{array}{c}{B}_{ver}=B_1\sin 45°+B_2\sin 45°-B_3\\ =\frac{{\mu }_{0}I}{2\pi \left(a\sqrt{2}\right)}\sin 45°+\frac{{\mu }_{0}I}{2\pi \left(a\sqrt{2}\right)}\sin 45°-\frac{{\mu }_{0}I}{2\pi \left(a\right)}\\ =\frac{{\mu }_{0}I}{2\pi a}\left(\sqrt{2}\sin 45°-1\right)\\ =0\end{array}$

Conclusion: The magnetic field at point $C$ will be zero.