Chapter 30, Problem 76PQ

(a)

To determine

The magnitude and direction of the magnetic field at point $A$.

Answer to Problem 76PQ

The magnitude of magnetic field at point $A$ is $\underset{\_}{2.4\times {10}^{-6}\text{T}}$ and the direction is positive X-axis.

Explanation of Solution

Write the expression for the magnetic field due to a current carrying wire as.

  $B=\left(\frac{{\mu }_0}{2\pi }\right)\frac{I}{r}$                                                                                                   (I)

Here, $B$ is the magnetic field, $I$ is the current flowing and $r$ is the distance of the point from the wire.

The direction of magnetic field on the point is given by the Right hand rule. When thumb is kept in the direction of the current then the direction of curling of the finger gives the direction of the magnetic field at that point.

The magnetic field on the point $A$ due to the wire having current $I_1$ is in negative X-axis direction and the wire having current $I_2$ is in positive X-axis direction.

Write the expression for the net field at the point $A$ as.

  $B_{net}=B_2-B_1$                                                                                                (II)

Here, $B_{net}$ is the net magnetic field, $B_2$  is the field due to wire $2$ , and $B_1$ is the field due to wire $1$.

Conclusion:

Substitute $4\pi \times {10}^{-7}\text{H/m}$ for ${\mu }_0$, $B_1$ for $B$,$2.00\text{A}$ for $I$ and $0.5\text{m}$ for $r$ in equation (I).

  $\begin{array}{c}{B}_1=\left(\frac{{\mu }_0}{2\pi }\right)\frac{2.00\text{A}}{0.5\text{m}}\left(-\widehat{i}\right)\\ =\left(\frac{4\pi \times {10}^{-7}\text{H/m}}{2\pi }\right)\frac{2.00\text{A}}{0.5\text{m}}\left(-\widehat{i}\right)\\ =8\times {10}^{-7}\text{}\text{T}\left(-\widehat{i}\right)\end{array}$

Substitute $4\pi \times {10}^{-7}\text{H/m}$ for ${\mu }_0$, $B_2$ for $B$,$8.00\text{A}$ for $I$ and $0.5\text{m}$ for $r$ in equation (I).

  $\begin{array}{c}{B}_2=\left(\frac{4\pi \times {10}^{-7}\text{H/m}}{2\pi }\right)\frac{8.00\text{A}}{0.5\text{m}}\left(\widehat{i}\right)\\ =3.2\times {10}^{-6}\text{}\text{T}\left(\widehat{i}\right)\end{array}$

Substitute $3.2\times {10}^{-6}\text{}\text{T}\left(\widehat{i}\right)$ for $B_2$ and $8\times {10}^{-7}\text{}\text{T}\left(-\widehat{i}\right)$ for $B_1$ in equation (II).

  $\begin{array}{c}{B}_{net}=B_2+B_1\\ =\left(3.2\times {10}^{-6}\text{}\text{T}\right)\left(\widehat{i}\right)+\left(8\times {10}^{-7}\text{}\text{T}\right)\left(-\widehat{i}\right)\\ =2.4\times {10}^{-6}\text{T }\left(\widehat{i}\right)\end{array}$

Thus, the magnitude of magnetic field at point $A$ is $\underset{\_}{2.4\times {10}^{-6}\text{T}}$ and the direction is positive X-axis.

(b)

To determine

The magnitude and direction of the magnetic field at point $B$.

Answer to Problem 76PQ

The magnitude of magnetic field at point $B$ is $1.44\text{μT}$ and direction is $56.3°\text{ below the horizontal}$.

Explanation of Solution

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

  $B=\left(\frac{{\mu }_0}{2\pi }\right)\frac{I}{r}$                                                                                                 (III)

Here, $B$ is the magnetic field, $I$ is the current flowing and $r$ is the distance of the point from the wire.

The direction of magnetic field on the point is given by the simple Right hand rule. When the thumb is kept in the direction of the current and the palm faces the point then the direction of curling of the finger gives the direction of the magnetic field at that point.

The magnetic field on the point $B$ due to the wire having current $I_1$ is in negative Y-axis direction and the wire having current $I_2$ is at ${45}^{\circ }$ from the positive X-axis direction.

The field $B_2$ is resolved as.

The net field at the point $B$ is given as.

  $B_{net}=\sqrt{B_x^2+B_y^2}$                                                                                         (IV)

Here, $B_x$ is the field at the point B due to x axis and $B_y$ is the field at the point B due to y axis.

The direction of the magnetic field at point B is,

  $\theta ={\tan }^{-1}\left(\frac{\left|B_y\right|}{B_x}\right)$                                                                                          (V)

Conclusion:

Substitute $4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}$ for ${\mu }_0$, $B_1$ for $B$, $2.00\text{A}$ for $I$ and $1.00\text{m}$ for $r$ in equation (III).

  $\begin{array}{c}{B}_1=\left(\frac{4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}}{2\pi }\right)\frac{2.00\text{A}}{1\text{m}}\\ =0.400\times {10}^{-6}\text{}\text{T}\left(\frac{{10}^6\text{μT}}{1\text{T}}\right)\\ =0.400\text{μT}\end{array}$

Substitute $4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}$ for ${\mu }_0$, $B_2$ for $B$, $8.00\text{A}$ for $I$ and $\left(1.00\text{m}\right)\sqrt{2}$ for $r$ in equation (III).

  $\begin{array}{c}{B}_2=\left(\frac{4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}}{2\pi }\right)\frac{8.00\text{A}}{\left(1.00\text{m}\right)\sqrt{2}}\\ =1.13\times {10}^{-6}\text{T}\left(\frac{{10}^6\text{μT}}{1\text{T}}\right)\\ =1.13\text{μT}\end{array}$

The net field at point B is resolved into following components.

  $\begin{array}{c}{B}_x=B_2\cos {45}^{\circ }\\ B_y=-B_1-B_2\cos {45}^{\circ }\end{array}$                                                                          (VI)

Substitute $1.13\text{μT}$ for $B_2$ and $0.400\text{μT}$ for $B_1$ in equation (VI).

  $\begin{array}{c}{B}_x=\left(1.13\text{μT}\right)\cos {45}^{\circ }\\ =0.800\text{μT}\\ B_y=-\left(0.400\text{μT}\right)-\left(1.13\text{μT}\right)\cos {45}^{\circ }\\ =-1.20\text{μT}\end{array}$

Substitute $0.800\text{μT}$ for $B_x$ and $-1.20\text{μT}$ for $B_y$ in equation (IV) to find $B_{net}$.

  $\begin{array}{c}{B}_{net}=\sqrt{{\left(0.800\text{μT}\right)}^2+{\left(-1.20\text{μT}\right)}^2}\\ =1.44\text{μT}\end{array}$

Substitute $0.800\text{μT}$ for $B_x$ and $1.20\text{μT}$ for $\left|B_y\right|$ in equation (V) to find $\theta$.

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{1.20\text{μT}}{0.800\text{μT}}\right)\\ =56.3°\text{ below the horizontal}\end{array}$

Thus, the magnitude of magnetic field at point $B$ is $1.44\text{μT}$ and direction is $56.3°\text{ below the horizontal}$.