Chapter 34, Problem 6P

(a)

To determine

The electric field the rod creates at the point $\left(x=0,y=20.0\text{cm},z=0\right)$.

Answer to Problem 6P

The electric field the rod creates at the point $\left(x=0,y=20.0\text{cm},z=0\right)$ is $\left(3.148\widehat{\text{j}}\right)\text{kV}/\text{m}$.

Explanation of Solution

Given info: The linear density of the rod is $35.0\text{nC}/\text{m}$ and the speed is $1.50\times {10}^7\text{m}/\text{s}$.

The value of permittivity of free space is $8.85\times {10}^{-12}\text{F}/\text{m}$.

The figure given below shows the location of the thin rod with respect to axis.

Figure (1)

The formula for the electric field due to long wire is,

$E=\frac{\lambda }{\text{2π}r{\epsilon }_0}$

Here,

${\epsilon }_0$ is the permittivity of free space.

$\lambda$ is the linear density of the electric charge.

$r$ is the distance of the electric field from the origin at $y$ axis.

Substitute $35.0\text{nC}/\text{m}$ for $\lambda$, $8.85\times {10}^{-12}\text{F}/\text{m}$ for ${\epsilon }_0$ and $20.0\text{cm}$ for $r$ in above equation to find $E$.

$\begin{array}{c}E=\frac{35.0\text{nC}\left(\frac{{10}^{-9}\text{C}}{1\text{nC}}\right)/\text{m}}{\text{2π}\left(20.0\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\left(8.85\times {10}^{-12}\text{F}/\text{m}\right)}\\ =\left(3.148\widehat{\text{j}}\right)\text{kV}/\text{m}\end{array}$

Conclusion:

Therefore, the electric field the rod creates at the point $\left(x=0,y=20.0\text{cm},z=0\right)$ is $\left(3.148\widehat{\text{j}}\right)\text{kV}/\text{m}$.

(b)

To determine

The magnetic field the rod creates at the point $\left(x=0,y=20.0\text{cm},z=0\right)$.

Answer to Problem 6P

The magnetic field the rod creates at the point $\left(x=0,y=20.0\text{cm},z=0\right)$ is $\left(5.25\times {10}^{-7}\widehat{\text{k}}\right)\text{T}$.

Explanation of Solution

Given info: The linear density of the rod is $35.0\text{nC}/\text{m}$ and the speed is $1.50\times {10}^7\text{m}/\text{s}$.

The value of the permeability constant is $\text{4π}\times {10}^{-7}\text{T}\cdot \text{m}/\text{A}$

The expression for the current in the wire is,

$I=\lambda v$

Here,

$v$ is the speed.

Substitute $35.0\text{nC}/\text{m}$ for $\lambda$ and $1.50\times {10}^7\text{m}/\text{s}$ for $v$ in above equation to find $I$.

$\begin{array}{c}I=\left(35.0\text{nC}\left(\frac{{10}^{-9}\text{C}}{1\text{nC}}\right)/\text{m}\right)\left(1.50\times {10}^7\text{m}/\text{s}\right)\\ =0.525\text{A}\end{array}$

Thus, the current in the wire is $0.525\text{A}$.

The formula for the magnetic flux due to wire is,

$B=\left(\frac{{\mu }_{0}I}{2\text{π}r}\right)$

Here,

${\mu }_0$ is the permeability constant.

Substitute $20.0\text{cm}$ for $r$, $\text{4π}\times {10}^{-7}\text{T}\cdot \text{m}/\text{A}$ for ${\mu }_0$ and $0.525\text{A}$ for $I$ in above equation to find $B$.

$\begin{array}{c}B=\left(\frac{\left(0.525\text{A}\right)\left(\text{4π}\times {10}^{-7}\text{H}/\text{m}\right)}{2\text{π}\left(20.0\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}\right)\\ =\left(5.25\times {10}^{-7}\widehat{\text{k}}\right)\text{T}\end{array}$

Conclusion:

Therefore, the magnetic field the rod creates at the point $\left(x=0,y=20.0\text{cm},z=0\right)$ is $\left(5.25\times {10}^{-7}\widehat{\text{k}}\right)\text{T}$.

(c)

To determine

The force exerted on an electron at point $\left(x=0,y=20.0\text{cm},z=0\right)$, moving with velocity $\left(2.40\times {10}^8\right)\widehat{\text{i}}\text{m}/\text{s}$.

Answer to Problem 6P

The force exerted on an electron at point $\left(x=0,y=20.0\text{cm},z=0\right)$, moving with velocity $\left(2.40\times {10}^8\right)\widehat{\text{i}}\text{m}/\text{s}$ is $4.83\times {10}^{-16}\left(-\widehat{\text{j}}\right)\text{N}$.

Explanation of Solution

Given info: The linear density of the rod is $35.0\text{nC}/\text{m}$ and the speed is $1.50\times {10}^7\text{m}/\text{s}$.

The charge on an electron is $-1.60\times {10}^{-19}\text{C}$

The Lorentz force on the electron is,

$\overrightarrow{\text{F}}=q\overrightarrow{\text{E}}+q\overrightarrow{\text{v}}\times \overrightarrow{\text{B}}$

Here,

$q$ is the charge on an electron.

Substitute $-1.60\times {10}^{-19}\text{C}$ for $q$, $\left(3.148\times {10}^3\widehat{\text{j}}\right)\text{V}/\text{m}$ for $\overrightarrow{\text{E}}$, $\left(2.40\times {10}^8\right)\widehat{\text{i}}\text{m}/\text{s}$ for $\overrightarrow{\text{v}}$ and $\left(5.25\times {10}^{-7}\widehat{\text{k}}\right)\text{T}$ for $\overrightarrow{\text{B}}$ in above equation to find $\overrightarrow{\text{F}}$.

$\begin{array}{c}\overrightarrow{\text{F}}=\left[\begin{array}{l}\left(-1.60\times {10}^{-19}\text{C}\right)\left(\left(3.148\times {10}^3\widehat{\text{j}}\right)\text{V}/\text{m}\right)+\left(-1.60\times {10}^{-19}\text{C}\right)\left(\left(2.40\times {10}^8\right)\widehat{\text{i}}\text{m}/\text{s}\right)\\ \times \left(\left(5.25\times {10}^{-7}\widehat{\text{k}}\right)\text{T}\right)\end{array}\right]\\ =5.04\times {10}^{-16}\left(-\widehat{\text{j}}\right)\text{N}+2.06\times {10}^{-17}\left(\widehat{\text{j}}\right)\text{N}\\ =4.83\times {10}^{-16}\left(-\widehat{\text{j}}\right)\text{N}\end{array}$

Conclusion:

Therefore, the force exerted on an electron at point $\left(x=0,y=20.0\text{cm},z=0\right)$, moving with velocity $\left(2.40\times {10}^8\right)\widehat{\text{i}}\text{m}/\text{s}$ is $4.83\times {10}^{-16}\left(-\widehat{\text{j}}\right)\text{N}$.