Chapter 27, Problem 13P

(a)

To determine

The magnetic field at origin.

Answer to Problem 13P

The magnetic field at origin is $-\left(9.0\text{pT}\right)\widehat{k}$ .

Explanation of Solution

Given:

The charge is $q=12\text{μC}$ .

The velocity is $\overrightarrow{v}=30\text{m}/\text{s}\widehat{i}$ .

Formula used:

The expression for magnetic field is given by,

  $\overrightarrow{B}=\frac{{\mu }_0}{4\pi }\frac{q\overrightarrow{v}\times \widehat{r}}{r^2}$

Calculation:

The magnetic field at origin is calculated as,

  $\begin{array}{c}\overrightarrow{B}=\frac{{\mu }_0}{4\pi }\frac{q\overrightarrow{v}\times \widehat{r}}{r^2}\\ =\left(\frac{4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}}{4\pi }\right)\left(\frac{\left(\left(12\text{μC}\right)\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right)\left(\left(30\text{m}/\text{s}\widehat{i}\right)\times \widehat{r}\right)}{{\left(0\widehat{i}-2.0\widehat{j}\right)}^2}\right)\\ =\left(\left(36\times {10}^{-12}\text{T}\cdot {\text{m}}^{\text{2}}\right)\left(\frac{{10}^{12}\text{pT}}{1\text{T}}\right)\right)\left(\frac{\left(\widehat{i}\times \left(-\widehat{j}\right)\right)}{{\left(2.0\text{m}\right)}^2}\right)\\ =-\left(9.0\text{pT}\right)\widehat{k}\end{array}$

Conclusion:

Therefore, the magnetic field at origin is $-\left(9.0\text{pT}\right)\widehat{k}$ .

(b)

To determine

The magnetic field at $x=0,y=1.0\text{m}$ .

Answer to Problem 13P

The magnetic field is $-\left(36.0\text{pT}\right)\widehat{k}$ .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

  $\begin{array}{c}\overrightarrow{B}=\frac{{\mu }_0}{4\pi }\frac{q\overrightarrow{v}\times \widehat{r}}{r^2}\\ =\left(\frac{4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}}{4\pi }\right)\left(\frac{\left(\left(12\text{μC}\right)\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right)\left(\left(30\text{m}/\text{s}\widehat{i}\right)\times \widehat{r}\right)}{{\left(0\widehat{i}+1.0\widehat{j}-0\widehat{i}-2.0\widehat{j}\right)}^2}\right)\\ =\left(\left(36\times {10}^{-12}\text{T}\cdot {\text{m}}^{\text{2}}\right)\left(\frac{{10}^{12}\text{pT}}{1\text{T}}\right)\right)\left(\frac{\left(\widehat{i}\times \left(-\widehat{j}\right)\right)}{{\left(1.0\text{m}\right)}^2}\right)\\ =-\left(36.0\text{pT}\right)\widehat{k}\end{array}$

Conclusion:

Therefore, the magnetic field is $-\left(36.0\text{pT}\right)\widehat{k}$ .

(c)

To determine

The magnetic field at $x=0,y=3.0\text{m}$ .

Answer to Problem 13P

The magnetic field is $\left(36.0\text{pT}\right)\widehat{k}$ .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

  $\begin{array}{c}\overrightarrow{B}=\frac{{\mu }_0}{4\pi }\frac{q\overrightarrow{v}\times \widehat{r}}{r^2}\\ =\left(\frac{4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}}{4\pi }\right)\left(\frac{\left(\left(12\text{μC}\right)\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right)\left(\left(30\text{m}/\text{s}\widehat{i}\right)\times \widehat{r}\right)}{{\left(0\widehat{i}+3.0\widehat{j}-0\widehat{i}-2.0\widehat{j}\right)}^2}\right)\\ =\left(\left(36\times {10}^{-12}\text{T}\cdot {\text{m}}^{\text{2}}\right)\left(\frac{{10}^{12}\text{pT}}{1\text{T}}\right)\right)\left(\frac{\left(\widehat{i}\times \left(\widehat{j}\right)\right)}{{\left(1.0\text{m}\right)}^2}\right)\\ =\left(36.0\text{pT}\right)\widehat{k}\end{array}$

Conclusion:

Therefore, the magnetic field is $\left(36.0\text{pT}\right)\widehat{k}$ .

(d)

To determine

The magnetic field at $x=0,y=4.0\text{m}$ .

Answer to Problem 13P

The magnetic field is $\left(9.0\text{pT}\right)\widehat{k}$ .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

  $\begin{array}{c}\overrightarrow{B}=\frac{{\mu }_0}{4\pi }\frac{q\overrightarrow{v}\times \widehat{r}}{r^2}\\ =\left(\frac{4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}}{4\pi }\right)\left(\frac{\left(\left(12\text{μC}\right)\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right)\left(\left(30\text{m}/\text{s}\widehat{i}\right)\times \widehat{r}\right)}{{\left(0\widehat{i}+4.0\widehat{j}-0\widehat{i}-2.0\widehat{j}\right)}^2}\right)\\ =\left(\left(36\times {10}^{-12}\text{T}\cdot {\text{m}}^{\text{2}}\right)\left(\frac{{10}^{12}\text{pT}}{1\text{T}}\right)\right)\left(\frac{\left(\widehat{i}\times \left(\widehat{j}\right)\right)}{{\left(2.0\text{m}\right)}^2}\right)\\ =\left(9.0\text{pT}\right)\widehat{k}\end{array}$

Conclusion:

Therefore, the magnetic field is $\left(9.0\text{pT}\right)\widehat{k}$ .