Chapter 27, Problem 14P

(a)

To determine

The magnetic field at $x=1.0\text{m},y=3.0\text{m}$ .

Answer to Problem 14P

The magnetic field is $\left(12.72\text{pT}\right)\widehat{k}$ .

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 is calculated as,

  $\begin{array}{c}\overrightarrow{B}=\frac{{\mu }_0}{4\pi }\frac{q\overrightarrow{v}\times \widehat{r}}{r^2}\\ =\left(\frac{4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}}{4\pi }\right)\left(\frac{\left(\left(12\text{μC}\right)\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right)\left(\left(30\text{m}/\text{s}\widehat{i}\right)\times \widehat{r}\right)}{{\left(\left(1.0\text{m}-0\text{m}\right)\widehat{i}+\left(3.0\text{m}-2.0\text{m}\right)\widehat{j}\right)}^2}\right)\\ =\left(\left(36\times {10}^{-12}\text{T}\cdot {\text{m}}^{\text{2}}\right)\left(\frac{{10}^{12}\text{pT}}{1\text{T}}\right)\right)\left(\frac{\left(\widehat{i}\times \left(\frac{1}{\sqrt{{\left(1\text{m}\right)}^2+{\left(1\text{m}\right)}^2}}\widehat{i}+\frac{1}{\sqrt{{\left(1\text{m}\right)}^2+{\left(1\text{m}\right)}^2}}\widehat{j}\right)\right)}{{\left(\sqrt{{\left(1\text{m}\right)}^2+{\left(1\text{m}\right)}^2}\right)}^2}\right)\\ =\left(12.72\text{pT}\right)\widehat{k}\end{array}$

Conclusion:

Therefore, the magnetic field at origin is $\left(12.72\text{pT}\right)\widehat{k}$ .

(b)

To determine

The magnetic field at $x=2.0,y=2.0\text{m}$ .

Answer to Problem 14P

The magnetic field is $0$ .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

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

Conclusion:

Therefore, the magnetic field is $0$ .

(c)

To determine

The magnetic field at $x=2.0,y=3.0\text{m}$ .

Answer to Problem 14P

The magnetic field is $\left(3.22\text{pT}\right)\widehat{k}$ .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

  $\begin{array}{c}\overrightarrow{B}=\frac{{\mu }_0}{4\pi }\frac{q\overrightarrow{v}\times \widehat{r}}{r^2}\\ =\left(\frac{4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}}{4\pi }\right)\left(\frac{\left(\left(12\text{μC}\right)\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right)\left(\left(30\text{m}/\text{s}\widehat{i}\right)\times \widehat{r}\right)}{{\left(\left(2.0\text{m}-0\text{m}\right)\widehat{i}+\left(3.0\text{m}-2.0\text{m}\right)\widehat{j}\right)}^2}\right)\\ =\left(\left(36\times {10}^{-12}\text{T}\cdot {\text{m}}^{\text{2}}\right)\left(\frac{{10}^{12}\text{pT}}{1\text{T}}\right)\right)\left(\frac{\left(\widehat{i}\times \left(\widehat{i}\right)\right)}{{\left(\sqrt{\left({\left(2.0\text{m}\right)}^2+{\left(1.0\text{m}\right)}^2\right)}\right)}^2}\right)\\ =\left(3.22\text{pT}\right)\widehat{k}\end{array}$

Conclusion:

Therefore, the magnetic field is $\left(3.22\text{pT}\right)\widehat{k}$ .