Chapter 27, Problem 15P

(a)

To determine

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

Answer to Problem 15P

The magnetic field is $0$ .

Explanation of Solution

Given:

The velocity is $\overrightarrow{v}=1.0\times {10}^2\text{m}/\text{s}\widehat{i}+2.0\times {10}^2\text{m}/\text{s}\widehat{j}$ .

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,

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

  $=\left(\frac{4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}}{4\pi }\right)\left(\frac{\left(\left(\text{1}\text{.602}\times {\text{10}}^{-19}\text{C}\right)\right)\left(\left(1.0\times {10}^2\text{m}/\text{s}\widehat{i}+2.0\times {10}^2\text{m}/\text{s}\widehat{j}\right)\times \widehat{r}\right)}{{\left(\left(2.0\text{m}-3.0\text{m}\right)\widehat{i}+\left(2.0\text{m}-4.0\text{m}\right)\widehat{j}\right)}^2}\right)$

  $=\left[\begin{array}{l}\left(\left(36\times {10}^{-12}\text{T}\cdot {\text{m}}^{\text{2}}\right)\left(\frac{{10}^{12}\text{pT}}{1\text{T}}\right)\right)\\ \left(\frac{\left(\widehat{i}\times \left(-\frac{1}{\sqrt{{\left(-1.0\text{m}\right)}^2+{\left(-2.0\text{m}\right)}^2}}\widehat{i}-\frac{2}{\sqrt{{\left(-1.0\text{m}\right)}^2+{\left(-2.0\text{m}\right)}^2}}\widehat{j}\right)\right)}{{\left(\sqrt{{\left(-1.0\text{m}\right)}^2+{\left(-2.0\text{m}\right)}^2}\right)}^2}\right)\end{array}\right]$

  $=0$

Conclusion:

Therefore, the magnetic field at origin is $0$ .

(b)

To determine

The magnetic field at $x=6.0,y=4.0\text{m}$ .

Answer to Problem 15P

The magnetic field is $\left(-3.6\times {10}^{-23}\text{T}\right)\widehat{k}$ .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

  $\begin{array}{c}\overrightarrow{B}=\frac{{\mu }_0}{4\pi }\frac{q\overrightarrow{v}\times \widehat{r}}{r^2}\\ =\left(\frac{4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}}{4\pi }\right)\left(\frac{\left(\left(\text{1}\text{.602}\times {\text{10}}^{-19}\text{C}\right)\right)\left(\left(1.0\times {10}^2\text{m}/\text{s}\widehat{i}+2.0\times {10}^2\text{m}/\text{s}\widehat{j}\right)\times \widehat{r}\right)}{{\left(\left(6.0\text{m}-3.0\text{m}\right)\widehat{i}-\left(4.0\text{m}-4.0\text{m}\right)\widehat{j}\right)}^2}\right)\\ =\left(\left(36\times {10}^{-12}\text{T}\cdot {\text{m}}^{\text{2}}\right)\left(\frac{{10}^{12}\text{pT}}{1\text{T}}\right)\right)\left(\frac{\left(\left(1.0\times {10}^2\text{m}/\text{s}\widehat{i}+2.0\times {10}^2\text{m}/\text{s}\widehat{j}\right)\times r\left(\frac{\left(3.0\text{m}\right)\widehat{i}}{3.0\text{m}}\right)\right)}{{\left(3.0\right)}^2}\right)\\ =\left(-3.6\times {10}^{-23}\text{T}\right)\widehat{k}\end{array}$

Conclusion:

Therefore, the magnetic field is $\left(-3.6\times {10}^{-23}\text{T}\right)\widehat{k}$ .

(c)

To determine

The magnetic field at $x=3.0,y=6.0\text{m}$ .

Answer to Problem 15P

The magnetic field is $\left(4.6\times {10}^{-23}\text{T}\right)\widehat{k}$ .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

  $\begin{array}{c}\overrightarrow{B}=\frac{{\mu }_0}{4\pi }\frac{q\overrightarrow{v}\times \widehat{r}}{r^2}\\ =\left(\frac{4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}}{4\pi }\right)\left(\frac{\left(\left(\text{1}\text{.602}\times {\text{10}}^{-19}\text{C}\right)\right)\left(\left(1.0\times {10}^2\text{m}/\text{s}\widehat{i}+2.0\times {10}^2\text{m}/\text{s}\widehat{j}\right)\times \widehat{r}\right)}{{\left(\left(3.0\text{m}-3.0\text{m}\right)\widehat{i}-\left(6.0\text{m}-4.0\text{m}\right)\widehat{j}\right)}^2}\right)\\ =\left(\left(36\times {10}^{-12}\text{T}\cdot {\text{m}}^{\text{2}}\right)\left(\frac{{10}^{12}\text{pT}}{1\text{T}}\right)\right)\left(\frac{\left(\left(1.0\times {10}^2\text{m}/\text{s}\widehat{i}+2.0\times {10}^2\text{m}/\text{s}\widehat{j}\right)\times r\left(\frac{\left(2.0\text{m}\right)\widehat{i}}{2.0\text{m}}\right)\right)}{{\left(2.0\right)}^2}\right)\\ =\left(4.6\times {10}^{-23}\text{T}\right)\widehat{k}\end{array}$

Conclusion:

Therefore, the magnetic field is $\left(4.6\times {10}^{-23}\text{T}\right)\widehat{k}$ .