Chapter 30, Problem 20P

(a)

To determine

The proof that $B=\frac{{\mu }_{0}Ia}{2\pi R}\frac{1}{\sqrt{R^2+a^2}}$ .

Answer to Problem 20P

It is proved that $B=\frac{{\mu }_{0}Ia}{2\pi R}\frac{1}{\sqrt{R^2+a^2}}$ .

Explanation of Solution

Formula used:

The expression for magnetic field is given by,

  $B=\frac{{\mu }_{0}Ia}{4\pi R}\left(\sin {\theta }_1+\sin {\theta }_2\right)$

The expression for angle is given by,

  $\sin \theta =\frac{a}{\sqrt{R^2+a^2}}$

Calculation:

The magnetic field is calculated as,

  $\begin{array}{c}B=\frac{{\mu }_{0}Ia}{4\pi R}\left(\sin {\theta }_1+\sin {\theta }_2\right)\\ =\frac{{\mu }_{0}Ia}{4\pi R}\left(\sin \theta +\sin \theta \right)\\ =\frac{{\mu }_{0}Ia}{4\pi R}\left(\frac{a}{\sqrt{R^2+a^2}}+\frac{a}{\sqrt{R^2+a^2}}\right)\\ =\frac{{\mu }_{0}Ia}{2\pi R}\frac{1}{\sqrt{R^2+a^2}}\end{array}$

Conclusion:

Therefore, it is proved that $B=\frac{{\mu }_{0}Ia}{2\pi R}\frac{1}{\sqrt{R^2+a^2}}$ .

(b)

To determine

The proof that $E_{\text{x}}dA=\frac{Q}{{\in }_0{\left(r^2+a^2\right)}^{\frac{3}{2}}}\pi dr$ .

Answer to Problem 20P

It is proved that $E_{\text{x}}dA=\frac{Q}{{\in }_0{\left(r^2+a^2\right)}^{\frac{3}{2}}}\pi dr$ .

Explanation of Solution

Formula used:

The expression for electric field is given by,

  $E_{\text{x}}=\frac{1}{4\pi {\in }_0}\frac{2Qa}{{\left(r^2+a^2\right)}^{\frac{3}{2}}}$

The expression for flux is given by,

  $E_{\text{x}}dA=E_{\text{x}}\left(2\pi rdr\right)$

Calculation:

The expression for flux is calculated as,

  $\begin{array}{c}{E}_{\text{x}}dA=E_{\text{x}}\left(2\pi rdr\right)\\ =\frac{1}{4\pi {\in }_0}\frac{2Qa}{{\left(r^2+a^2\right)}^{\frac{3}{2}}}\left(2\pi rdr\right)\\ =\frac{Qa}{{\in }_0{\left(r^2+a^2\right)}^{\frac{3}{2}}}rdr\\ =\frac{Q}{{\in }_0{\left(r^2+a^2\right)}^{\frac{3}{2}}}\pi dr\end{array}$

Conclusion:

Therefore, it is proved that $E_{\text{x}}dA=\frac{Q}{{\in }_0{\left(r^2+a^2\right)}^{\frac{3}{2}}}\pi dr$ .

(c)

To determine

The proof that ${\varphi }_{\text{e}}=\frac{Q}{{\in }_0}\left(1-\frac{a}{\sqrt{a^2+R^2}}\right)$ .

Answer to Problem 20P

It is proved that ${\varphi }_{\text{e}}=\frac{Q}{{\in }_0}\left(1-\frac{a}{\sqrt{a^2+R^2}}\right)$ .

Explanation of Solution

Formula used:

The expression for flux from (b) is given by,

  ${\in }_{0}d{\varphi }_{\text{e}}=\frac{Qa}{{\in }_0{\left(r^2+a^2\right)}^{\frac{3}{2}}}\left(rdr\right)$ …… (1)

Calculation:

Integrate equation (1) from $0$ to $R$ .

  $\begin{array}{c}d{\varphi }_{\text{e}}=\frac{1}{{\in }_0}{\displaystyle {\int }_0^R\frac{Qa}{{\in }_0{\left(r^2+a^2\right)}^{\frac{3}{2}}}\left(rdr\right)}\\ =\frac{Qa}{{\in }_0}\left(\frac{-1}{\sqrt{R^2+a^2}}+\frac{1}{a}\right)\\ =\frac{Q}{{\in }_0}\left(1-\frac{a}{\sqrt{R^2+a^2}}\right)\end{array}$

Conclusion:

Therefore, it is proved that ${\varphi }_{\text{e}}=\frac{Q}{{\in }_0}\left(1-\frac{a}{\sqrt{a^2+R^2}}\right)$ .

(d)

To determine

The proof that $I+I_{\text{d}}=I\frac{a}{\sqrt{a^2+R^2}}$ .

Answer to Problem 20P

It is proved that $I+I_{\text{d}}=I\frac{a}{\sqrt{a^2+R^2}}$ .

Explanation of Solution

Formula used:

The expression for displacement current is given by,

  $I_{\text{d}}=-I\left(1-\frac{a}{\sqrt{R^2+a^2}}\right)$

Calculation:

The summation of current and displacement current is calculated as,

  $\begin{array}{c}I+I_{\text{d}}=I+\left(-I\left(1-\frac{a}{\sqrt{R^2+a^2}}\right)\right)\\ =I-I\left(1-\frac{a}{\sqrt{R^2+a^2}}\right)\\ =I-I+I\frac{a}{\sqrt{R^2+a^2}}\\ =I\frac{a}{\sqrt{R^2+a^2}}\end{array}$

Conclusion:

Therefore, it is proved that $I+I_{\text{d}}=I\frac{a}{\sqrt{a^2+R^2}}$ .

(e)

To determine

The proof that result from generalized ampere’s law is similar to part (a).

Answer to Problem 20P

It is proved that result from generalized ampere’s law is similar to part (a).

Explanation of Solution

Formula used:

The expression for generalized form of ampere’s law is given by,

  ${\displaystyle \int B\cdot dl}={\mu }_0\left(I+I_{\text{d}}\right)$

Calculation:

The magnetic field is calculated as,

  $\begin{array}{c}{\displaystyle \int B\cdot dl}={\mu }_0\left(I+I_{\text{d}}\right)\\ B\left(2\pi R\right)={\mu }_{0}I\frac{a}{\sqrt{R^2+a^2}}\\ B=\frac{{\mu }_{0}I}{2\pi R}\frac{a}{\sqrt{R^2+a^2}}\\ =\frac{{\mu }_{0}Ia}{2\pi R}\frac{1}{\sqrt{R^2+a^2}}\end{array}$

Conclusion:

Therefore, it is proved that result from generalized ampere’s law is similar to part (a).