Q32P - Introduction to Electrodynamics

Question

Two tiny wire loops, with areas and , are situated a displacement apart (Fig. 7 .42). FIGURE7.42 

 

 

(a) Find their mutual inductance. [Hint: Treat them as magnetic dipoles, and use Eq. 5.88.] Is your formula consistent with Eq. 7.24? 

 

(b) Suppose a current is flowing in loop 1, and we propose to turn on a current in loop 2. How much work must be done, against the mutually induced emf, to keep the current flowing in loop 1? In light of this result, comment on Eq. 6.35.

Step-by-Step Solution

Answer

(a) The mutual inductance of the wire loops is $M=\frac{{\mu }_0}{4{\mathrm{\pi r}}^3}\left[3\left(a_1.\hat{r}\right)\left(a_2.\hat{r}\right)-a_1.a_2\right]$.

 

(b) The work done is $w=\frac{{\mu }_0}{4{\mathrm{\pi r}}^3}\left[3\left(m_1.\hat{r}\right)\left(m_2.\hat{r}\right)-m_1.m_2\right]$.

Step 1: Given information

The area of the first wire loop is, $a_1$ .

The area of the second wire loop is, $a_2$.

The distance between the centres of both wire loops is,r .

The current flowing the first loop is, $l_2$.

The current flowing the second loop is,$l_2$ .

Step 2: Mutual inductance

 

Consider for two conducting wire loops are kept near to each other and current is supplied to one wire loop then because of ‘mutual inductance’ a certain amount of emf is produced in the second loop.

 

In case of mutual inductance in two wire loops, the work done against the induced emf in one loop to keep the current flowing is always negative.

 

Step 3(a): Determine the mutual inductance of the wire loops

According to the Eq. 5.88, the formula for the magnetic field in the polar coordinate form is given by,

$$\begin{gathered} B_{dip}\left(r\right)=\nabla \times A \\ {B}_{dip}\left(r\right)=\frac{{\mu }_{0}m}{4{\mathrm{\pi r}}^3}\left(2 \cos \theta \hat{r} +\sin \theta \hat{\theta }\right) \end{gathered}$$

The magnetic field of a dipole can be written in coordinate free form as,

 $B_{dip}\left(r\right)=\frac{{\mu }_0 1}{4{\mathrm{\pi r}}^3}\left[3\left(m.\hat{r}\right)\hat{r}-m\right]$

For wire loop having area , the magnetic field in the first loop is given by,

$B_1=\frac{{\mu }_0 1}{4{\mathrm{\pi r}}^3}\left[3\left(m_1.\hat{r}\right)\hat{r}-m_1\right]$

Putting magnetic dipole moment value  in expression,

$$\begin{gathered} B_1=\frac{{\mu }_0 1}{4{\mathrm{\pi r}}^3}\left[3\left(l_1{a}_1.\hat{r}\right)\hat{r}-m_1\right] \\ {B}_1=\frac{{\mu }_0 1}{4{\mathrm{\pi r}}^3}\left[3\left(a_1.\hat{r}\right)\hat{r}-a_1\right] \end{gathered}$$

The formula for the magnetic flux produced in the second wire loop having area , due to the magnetic field in the first loop is given by,

 $$\begin{gathered} {\Phi }_2=B_1.a_2 \\ {\Phi }_2=\frac{{\mu }_0}{4\mathrm{\pi }}\frac{1}{r^3}{l}_1\left[3\left(a_1.\hat{r}\right)\hat{r}-a_1\right].a_2 \\ {\Phi }_2=\frac{{\mu }_0}{4\mathrm{\pi }}\frac{1}{r^3}{l}_1\left[3\left(a_1.\hat{r}\right)\hat{r}-a_1.a_2\right] \end{gathered}$$

Also, the formula for the mutual inductance in the second loop is given by,

${\Phi }_2=M{l}_1$ 

Equating both values,

 $$\begin{gathered} M{l}_1=\frac{{\mu }_0{l}_1}{4{\mathrm{\pi r}}^3}\left[3\left(a_1.\hat{r}\right)\left(a_1.\hat{r}\right)-a_1.a_2\right] \\ M{l}_1=\frac{{\mu }_0{l}_1}{4{\mathrm{\pi r}}^3}\left[3\left(a_1.\hat{r}\right)\left(a_1.\hat{r}\right)-a_1.a_2\right] \end{gathered}$$

Hence, the mutual inductance of the wire loops is 

$M{l}_1=\frac{{\mu }_0{l}_1}{4{\mathrm{\pi r}}^3}\left[3\left(a_1.\hat{r}\right)\left(a_1.\hat{r}\right)-a_1.a_2\right]$

Step 4(b): Work done against the mutual emf

The formula for the amount of emf induced in the first loop due to mutual inductance is given by,

${\epsilon }_1=-M\frac{d{l}_2}{dt}$ 

The formula for the amount of emf induced in the first loop due to mutual inductance is given by,

 

$\left(\frac{dW}{dt}\right)=-{\epsilon }_1{l}_1$

Negative sign indicates that work done is against the mutual emf in the first loop.

$$\begin{gathered} {\left(\frac{dW}{dt}\right)}_1=-\left(M{l}_1\right)\frac{d{l}_2}{dt} \\ {\left(\frac{dW}{dt}\right)}_1=-M{l}_1\frac{d{l}_2}{dt} \end{gathered}$$

Integrating both sides,

$W_1=M{l}_1{l}_2$

Putting the value of from equation (1),

$$\begin{gathered} W_1=\frac{{\mu }_0}{4{\mathrm{\pi r}}^3}\left[3\left(a_1.\hat{r}\right)\left(a_1.\hat{r}\right)-a_1.a_2\right]l_1{l}_2 \\ {W}_1=\frac{{\mu }_0}{4{\mathrm{\pi r}}^3}\left[3\left(l_1{a}_1.\hat{r}\right)\left(l_1{a}_1.\hat{r}\right)-l_1{a}_1.l_2{a}_2\right] \end{gathered}$$

Putting $l_1{a}_1=m$ and $l_2{a}_2=m_2$in the expression,

$W_1=\frac{{\mu }_0}{4{\mathrm{\pi r}}^3}\left[3\left(m_1.\hat{r}\right)\left(m_2.\hat{r}\right)-m_1.m_2\right]$

Hence, the work done against the mutually induced emf in the loop 1 is 

$W_1=\frac{{\mu }_0}{4{\mathrm{\pi r}}^3}\left[3\left(m_1.\hat{r}\right)\left(m_2.\hat{r}\right)-m_1.m_2\right]$

The Eq. 6.35 is given as, the energy required for loop 1 is given by,

 $U=\frac{{\mu }_0}{4{\mathrm{\pi r}}^3}\left[m_1.m_2-3\left(m_2.\hat{r}\right)\left(m_2.\hat{r}\right)\right]$

$Comparing the equation for the work done with Eq. 6.35,$

$U=-W$

The energy required to keep the current flowing in the first loop is equal to the negative value of work done against the mutually induced emf in the first loop