- 1,444

- 0

I'm only stuck at the very end of this quesstion but ill wirte out the whole thing anyway.

Two tiny wire loops, with (vector) areas [itex]\vec{A_1},\vec{A_2}[/itex] respectively are seperated by a p[osition vector [itex]\vec{r}[/itex] with [itex]r^2>>A_1,A_2[/itex]

a) Treating the loops as magnetic dipoles, find their mutual inductance. Confirm that it is symmetric under exchange of loops.

b) A current I1 is flowing in loops 1 and a current I2 is then slwoly built up in loop two. How much work must be done against the mutually induced emf to keep the current I2 flowing? How does this result compare with that in magnetostatics for the work requiredto bring up a dipole of moment [itex]\vec{m}[/itex] from infinity in a magnetic field [itex]\vec{B}[/itex]

I have succesfully done a) and checked it with my tutor.

The mutual inductance was [itex]M=\frac{\mu_0}{4 \pi} \frac{\left[3(\vec{A_1} \cdot \vec{r})(\vec{A_2} \cdot \vec{r}) - \vec{A_1} \cdot \vec{A_2} r^2\right]}{r^5}[/itex]

clearly this is symmetric under relabbeling of [itex]1 \rightarrow 2[/itex] as required and so we have just said [itex]M_{12}=M_{21}=M[/itex]

As for part b) I have found the work that must be done as follows:

[itex]\varepsilon_1=-\frac{d \Phi_1}{dt}[/itex] but [itex]\Phi_1=I_2 M[/itex]

and so [itex]\varepsilon_1=-M \frac{d I_2}{dt}[/itex]

we know that [itex]\frac{dW}{dt}=- \varepsilon_1 I_1=M I_1 \frac{d I_2}{dt} \Rightarrow W=MI_1I_2[/itex]

AS FOR THE PART about bringing a dipole from infinity. so far I have:

[itex]\vec{F}=\nabla (\vec{m} \cdot \vec{B})[/itex] and so

[itex]W=\int_{\infty}^{\vec{r}} \vec{F} \cdot \vec{dr} = \vec{m} \cdot \vec{B}[/itex]

the field from a dipole is [itex]\vec{B}=\frac{\mu_0}{4 \pi} \frac{3(\vec{m} \cdot \vec{r})-\vec{m} r^2}{r^5}[/itex]

but substituting that in should get me back to [itex]W=MI_1I_2[/itex] should it not? i must be doing something wrong here.

would it be ok to say that the arbitrary field [itex]\vec{B}[/itex] is generated by a dipole of moment [itex]\vec{m_1}=I_1 \vec{A_1}[/itex] and that the dipole referred to in the question has moment [itex]\vec{m_2}=I_2 \vec{A_2}[/itex] as then this does match up with the answer from the first part.

also am i correct in saying [itex]\varepsilon_1[/itex] is aback emf?

Two tiny wire loops, with (vector) areas [itex]\vec{A_1},\vec{A_2}[/itex] respectively are seperated by a p[osition vector [itex]\vec{r}[/itex] with [itex]r^2>>A_1,A_2[/itex]

a) Treating the loops as magnetic dipoles, find their mutual inductance. Confirm that it is symmetric under exchange of loops.

b) A current I1 is flowing in loops 1 and a current I2 is then slwoly built up in loop two. How much work must be done against the mutually induced emf to keep the current I2 flowing? How does this result compare with that in magnetostatics for the work requiredto bring up a dipole of moment [itex]\vec{m}[/itex] from infinity in a magnetic field [itex]\vec{B}[/itex]

I have succesfully done a) and checked it with my tutor.

The mutual inductance was [itex]M=\frac{\mu_0}{4 \pi} \frac{\left[3(\vec{A_1} \cdot \vec{r})(\vec{A_2} \cdot \vec{r}) - \vec{A_1} \cdot \vec{A_2} r^2\right]}{r^5}[/itex]

clearly this is symmetric under relabbeling of [itex]1 \rightarrow 2[/itex] as required and so we have just said [itex]M_{12}=M_{21}=M[/itex]

As for part b) I have found the work that must be done as follows:

[itex]\varepsilon_1=-\frac{d \Phi_1}{dt}[/itex] but [itex]\Phi_1=I_2 M[/itex]

and so [itex]\varepsilon_1=-M \frac{d I_2}{dt}[/itex]

we know that [itex]\frac{dW}{dt}=- \varepsilon_1 I_1=M I_1 \frac{d I_2}{dt} \Rightarrow W=MI_1I_2[/itex]

AS FOR THE PART about bringing a dipole from infinity. so far I have:

[itex]\vec{F}=\nabla (\vec{m} \cdot \vec{B})[/itex] and so

[itex]W=\int_{\infty}^{\vec{r}} \vec{F} \cdot \vec{dr} = \vec{m} \cdot \vec{B}[/itex]

the field from a dipole is [itex]\vec{B}=\frac{\mu_0}{4 \pi} \frac{3(\vec{m} \cdot \vec{r})-\vec{m} r^2}{r^5}[/itex]

but substituting that in should get me back to [itex]W=MI_1I_2[/itex] should it not? i must be doing something wrong here.

would it be ok to say that the arbitrary field [itex]\vec{B}[/itex] is generated by a dipole of moment [itex]\vec{m_1}=I_1 \vec{A_1}[/itex] and that the dipole referred to in the question has moment [itex]\vec{m_2}=I_2 \vec{A_2}[/itex] as then this does match up with the answer from the first part.

also am i correct in saying [itex]\varepsilon_1[/itex] is aback emf?

Last edited: