Direction of the magnetic field around a solenoid

[WeiShan Ng]

Homework Statement

Example 5.9 in Griffiths's Introduction to Electrodynamics 4th shows us how to find B of a very long solenoid, consisting of n closely wound turns per unit length on a cylinder of radius R, each carrying a steady current I. In the solution, he goes on to explain why we don't have the radial and "circumferential" component of B. I get the explanation on the radial part but not quite understand the "circumferential" part...

The book draws an Amperian loop around the solenoid and writes

\oint \textbf{B} \cdot d\textbf{l} = B_\phi (2\pi s) = \mu_0 I_{enc}=0

Why is the current enclosed equals to zero? I thought the Amperian loop enclose the winding which carries current with it? And why do we assume $B_\phi$ would be constant around the Amperian loop?

 

[ZapperZ]

Homework Statement

Example 5.9 in Griffiths's Introduction to Electrodynamics 4th shows us how to find B of a very long solenoid, consisting of n closely wound turns per unit length on a cylinder of radius R, each carrying a steady current I. In the solution, he goes on to explain why we don't have the radial and "circumferential" component of B. I get the explanation on the radial part but not quite understand the "circumferential" part...

The book draws an Amperian loop around the solenoid and writes

\oint \textbf{B} \cdot d\textbf{l} = B_\phi (2\pi s) = \mu_0 I_{enc}=0

Why is the current enclosed equals to zero? I thought the Amperian loop enclose the winding which carries current with it? And why do we assume $B_\phi$ would be constant around the Amperian loop?

OK, let's do an Amperian loop around the solenoid:

Assuming that this is an ideal solenoid, i.e. the coils are tightly wound, what is I_encl, i.e. the net flow of current through the surface bounded by the Amperian loop?

Zz.

 

[Charles Link]

To add to @ZapperZ 's explanation, evaluate $B_{outside}$ in the formula $\oint B_{outside} \cdot dl =B_{outside}(2 \pi r)=\mu_o I$, so that $B_{outside}=\frac{\mu_o I}{2 \pi r}$ $\\$ Compare the factor $\frac{1}{2 \pi r}$ to the factor $n$, $\\$ where $n$ =number of turns per unit length. $\\$ $B_{z \, inside}=n \mu_o I$. $\\$ A very loosely packed solenoid has $n \approx 10/cm$, while $\frac{1}{2 \pi r}$ for $r> 3.2 \, cm$ is a factor that is less than $.05/cm$. For this very conservative case, the magnetic field inside the solenoid is 200 x greater.

 

[WeiShan Ng]

OK, let's do an Amperian loop around the solenoid:

View attachment 223484

Assuming that this is an ideal solenoid, i.e. the coils are tightly wound, what is I_encl, i.e. the net flow of current through the surface bounded by the Amperian loop?

Zz.

Isn't the I_{enc}be a net I flowing upward/downward through the surface?

 

[WeiShan Ng]

To add to @ZapperZ 's explanation, evaluate $B_{outside}$ in the formula $\oint B_{outside} \cdot dl =B_{outside}(2 \pi r)=\mu_o I$, so that $B_{outside}=\frac{\mu_o I}{2 \pi r}$ $\\$ Compare the factor $\frac{1}{2 \pi r}$ to the factor $n$, $\\$ where $n$ =number of turns per unit length. $\\$ $B_{z \, inside}=n \mu_o I$. $\\$ A very loosely packed solenoid has $n \approx 10/cm$, while $\frac{1}{2 \pi r}$ for $r> 3.2 \, cm$ is a factor that is less than $.05/cm$. For this very conservative case, the magnetic field inside the solenoid is 200 x greater.

So you saying we still have $B_{\phi}$, but it is negligible compared to $B_{inside}$, so we just take it as zero?

 

[Charles Link]

So you saying we still have $B_{\phi}$, but it is negligible compared to $B_{inside}$, so we just take it as zero?

Yes. That is correct.

 

[vela]

Homework Statement

Example 5.9 in Griffiths's Introduction to Electrodynamics 4th shows us how to find B of a very long solenoid, consisting of n closely wound turns per unit length on a cylinder of radius R, each carrying a steady current I. In the solution, he goes on to explain why we don't have the radial and "circumferential" component of B. I get the explanation on the radial part but not quite understand the "circumferential" part...

The book draws an Amperian loop around the solenoid and writes

\oint \textbf{B} \cdot d\textbf{l} = B_\phi (2\pi s) = \mu_0 I_{enc}=0

Why is the current enclosed equals to zero? I thought the Amperian loop enclose the winding which carries current with it? And why do we assume $B_\phi$ would be constant around the Amperian loop?

The symmetry of the situation should convince you $B_\phi$ is constant.

Anyway, in my copy of Griffiths, he explicitly addressed your questions. Is there something about his explanation you didn't understand?

 

[WeiShan Ng]

The symmetry of the situation should convince you $B_\phi$ is constant.

Anyway, in my copy of Griffiths, he explicitly addressed your questions. Is there something about his explanation you didn't understand?

Everything is good now. Thank you！