- #1
palaphys
- 266
- 17
- Homework Statement
- A thin cylindrical shell of radius
𝑅
R carries a steady current
𝐼
I along its surface (the current flows parallel to the axis of the cylinder, i.e. into or out of the page).
The cylinder is cut lengthwise into two equal halves (each being a semicylindrical shell) and the two halves are kept close together, parallel, as shown.
Find the magnitude of the magnetic force between the two halves per unit length of the cylinder.
- Relevant Equations
- Magnetism Equations
Here is my solution:
Consider a thin cylindrical shell of radius ##R## carrying total current ##I##, cut lengthwise into two identical semicylindrical halves of length ##l##.
We can treat the flat cut face between the halves as a surface experiencing magnetic pressure
## p=\frac{B^2}{2\mu_0}, ##
where ##B## is the magnetic field just outside the shell. For a thin cylindrical shell (far from the ends) the field at the surface is approximately given by
## B= \frac{\mu_0 I}{2\pi R}. ##
The projected (flat) area of the cut face is
## A_{\rm cut}=(\text{width})\times(\text{height})=2R\cdot\ell. ##
Thus the force (attractive) between the two halves is
## F_{\rm}=p\;A_{\rm cut}
=\frac{B^2}{2\mu_0}\,(2R\ell)
=\frac{1}{2\mu_0}\left(\frac{\mu_0 I}{2\pi R}\right)^2(2R\ell). ##
Simplifying:
## \boxed{\,F_{\rm approx}=\frac{\mu_0 I^2\ell}{4\pi^2 R}\,}
\qquad\Longrightarrow\qquad
\boxed{\,\frac{F_{\rm}}{\ell}=\frac{\mu_0 I^2}{4\pi^2 R}\,.} ##
and this is the expected answer as well.
However, I am thinking of another approach, which involves considering small elemental wires carrying an infinitesimal current given by ##di##.
Let the thin cylindrical shell of radius ##R## carry total current ##I##. Parameterising the surface by the azimuthal angle ##\theta##, An elemental current on the right-hand semicylinder is
## dI(\theta)=\frac{I}{2\pi}\,d\theta. ##
we can see that the elemental current ##dI## experiences a magnetic force when placed in a magnetic field ##B## and also we need to take the horizontal (effective) component via a factor ##\cos\theta##, as the other component would cancel out fully. Then the elemental horizontal force is written as
## dF(\theta) = dI(\theta)\;\ell\;B\;\cos\theta. ##
Integrating ##dF## over the right semicylinder ##\theta\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}]##:
##
\begin{aligned}
F &= \int_{-\pi/2}^{\pi/2} dF(\theta)
= \ell B \int_{-\pi/2}^{\pi/2} \frac{I}{2\pi}\cos\theta\,d\theta \\[6pt]
&= \ell B\left(\frac{I}{2\pi}\right)\big[ \sin\theta \big]_{-\pi/2}^{\pi/2}
= \ell B\left(\frac{I}{2\pi}\right)(1-(-1)) \\[6pt]
&= \ell B\left(\frac{I}{\pi}\right).
\end{aligned}
##
if we use the entire field caused by the whole cylinder, for the value of B
## B = \frac{\mu_0 I}{2\pi R}, ##
then substituting gives
##
\begin{aligned}
F &= \ell\left(\frac{I}{\pi}\right)\left(\frac{\mu_0 I}{2\pi R}\right)
= \frac{\mu_0 I^2 \ell}{2\pi^2 R}.
\end{aligned}
##
Which seems to differ from the correctly derived answer by a factor of 1/2 . I feel the mistake lies in substituting the incorrect value of B. But I am unsure how to correct this.
Consider a thin cylindrical shell of radius ##R## carrying total current ##I##, cut lengthwise into two identical semicylindrical halves of length ##l##.
We can treat the flat cut face between the halves as a surface experiencing magnetic pressure
## p=\frac{B^2}{2\mu_0}, ##
where ##B## is the magnetic field just outside the shell. For a thin cylindrical shell (far from the ends) the field at the surface is approximately given by
## B= \frac{\mu_0 I}{2\pi R}. ##
The projected (flat) area of the cut face is
## A_{\rm cut}=(\text{width})\times(\text{height})=2R\cdot\ell. ##
Thus the force (attractive) between the two halves is
## F_{\rm}=p\;A_{\rm cut}
=\frac{B^2}{2\mu_0}\,(2R\ell)
=\frac{1}{2\mu_0}\left(\frac{\mu_0 I}{2\pi R}\right)^2(2R\ell). ##
Simplifying:
## \boxed{\,F_{\rm approx}=\frac{\mu_0 I^2\ell}{4\pi^2 R}\,}
\qquad\Longrightarrow\qquad
\boxed{\,\frac{F_{\rm}}{\ell}=\frac{\mu_0 I^2}{4\pi^2 R}\,.} ##
and this is the expected answer as well.
However, I am thinking of another approach, which involves considering small elemental wires carrying an infinitesimal current given by ##di##.
Let the thin cylindrical shell of radius ##R## carry total current ##I##. Parameterising the surface by the azimuthal angle ##\theta##, An elemental current on the right-hand semicylinder is
## dI(\theta)=\frac{I}{2\pi}\,d\theta. ##
we can see that the elemental current ##dI## experiences a magnetic force when placed in a magnetic field ##B## and also we need to take the horizontal (effective) component via a factor ##\cos\theta##, as the other component would cancel out fully. Then the elemental horizontal force is written as
## dF(\theta) = dI(\theta)\;\ell\;B\;\cos\theta. ##
Integrating ##dF## over the right semicylinder ##\theta\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}]##:
##
\begin{aligned}
F &= \int_{-\pi/2}^{\pi/2} dF(\theta)
= \ell B \int_{-\pi/2}^{\pi/2} \frac{I}{2\pi}\cos\theta\,d\theta \\[6pt]
&= \ell B\left(\frac{I}{2\pi}\right)\big[ \sin\theta \big]_{-\pi/2}^{\pi/2}
= \ell B\left(\frac{I}{2\pi}\right)(1-(-1)) \\[6pt]
&= \ell B\left(\frac{I}{\pi}\right).
\end{aligned}
##
if we use the entire field caused by the whole cylinder, for the value of B
## B = \frac{\mu_0 I}{2\pi R}, ##
then substituting gives
##
\begin{aligned}
F &= \ell\left(\frac{I}{\pi}\right)\left(\frac{\mu_0 I}{2\pi R}\right)
= \frac{\mu_0 I^2 \ell}{2\pi^2 R}.
\end{aligned}
##
Which seems to differ from the correctly derived answer by a factor of 1/2 . I feel the mistake lies in substituting the incorrect value of B. But I am unsure how to correct this.