Guess why Cylindrical Shell wants To contract it's surface?

Electricity and Magnetism Level 4

A very long thin cylindrical shell of radius R R carries uniform ucrrent I 0 I_0 along its length. Calculate the pressure on the wall of the cylindrical shell due to the magnetic force.

Details and Assumptions :

  • I 0 = π × 10 Ampere I_0 = \sqrt{\pi} \times 10 \text{ Ampere} .
  • R = 1 mm R = 1\text{ mm} .
  • π = 3.14 \pi = 3.14 .
Try more here


The answer is 5.

Deepanshu Gupta by Deepanshu Gupta , 25, India

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4 solutions

Mvs Saketh
Sep 26, 2014

because the currents all being along same direction attrach each other

9 Helpful 0 Interesting 0 Brilliant 0 Confused

yeah...can you please post full solution of this problem..becoz i notice every time you solved questions with different approach....Thanks

Karan Shekhawat - 6 years, 8 months ago

Yeah, but I guess the correct answer is 5 !!

raj abhinav - 1 year ago
Kishore S. Shenoy
Dec 26, 2016

Relevant wiki: Lorentz Force Law (Magnetic and Mixed Fields)

Calculating magnetic fields, B in = 0 B out = μ 0 4 π 2 i R \mathbf{B}_{\text{in}} = 0\\\mathbf{B}_{\text{out}} = \dfrac{\mu_0}{4\pi}\dfrac{2i}{R}

So, the average value becomes, B avg = 1 2 ( B in + B out ) = μ 0 4 π i R \mathbf{B}_{\text{avg}} = \dfrac12\left(\mathbf{B}_{\text{in}}+\mathbf{B}_{\text{out}}\right) = \dfrac{\mu_0}{4\pi}\dfrac{i}{R}

Now, pressure f = J × B = i 2 π R B \begin{aligned}\mathbf{f}&=\mathbf{J}\times\mathbf{B}\\& = \dfrac{i}{2\pi R} B\end{aligned} Note that, here B = B avg \mathbf{B} = \mathbf{B}_{\text{avg}}

So, ∣ f ∣ = μ 0 4 π i 2 2 π R 2 = 5 \left|\mathbf{f}\right| =\dfrac{\mu_0}{4\pi}\dfrac{i^2}{2\pi R^2}=\color{#3D99F6}{\boxed{5}}

5 Helpful 2 Interesting 0 Brilliant 0 Confused

Why do we consider Bavg??

Rishav Dugar - 2 years, 10 months ago

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We do that because we want to find out the external magnetic field, B ext \mathbf{B}_{\text{ext}} .

This comes from the fact that, for all fields caused by moving charges we deal with, ∇ ⋅ B ⃗ = 0 \nabla\cdot\vec{\mathbf{B}} = 0 from which, B ⃗ up, normal = − B ⃗ down, normal \vec{\mathbf{B}}_{\text{up, normal}} = -\vec{\mathbf{B}}_{\text{down, normal}}

So, superpositioning to get the total magnetic field, B ⃗ up, total = B ⃗ up, normal + B ⃗ ext B ⃗ down, total = B ⃗ down, normal + B ⃗ ext B ⃗ avg = 1 2 ( B ⃗ up, total + B ⃗ down, total ) = B ⃗ ext \begin{aligned}\vec{\mathbf{B}}_{\text{up, total}} &= \vec{\mathbf{B}}_{\text{up, normal}} + \vec{\mathbf{B}}_{\text{ext}}\\ \vec{\mathbf{B}}_{\text{down, total}} &= \vec{\mathbf{B}}_{\text{down, normal}} + \vec{\mathbf{B}}_{\text{ext}}\\ \vec{\mathbf{B}}_{\text{avg}} &= \dfrac{ 1}{2}\left(\vec{\mathbf{B}}_{\text{up, total}} + \vec{\mathbf{B}}_{\text{down, total}}\right)\\ &= \vec{\mathbf{B}}_{\text{ext}}\end{aligned}

Kishore S. Shenoy - 2 years, 10 months ago

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Thanks for this information. :D

Md Zuhair - 2 years, 10 months ago

And when to consider Bavg and when to Consider only the B(outside)?

Md Zuhair - 2 years, 10 months ago
Jafar Badour
Mar 23, 2015

we calculate the magnetic field in the cylindrical shell B=muI/2piR then the pressure is the average force between the force inside (zero) and the force out side ILB we yield P=1/2F/(SL) S is the arc length P=4pi^2I^2/(4pi^2*R^2)

2 Helpful 0 Interesting 0 Brilliant 0 Confused

taking 2 current elements of length dx can be take and you must know the double integral method (if you want to do it my way !) to calculate the current force the elements will apply o each other !

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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