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B field at the centre of a current-carrying semi-circular loop


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#1 Sriman Dutta

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Posted 23 December 2016 - 06:01 AM

Hello everyone,

As shown in the diagram, there is a semi-circular loop of current carrying wire, such that point P is the centre of the semi-circle and r is the radius. What shall be the magnetic flux density at the centre P?

I presume that the diagram must be treated by the Ampere's Law. So,

 B=\frac{\mu_0 I}{2\pi r}

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Edited by Sriman Dutta, 23 December 2016 - 06:02 AM.

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#2 Bender

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Posted 23 December 2016 - 01:33 PM

Half a loop, so half the flux density of a full loop.


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#3 Sriman Dutta

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Posted 23 December 2016 - 03:19 PM

So, the denominator should be 4\pi r


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#4 Bender

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Posted 23 December 2016 - 09:44 PM

On my phone, right now, but I think the pi is canceled out.
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#5 Sriman Dutta

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Posted 24 December 2016 - 05:18 AM

On my phone, right now, but I think the pi is canceled out.

Why and how ?


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#6 Bender

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Posted 24 December 2016 - 08:15 AM

Why and how ?

Because you integrate over half a circle, which in this case comes down to a multiplication with \pi R

 

Derivation for full loop


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#7 Sriman Dutta

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Posted 24 December 2016 - 09:29 AM

Thanks Bender......... :-)

 

Here I summarize all the formulae for four different situations.

 

For a straight wire carrying current

 B=\frac{\mu_0 \mu_r I}{2 \pi r}

For a single loop of wire

 B=\frac{\mu_0 \mu_r I}{2r}

For a solenoid of length l and having n turns

 B=\frac{\mu_0 \mu_r nI}{l}

For a toroid of single turn and of radius r

 B=\frac{\mu_0 \mu_r nI}{2 \pi r}

 

Are they correct??


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Excellence is then not an act but a habit.
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#8 Bender

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Posted 25 December 2016 - 11:12 PM

Yes. I double checked with the hand book I use (Giancoli )
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#9 Sriman Dutta

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Posted 26 December 2016 - 06:10 AM

Thanks.


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