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Sriman Dutta

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

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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,

[math] B=\frac{\mu_0 I}{2\pi r}[/math]

post-119781-0-71343500-1482472642_thumb.jpg

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

Why and how ?

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

 

Here I summarize all the formulae for four different situations.

 

For a straight wire carrying current

[math] B=\frac{\mu_0 \mu_r I}{2 \pi r}[/math]

For a single loop of wire

[math] B=\frac{\mu_0 \mu_r I}{2r}[/math]

For a solenoid of length [math]l[/math] and having [math]n[/math] turns

[math] B=\frac{\mu_0 \mu_r nI}{l}[/math]

For a toroid of single turn and of radius [math]r[/math]

[math] B=\frac{\mu_0 \mu_r nI}{2 \pi r}[/math]

 

Are they correct??

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