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

[Sriman Dutta]

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]

Edited December 23, 2016 by Sriman Dutta

[Bender]

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

[Sriman Dutta]

So, the denominator should be [math]4\pi r [/math]

[Bender]

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

[Sriman Dutta]

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

Why and how ?

[Bender]

Why and how ?

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

 

Derivation for full loop

[Sriman Dutta]

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??

[Bender]

Yes. I double checked with the hand book I use (Giancoli )

[Sriman Dutta]

Thanks.