# 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,

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

Edited by Sriman Dutta

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Bender    132

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

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So, the denominator should be $4\pi r$

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Bender    132

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

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

Why and how ?

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Bender    132

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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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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Bender    132

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

Thanks.

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