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Posted (edited)

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}

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

Edited by Sriman Dutta
Posted

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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