Sriman Dutta Posted December 23, 2016 Posted December 23, 2016 (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} Edited December 23, 2016 by Sriman Dutta
Bender Posted December 23, 2016 Posted December 23, 2016 Half a loop, so half the flux density of a full loop.
Sriman Dutta Posted December 23, 2016 Author Posted December 23, 2016 So, the denominator should be 4\pi r
Bender Posted December 23, 2016 Posted December 23, 2016 On my phone, right now, but I think the pi is canceled out.
Sriman Dutta Posted December 24, 2016 Author Posted December 24, 2016 On 12/23/2016 at 9:44 PM, Bender said: On my phone, right now, but I think the pi is canceled out. Why and how ?
Bender Posted December 24, 2016 Posted December 24, 2016 On 12/24/2016 at 5:18 AM, Sriman Dutta said: 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
Sriman Dutta Posted December 24, 2016 Author Posted December 24, 2016 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??
Bender Posted December 25, 2016 Posted December 25, 2016 Yes. I double checked with the hand book I use (Giancoli )
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