1.

A thin insulated wire forms a plane spiral of N = 100 tight turns carrying a current I = 8 mA. The radii of inside and outside turns are equal to a = 50 mm and b = 100 mm. Find the magnetic moment of the spiral with a given current

Answer»

`15 Am^(2)`
`10 Am^(2)`
`30 Am^(2)`
`5 Am^(2)`

Solution :Magnetic moment for one turn = iA
Magnetic moment of a tum of radius R is `M = IPIR^(2)`
Number of TURNS in next dr element `dN = N/(B-a)dr`
`P = int M dN = int_(a)^(b) ipir^(2)N/(b-a)dr=(piiN(b^(3)-a^(2)))/(3(b-a))=15Am^(2)`


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