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Calculating the magnetic moment ( in Am^...

Calculating the magnetic moment ( in `Am^2`) of a thin wire with a current I=0.8A, wound tightlly on a half a tore (see figure). The diameter of the cross section of the tore is equal to d=5cm, and the number of turns is N=500.

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The correct Answer is:
5
The magnetic moment of circular current is given by AI, I being
the curculating current and A is the area of cross-section, the
direction is perpendicular to the plane of current. Now, for an
element of toroid of length rdq, its magnetic moment is along the
direction of arrow as shown, of magnitude (perpendicular to its
cross-section)=currentxareaxnumber of turns in the length rdq.

`dM=I(pid^2)/4(N/(pir)r d theta)=N/4 d^2Id theta`
Resolving dM into components along x and y, we get `dMsin theta`
and `dM cos theta`, components along y form neightbouring elements
cancel out to zero, and components along x are added. so
`M=intdM sin theta=int_0^(pi)N/4,d^2Isin thetad theta=(Nd^2I)/2`
putting the given values we get `M=5Am^2`.
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