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Question
A uniform conducting wire of length 12a and resistance R is wound up as a current-carrying coil in the shape of (i) an equilateral triangle of side a; (ii) a square of sides a and, (iii) a regular hexagon of sides a. The coil is connected to a voltage source V0. Find the magnetic moment of the coils in each case.
Solution
In this problem, different shapes form figures of different area and the number of loops in each case is different hence, their magnetic moments varies.
Magnetic moment is m = nlA.
Since, the same wire is used in three cases with the same potentials, therefore, same current flows in three cases.
(i) For an equilateral triangle of side a,
As the total wire of length = 12a, so, the no. of loops n = `(12a)/(3a)` = 4
Magnetic moment of the coils m = nlA
As area of triangle is A = `sqrt(3)/4 a^2`
= `4I (sqrt(3)/4 a^2)`
∴ m = `Ia^2 sqrt(3)`
(ii) For a square of sides a,
A = a2
No. of loops n = `(12a)/(4a)` = 3
Magnetic moment of the coils m = nlA - `3I(a^2) = 3Ia^2`
(iii) For a regular hexagon of sides a,
No. of loops n = `(12a)/(6a)` = 2
Area, A = `(6sqrt(3))/4 a^2`
Magnetic moment of the coils m = nlA
⇒ m = `2I((6sqrt(3))/4 a^2)`
⇒ m = `3sqrt(3)a^2I`, m is in a geometric series.
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