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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.

Answers (1)

(i) For equilateral triangle of side a,

As the total wire length = 12a, so the no. of the loops

 n = \frac{12a}{3a}=4

The magnetic moment of the coils m = nIA

As the area of the triangle is

 A=\frac{\sqrt{3}}{4}a^{2}

=4I\left (\frac{\sqrt{3}}{4}a^{2} \right )

\therefore m=Ia^{2}\sqrt{3}

Iii) For a square of sides a, 

A=a^{2}

No. of loops n =

 \frac{12a}{4a}=3

Magnetic moment of the coils m =nIA= 3I(a2) = 3Ia2

(iii) For regular hexagon of sides a,

No. of loops n = \frac{12a}{6a}=2

Aera,

 A=\frac{6 \sqrt{3}}{4}a^{2}

Magnetic moment of the coils m = nIA

\Rightarrow m=2I=\left (\frac{6 \sqrt{3}}{4}a^{2} \right )I

\Rightarrow m=3\sqrt{3}a^{2}I, m is in a geometric series.

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