Question

A uniform conducting wire of length $12a$ and resistance $R$ is wound up as a current carrying coil in the shape of -

$\left(i\right)$ an equilateral triangle of side $a$.

$\left(ii\right)$ a square of side $a$.

The magnetic dipole moments of the coils in each case respectively are -

$I$ is the current through the coils.

• A. $4I{a}^2$ and $3I{a}^2$
• B. $\sqrt{3}I{a}^2$ and $3I{a}^2$
• C. $3I{a}^2$ and $I{a}^2$
• D. $3I{a}^2$ and $4I{a}^2$

Answer 1

Answer: (B)

Step 1: Given that:

Length of the uniform conducting wire = $12a$

Resistance of the wire = R

Current through the coils = I

Step 2: Formula used:

The magnetic dipole moment of the coil is given as;

$\mu =NiA$

Where; $\mu$ is the magnetic moment, $N$ is the total number of turns in the wire,$\text{i}$ is the current in the coil and $\text{A}$ is the area of cross-section.

$N=\frac{Totallengthofthewire}{lengthofeachturn}$

Step 3: a) Calculation of the magnetic dipole moment when the current carrying coil is in the shape of an equilateral triangle:

Side of equilateral triangle = $a$

The length of the equilateral triangular turn = $a+a+a=3a$

Number of total triangular turns in the wire;

$N=\frac{12a}{3a}=4$

Area(A) of equilateral triangle = $\frac{\sqrt{3}}{4}\times {\left(side\right)}^2$

Thus,

Magnetic dipole moment($\mu$) = $4\times I\times \frac{\sqrt{3}}{4}\times a^2$

= $\sqrt{3}I{a}^2$

Step 4: b) Calculation of the magnetic dipole moment when the current carrying coil is in the shape of a square:

Side of square = $a$

The length of square coil = $a+a+a+a=4a$

Number of total square turns in the wire;

$N=\frac{12a}{4a}=3$

Area(A) of a square = ${\left(side\right)}^2$

Thus,

Magnetic dipole moment($\mu$) = $3\times I\times a^2$

= $3I{a}^2$

Thus,

The magnetic moment in first case = $\sqrt{3}I{a}^2$

magnetic moment in second case = $3I{a}^2$