Question

A thin wire of length L and uniform linear mass density rho is bent into a circular coil The MI of the coil about a tangential axis in its plane is

• A. $\frac{3\rho L^3}{8{\pi }^2}$
• B. $\frac{8{\pi }^2}{3\rho L^3}$
• C. $\frac{3\pi L^3}{8{\rho }^2}$
• D. $\frac{8{\pi }^2}{8\rho L^2}$

Answer 1

The correct option is A $\frac{3\rho L^3}{8{\pi }^2}$

Given that,

Length of the wire =$L$

Mass density = $\rho$

We know that,

$\rho =\frac{M}{L}$

$M=\rho L$

We know that,

$L=2\pi R$

$R=\frac{L}{2\pi }$

Now, the moment of inertia of a circular coil about a tangential axis in the plane of the coil

$I=\frac{3}{2}M{R}^2$

Now, put the value of R and M

$I=\frac{3}{2}\times \rho L{\left(\frac{L}{2\pi }\right)}^2$

$I=\frac{3\rho L^3}{8{\pi }^2}$

Hence, the Moment of inertia of the coil about a tangential axis in its plane is $\frac{3\rho L^3}{8{\pi }^2}$