Q

# A circular coil of 16 turns and radius 10 cm carrying a current of 0.75 A rests with its plane normal to an external field of magnitude 5.0 into 10 ^ 2 T. The coil is free to turn about an axis in its plane perpendicular to the field direction.

5.9. A circular coil of 16 turns and radius 10 cm carrying a current of 0.75 A rests with its plane normal to an external field of magnitude $5.0 \times 10 ^{-2}$ The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of $2.0 s^{-1}$ . What is the moment of inertia of the coil about its axis of rotation?

Views

Given,

Number of turns, N = 16

Radius of the coil, r = 10 cm = 0.1 m

Current in the coil, I = 0.75 A

Magnetic field strength, B = 5.0 x $\dpi{100} 10^{-2}$ T

Frequency of oscillations of the coil, f = 2.0 $\dpi{100} s^{-1}$

Now, Cross-section of the coil, A = $\dpi{100} \pi r^2$ = $\dpi{100} \pi \times (0.1)^2 m^2$

We know, Magnetic moment, m = NIA

= (16)(0.75 A)($\dpi{100} \pi \times (0.1)^2 m^2$

= 0.377 $\dpi{100} JT^{-1}$

We know, frequency of oscillation in a magnetic field is:

$\dpi{100} f = \frac{1}{2\pi}\sqrt{\frac{MB}{I}}$(I = Moment of Inertia of the coil)

$\dpi{100} \implies I = \frac{MB}{4\pi^2f^2}$

$\dpi{100} \implies I = \frac{0.377\times5\times10^{-2}}{4\pi^22^2}$

$I = 1.19\times$ $\dpi{100} 10^{-4}$ $\dpi{100} kgm^2$

The moment of inertia of the coil about its axis of rotation is $1.19$ $\times$ $\dpi{100} 10^{-4}$ $\dpi{100} kgm^2$.

Exams
Articles
Questions