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A circular coil of 16 turns and radius $10 \mathrm{~cm}$ carrying a current of $0.75$ A rests with its plane normal to an external field of magnitude $5.0 \times 10^{-2} \mathrm{~T}$. 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 $2.0 \mathrm{~s}^{-1}$. What is the moment of inertia of the coil about its axis of rotation?
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Given: No. of turns $=16$
radius $\mathrm{a}=10 \mathrm{~cm}=10 \times 10^{-2}=0.1 \mathrm{~m}$
current $\mathrm{I}=0.75 \mathrm{~A}$
field $\mathrm{B}=5 \times 10^{-2} \mathrm{~T}$
frequency $v=2.0 \mathrm{~s}^{-1}$
To find: Moment of inertia $\mathrm{I}=$ ?
Formula used: $v=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{MB}}{\mathrm{I}}}$
The magnetic moment of the coil $\mathrm{M}=\mathrm{NIA}$
$=\mathrm{NI} \pi \mathrm{a}^2$
$=16 \times 0.75 \times \pi \times(0.1)^2$
$=12 \times 3.14 \times 10^{-2}=0.377 \mathrm{Am}^2$
Frequency $v=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{MB}}{\mathrm{I}}}$
$$
\Rightarrow \quad \mathrm{I}=\frac{\mathrm{MB}}{4 \pi^2 v^2}=\frac{0.377 \times 5 \times 10^{-2}}{4 \times(3.14)^2 \times 2^2}
$$
$=\frac{0.377 \times 5 \times 10^{-2}}{16 \times 9.87}=1.2 \times 10^{-4} \mathrm{~kg} \mathrm{~m}^2$
radius $\mathrm{a}=10 \mathrm{~cm}=10 \times 10^{-2}=0.1 \mathrm{~m}$
current $\mathrm{I}=0.75 \mathrm{~A}$
field $\mathrm{B}=5 \times 10^{-2} \mathrm{~T}$
frequency $v=2.0 \mathrm{~s}^{-1}$
To find: Moment of inertia $\mathrm{I}=$ ?
Formula used: $v=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{MB}}{\mathrm{I}}}$
The magnetic moment of the coil $\mathrm{M}=\mathrm{NIA}$
$=\mathrm{NI} \pi \mathrm{a}^2$
$=16 \times 0.75 \times \pi \times(0.1)^2$
$=12 \times 3.14 \times 10^{-2}=0.377 \mathrm{Am}^2$
Frequency $v=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{MB}}{\mathrm{I}}}$
$$
\Rightarrow \quad \mathrm{I}=\frac{\mathrm{MB}}{4 \pi^2 v^2}=\frac{0.377 \times 5 \times 10^{-2}}{4 \times(3.14)^2 \times 2^2}
$$
$=\frac{0.377 \times 5 \times 10^{-2}}{16 \times 9.87}=1.2 \times 10^{-4} \mathrm{~kg} \mathrm{~m}^2$
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