KCET · Physics · Oscillations
A magnetic dipole of magnetic moment \( 6 \times 10^{-2} \mathrm{Am}^{2} \) and moment of inertia
\( 12 \times 10^{-6} \mathrm{~kg} \mathrm{~m}^{2} \) performs oscillations in a magnetic field of \( 2 \times 10^{-2} \mathrm{~T} \). The time taken by
the dipole to complete \( 20 \) oscillations is \( (\pi \simeq 3) \)
- A \( 36 s \)
- B \( 06 s \)
- C \( 12 s \)
- D \( 18 s \)
Answer & Solution
Correct Answer
(C) \( 12 s \)
Step-by-step Solution
Detailed explanation
Given, magnetic moment, \( M=6 \times 10^{-2} \mathrm{Am}^{2} \), moment of inertia, \( I=12 \times 10^{-6} \mathrm{kgm}^{2} \); magnetic field,
\( B=2 \times 10^{-2} T \); number of oscillations \( =20 \).
Now, time period of one oscillation, \( t=2 \pi \sqrt{\frac{l}{B \times M}} \)
\( t=2 \pi \sqrt{\frac{12 \times 10^{-6}}{2 \times 10^{-2} \times 6 \times 10^{-2}}}=2 \Pi \times 10^{-1} \)
Therefore, time taken by dipole to complete \( 20 \) oscillations is
\( T=20 \times t=20 \times 2 \Pi \times 10^{-1}=20 \times 2 \times 3 \times \frac{1}{10}=12 s \)
\( B=2 \times 10^{-2} T \); number of oscillations \( =20 \).
Now, time period of one oscillation, \( t=2 \pi \sqrt{\frac{l}{B \times M}} \)
\( t=2 \pi \sqrt{\frac{12 \times 10^{-6}}{2 \times 10^{-2} \times 6 \times 10^{-2}}}=2 \Pi \times 10^{-1} \)
Therefore, time taken by dipole to complete \( 20 \) oscillations is
\( T=20 \times t=20 \times 2 \Pi \times 10^{-1}=20 \times 2 \times 3 \times \frac{1}{10}=12 s \)
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