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A magnetic needle has a magnetic moment of \( 5 \times 10^{-2} \mathrm{Am}^{2} \) and moment of inertia \( 8 \times \)
\( 10^{-6} \mathrm{kgm}^{2} \). It hasa period of oscillation of \( 2 \mathrm{~s} \) in a magnetic field \( \vec{B} \). The magnitude of magnetic
field is approximately
Options:
\( 10^{-6} \mathrm{kgm}^{2} \). It hasa period of oscillation of \( 2 \mathrm{~s} \) in a magnetic field \( \vec{B} \). The magnitude of magnetic
field is approximately
Solution:
2627 Upvotes
Verified Answer
The correct answer is:
\( 1.6 \times 10^{-4} \mathrm{~T} \)
(B)
\[
\begin{array}{l}
T=2 \pi \sqrt{\frac{I}{M B}} \\
2=2 \pi \sqrt{\frac{8 \times 10^{-6}}{5 \times 10^{-2} \times B}} \\
\text { Squaring } 1=\frac{\Pi^{2} \times 8 \times 10^{-6}}{5 \times 10^{-2} \times B} \\
\therefore B=\frac{3.14^{2} \times 8 \times 10^{-6}}{5}=1.6 \times 10^{-4} \mathrm{~T}
\end{array}
\]
\[
\begin{array}{l}
T=2 \pi \sqrt{\frac{I}{M B}} \\
2=2 \pi \sqrt{\frac{8 \times 10^{-6}}{5 \times 10^{-2} \times B}} \\
\text { Squaring } 1=\frac{\Pi^{2} \times 8 \times 10^{-6}}{5 \times 10^{-2} \times B} \\
\therefore B=\frac{3.14^{2} \times 8 \times 10^{-6}}{5}=1.6 \times 10^{-4} \mathrm{~T}
\end{array}
\]
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