KCET · Physics · Magnetic Effects of Current
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
- A \( 3.2 \times 10^{-4} \mathrm{~T} \)
- B \( 1.6 \times 10^{-4} \mathrm{~T} \)
- C \( 0.8 \times 10^{-4} \)
- D \( 0.4 \times 10^{-4} \)
Answer & Solution
Correct Answer
(B) \( 1.6 \times 10^{-4} \mathrm{~T} \)
Step-by-step Solution
Detailed explanation
\(T=2 \pi \sqrt{\frac{I}{M B}} \)
\(2=2 \pi \sqrt{\frac{8 \times 10^{-6}}{5 \times 10^{-2} \times B}} \)
\(\text {Squaring } 1=\frac{\Pi^{2} \times 8 \times 10^{-6}}{5 \times 10^{-2} \times B} \)
\(\therefore B=\frac{3.14^{2} \times 8 \times 10^{-6}}{5}=1.6 \times 10^{-4} \mathrm{~T}\)
\(2=2 \pi \sqrt{\frac{8 \times 10^{-6}}{5 \times 10^{-2} \times B}} \)
\(\text {Squaring } 1=\frac{\Pi^{2} \times 8 \times 10^{-6}}{5 \times 10^{-2} \times B} \)
\(\therefore B=\frac{3.14^{2} \times 8 \times 10^{-6}}{5}=1.6 \times 10^{-4} \mathrm{~T}\)
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