Q156 / 180 · KCET 2015 · Physics
Two coils have a mutual inductance \( 0.005 \mathrm{H} \). The current changes in the first coil according to
the equation \( i=i_{m} \sin \omega t \) where \( i_{m}=10 \mathrm{~A} \) and \( \omega=100 \Pi \mathrm{rad} \mathrm{s}^{-1} \). The maximum value of
the emf induced in the second coil is
the equation \( i=i_{m} \sin \omega t \) where \( i_{m}=10 \mathrm{~A} \) and \( \omega=100 \Pi \mathrm{rad} \mathrm{s}^{-1} \). The maximum value of
the emf induced in the second coil is
- A\( 2 \Pi \)
- Correct answerB( \( 5 \Pi\)
- C\( \Pi \)
- D\( 4 \Pi \)
Solution
Given, mutual inductance,
\( M=0.005 H ; i=i_{\omega} \sin \omega t ; i_{m}=10 A ; \omega=100 \pi r a d s^{-1} \)
Then emf induced is
\( \varepsilon=M \frac{d I}{d t}=M \frac{d}{d t}\left(i_{m} \sin \omega t\right)=M i_{m} \omega \cos \omega t \)
Maximum value of emf is obtained when \( \cos \omega t=1 \).
Therefore \( \varepsilon_{\max }=M \omega i_{m}=0.005 \times 10 \times 100 \Pi=5 \Pi \)
Thus, maximum value of emf induced in the second coil is \( 5 \pi \)
\( M=0.005 H ; i=i_{\omega} \sin \omega t ; i_{m}=10 A ; \omega=100 \pi r a d s^{-1} \)
Then emf induced is
\( \varepsilon=M \frac{d I}{d t}=M \frac{d}{d t}\left(i_{m} \sin \omega t\right)=M i_{m} \omega \cos \omega t \)
Maximum value of emf is obtained when \( \cos \omega t=1 \).
Therefore \( \varepsilon_{\max }=M \omega i_{m}=0.005 \times 10 \times 100 \Pi=5 \Pi \)
Thus, maximum value of emf induced in the second coil is \( 5 \pi \)
Hence, the correct answer is option: 'B'.