KCET · Physics · Electromagnetic Induction
The magnetic flux linked with a coil varies as \( \phi=3 t^{2}+4 t+9 . \) The magnitude of the emf
induced at \( t=2 \) seconds is
- A \( 8 \mathrm{~V} \)
- B \( 16 \mathrm{~V} \)
- C \( 32 \mathrm{~V} \)
- D \( 64 \mathrm{~V} \)
Answer & Solution
Correct Answer
(B) \( 16 \mathrm{~V} \)
Step-by-step Solution
Detailed explanation
Given, magnetic flux, \( \phi=3 t^{2}+4 t+9 ; t=2 s \)
Since, emf induced \( =\frac{d \phi}{d t} \)
Therefore, emf induced at \( t=2=\left.\frac{d \phi}{d t}\right|_{t=2} \)
Now, \( \frac{d \phi}{d t}=6 t+4 \)
\( \left.\Rightarrow \frac{d \phi}{d t}\right|_{t=2}=6 \times 2+4=12+4=16 \)
Thus, emf induced \( =16 \mathrm{~V} \)
Since, emf induced \( =\frac{d \phi}{d t} \)
Therefore, emf induced at \( t=2=\left.\frac{d \phi}{d t}\right|_{t=2} \)
Now, \( \frac{d \phi}{d t}=6 t+4 \)
\( \left.\Rightarrow \frac{d \phi}{d t}\right|_{t=2}=6 \times 2+4=12+4=16 \)
Thus, emf induced \( =16 \mathrm{~V} \)
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