MHT CET · Physics · Electromagnetic Induction
The magnetic flux linked with the coil with time as \(\Phi=4 \mathrm{t}^2+3 \mathrm{t}+7\). The magnitude of the induced e.m.f at \(t=2 \mathrm{~s}\) is
- A \(16 \mathrm{~V}\)
- B \(29 \mathrm{~V}\)
- C \(11 \mathrm{~V}\)
- D \(19 \mathrm{~V}\)
Answer & Solution
Correct Answer
(A) \(16 \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
Given that
Total magnetic flux \(\phi=3 t^2+4 t+7\)
We know that,
The induced e.m.f
\(\mathrm{e}=\left|\frac{-\mathrm{d} \phi}{\mathrm{dt}}\right|=\frac{\mathrm{d}\left(3 \mathrm{t}^2+4 \mathrm{t}+7\right.}{\mathrm{dt}}=6 \mathrm{t}+4\)
Now, the induced e.m.f at \(\mathrm{t}=2 \mathrm{sec}\)
\(e=6 \times 2+4=16\) volt
Hence, the magnitude of induced e.m.f is 16 volt
Total magnetic flux \(\phi=3 t^2+4 t+7\)
We know that,
The induced e.m.f
\(\mathrm{e}=\left|\frac{-\mathrm{d} \phi}{\mathrm{dt}}\right|=\frac{\mathrm{d}\left(3 \mathrm{t}^2+4 \mathrm{t}+7\right.}{\mathrm{dt}}=6 \mathrm{t}+4\)
Now, the induced e.m.f at \(\mathrm{t}=2 \mathrm{sec}\)
\(e=6 \times 2+4=16\) volt
Hence, the magnitude of induced e.m.f is 16 volt
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