JEE Mains · Physics · STD 12 -6. Electromagnetic induction
The magnetic flux through a coil perpendicular to its plane is varying according to the relation \(\phi=\) \(\left(5 t^{3}+4 t^2+2 t-5\right) \; Weber\). If the resistant of the coil is \(5 \; ohm\), then the induced current through the coil at \(t =2 \; sec\) will be \(....\,A\)
- A \(15.6\)
- B \(16.6\)
- C \(17.6\)
- D \(18.6\)
Answer & Solution
Correct Answer
(A) \(15.6\)
Step-by-step Solution
Detailed explanation
\(\phi=5 t ^{3}+4 t ^{2}+2 t -5\) \(|e|=\frac{d \phi}{d t}=15 t^{2}+8 t+2\) At \(t=2,|e|=15 \times 2^{2}+8 \times 2+2\) \(\Rightarrow e =78 \; V \Rightarrow I =\frac{ e }{ R }=\frac{78}{5}=15.60\)
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