JEE Mains · Physics · STD 12 -6. Electromagnetic induction
An elliptical loop having resistance \(R\), of semi major axis \(a,\) and semi inor axis \(b\) is placed in a magnetic field as shown in the figure. If the loop is rotated about the \(x-\) axis with angular frequency \(\omega\), the average power loss in the loop due to Joule heating is :
- A \(\frac{\pi^{2} a ^{2} b ^{2} B ^{2} \omega^{2}}{2 R }\)
- B Zero
- C \(\frac{\pi^{2} a^{2} b^{2} B^{2} \omega^{2}}{R}\)
- D \(\frac{\pi abB \omega}{ R }\)
Answer & Solution
Correct Answer
(A) \(\frac{\pi^{2} a ^{2} b ^{2} B ^{2} \omega^{2}}{2 R }\)
Step-by-step Solution
Detailed explanation
\(\in= NAB \omega \cos \omega t \quad[ N =1]\) \(P _{ avg }=<\frac{\epsilon^{2}}{ R }>=<\frac{( AB \omega \cos \omega t )^{2}}{ R }>\) \(=\frac{ A ^{2} B ^{2} \omega^{2}}{ R } \frac{1}{2}=\frac{\pi^{2} a ^{2} b ^{2} B ^{2} \omega^{2}}{2 R }\)
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