AP EAMCET · PHYSICS · Electromagnetic Induction
A long solenoid is carrying a current \(I=I_0 \sin (\omega t)\), having \(N\) turns per unit length and radius \(R\). A square loop is placed inside the solenoid with its plane perpendicular to the solenoid axis, and its corners touching the solenoid. Now the emf induced in the square coil.
- A \(\mu_0 N I_0 R^2 \sin (\omega t)\)
- B \(2 \mu_0 N l_0 R^2 \sin (\omega t)\)
- C \(2 \mu_0 N I_0 \dot{R}^2 \omega \cos (\omega t)\)
- D \(\mu_0 N I_0 R^2 \pi \omega \cos (\omega t)\)
Answer & Solution
Correct Answer
(C) \(2 \mu_0 N I_0 \dot{R}^2 \omega \cos (\omega t)\)
Step-by-step Solution
Detailed explanation
Given, \(I=I_0 \sin \omega t\) Magnetic field inside the solenoid, \(B=\mu_0 N I\) \[ \Rightarrow \quad B=\mu_0 N I_0 \sin \omega t \] When side of square loop be \(l\), then \[ \begin{array}{rlrl} & & 4 l & =2 \pi R \\ \Rightarrow \quad & l & =\frac{\pi R}{2} \end{array} \]…
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