A) \[{{f}_{c}}=9\sqrt{{{N}_{\max }}}\]
B) \[{{m}^{3}}\]
C) \[\therefore \]
D) \[{{N}_{\max }}=\frac{f_{c}^{1}}{81}=\frac{{{(10\times {{10}^{6}})}^{2}}}{81}\]
Correct Answer: B
Solution :
From Faradays law of electromagnetic induction the induced emf is equal to negative rate of change of magnetic flux. That is \[\frac{x\pi {{R}^{2}}{{n}_{1}}}{{{n}_{2}}}\] Flux induced \[\frac{x\pi {{R}^{2}}{{n}_{2}}}{{{n}_{1}}}\] where B is magnetic field, A is area. Given, \[\frac{\pi R{{n}_{1}}}{{{n}_{2}}}\] \[\pi {{R}^{2}}x\] \[\omega \] \[{{v}_{0}}\] \[\sqrt{3}{{v}_{0}}\] Circumference of a circle of radius r is \[2\pi r\]. \[3{{v}_{0}}\] Induced electric field E is \[9{{v}_{0}}\]
You need to login to perform this action.
You will be redirected in
3 sec
