Electro Magnetic Induction And Alternating Currents Question 196
Question: In a uniform magnetic field of induction B a wire in the form of a semicircle of radius r rotates about the diameter of the circle with an angular frequency $ \omega $ . The axis of rotation is perpendicular to the field. If the total resistance of the circuit is R, the mean power generated per period of rotation is
Options:
A) $ \frac{{{(B\pi r\omega )}^{2}}}{2R} $
B) $ \frac{{{(B\pi r^{2}\omega )}^{2}}}{8R} $
C) $ \frac{B\pi r^{2}\omega }{2R} $
D) $ \frac{{{(B\pi r{{\omega }^{2}})}^{2}}}{8R} $
Show Answer
Answer:
Correct Answer: B
Solution:
-
$ \phi =\overset{\to }{\mathop{B}},.\overset{\to }{\mathop{A}}, $ ;
$ \phi =BA,\cos ,\omega t $ $ \varepsilon =-\frac{d\phi }{dt}=\omega BA,\sin ,\omega t $ ;
$ i=\frac{\omega BA}{R},\sin ,\omega t $
$ P_{inst}=i^{2}R={{( \frac{\omega BA}{R} )}^{2}}\times R,{{\sin }^{2}},\omega t $
$ P_{avg}=\frac{\int\limits_{0}^{T}{P_{inst}\times dt}}{\int\limits_{0}^{T}{dt}}=\frac{{{(\omega ,B,A)}^{2}}}{R}\frac{\int\limits_{0}^{T}{{{\sin }^{2}},\omega ,tdt}}{\int\limits_{0}^{T}{dt}}=\frac{1}{2},\frac{{{(\omega ,B,A)}^{2}}}{R} $
$ \therefore P_{avg}=,\frac{{{(\omega ,B,\pi ,r^{2})}^{2}}}{8,R} $$ [ A=\frac{\pi ,r^{2}}{2} ] $
BETA
