In a uniform magnetic field of induction B, a wire,in the form of semicircle of radius r rotates about the diameter of the circle with angular frequency $ω$ If the total resistance of the circuit is R, the mean power generated per period of rotation is :

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{{Bπr}^{2} ω}{2 R}$

$\frac{{({Bπr}^{2} ω)}^{2}}{8 R}$

$\frac{{(Bπrω)}^{2}}{2 R}$

$\frac{{({Bπrω}^{2})}^{2}}{8 R}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$F l u x l i n k e d w i t h t h e c o i l ϕ = B A \cos ϕ = \frac{1}{2} π r^{2} B \cos ω t e = i n d u c e d e m f = - \frac{d ϕ}{d t} = \frac{1}{2} π r^{2} B ω \sin ω t i n s \tan \tan e o u s p o w e r = P = \frac{e^{2}}{R} = \frac{1}{4 R} π^{2} r^{4} B^{2} ω^{2} \sin^{2} ω t A v e r a g e p o w e r f o r t h e f u l l c y c l e = \frac{1}{4 R} π^{2} r^{4} B^{2} ω^{2} \frac{\int_{0}^{T} \sin^{2} ω t d t}{T} = \frac{1}{4 R} π^{2} r^{4} B^{2} ω^{2} \frac{1}{2} = \frac{1}{8 R} π^{2} r^{4} B^{2} ω^{2}$