# Magnetic field in a region is equal to B. A loop of resistance R and radius r is lying in the field with the plane perpendicular to the field if the loop is turned by 180 degree. Find the charge which flows through a given cross section of the loop

Dear Student,

Please find below the solution to the asked query:

The emf generated by in the loop due to magnetic field is $\epsilon =-\frac{d\varphi }{dt}=-\frac{d}{dt}BA\mathrm{cos}\left(\omega t\right)=BA\omega \mathrm{sin}\left(\omega t\right)$
Now if r is the radius of the loop then area is given by $A={\mathrm{\pi r}}^{2}$
If I is the current through the loop we, then we have $I=\frac{\epsilon }{R}=\frac{B\pi {r}^{2}\omega \mathrm{cos}\left(\omega t\right)}{R}$
So total charge flown through the circuit is given by
$q={\int }_{0}^{T}Idt={\int }_{0}^{T}\frac{B{\mathrm{\pi r}}^{2}\mathrm{\omega sin}\left(\mathrm{\omega t}\right)}{R}dt$
Here we are considering that the time taken to rotate 180 degree is T. Now we have
$\theta =\omega t\phantom{\rule{0ex}{0ex}}⇒d\theta =\omega dt$
Using above value we get
$q={\int }_{0}^{\mathrm{\pi }}\frac{B\pi {r}^{2}}{R}sin\theta \mathrm{d}\theta =\frac{B\pi {r}^{2}}{R}{\int }_{0}^{\mathrm{\pi }}sin\theta \mathrm{d}\theta =\frac{B\pi {r}^{2}}{R}2=\frac{2B\pi {r}^{2}}{R}$