First slide

Magnetic flux and induced emf

Question

A ring of mass m, radius R having charge q uniformly distributed over it and free to rotate about its own axis is placed in a region having a magnetic field B parallel to its axis. If the magnetic field is suddenly switched off, the angular velocity acquired by the ring is

Difficult

A

$\frac{q B}{m}$
B

$\frac{2 q B}{m}$
C

$\frac{q B}{2 m}$
D

none of these

Solution

$C o n s i d e r a s h a d e d r i n g o f t h i c k n e s s d r, h a v i n g s m a l l c h a r g e d q . d q = \frac{q}{π R^{2}} (2 π r d r) I n d u c e d e l e c t r i c f i e l d i s E (s a y) E (2 π r) = (π r^{2}) \frac{d B}{d t} \Rightarrow E = \frac{r}{2} \frac{d B}{d t} T h i s e l e c t r i c f i e l d w i l l a p p l y f o r c e o n c h a r g e d q, w h i c h w i l l g e n e r a t e t o r q u e d τ . τ = \int d τ = \int (r d F) = \int (r E d q) = \int r \times \frac{r}{2} \frac{d B}{d t} \times \frac{q}{π R^{2}} (2 π r d r) = \frac{q R^{2}}{4} \frac{△ B}{△ t} W e k n o w, A n g u l a r I m p u l s e = △ L = τ △ t \Rightarrow I △ ω = \frac{q R^{2}}{4} \frac{△ B}{△ t} \times △ t \Rightarrow (\frac{m R^{2}}{2}) △ ω = \frac{q R^{2}}{4} △ B \Rightarrow △ ω = \frac{q B}{2 m}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App