Saved Bookmarks
| 1. |
A circular conducting loop of radius r_(0) and having resistance per unit length lambda as shown in the figure is placed in a magnetic field B which is constant in space and time. The ends of the loop are crossed and pulled in opposite directions with a velocity v such that the loop always remains circular and the radius of the loop goes on decreasing, then |
|
Answer» radius of the loop changes with r as `r=r_(0)-(V t)/(pi)` `therefore (d)/(dt)(2pi r)=2v "" therefore (dr)/(dt)=(v)/(pi)` `therefore r=(r_(0)-(v)/(pi)t)` and `phi=B. pi r^(2)` `rArr epsilon = |(-d phi)/(dt)|=B. 2pi. r(dr)/(dt)` `therefore epsilon=2B pi(r_(0)-(v)/(pi)t)(v)/(pi)=2Bv (r_(0)-(v)/(pi)t)` `therefore I=(phi)/(R )=(2B br)/(lambda.2pi r)=(Bv)/(pi lambda)` |
|