Saved Bookmarks
| 1. |
There is one conducting circular loop of radius R, which carries current I. There is another conducting circular loop of radius r placed coaxially at a distance x from it. Here r lt lt R and xgt gt R. Magnetic flux linked smaller coil due to current in bigger coil is |
|
Answer» `(mu_(0)IR^(2)pir^(2))/(2x^(3))` `B= (mu_(0)IR)/(2(R^(2)+x^(2))^((3)/(2))` Here x is much greater than R hence we can NEGLECT R in denominator to write magnetic field as follows: `B= (mu_(0)IR)/(2x^(3))` Both loops are coaxial hence magnetic field due to bigger loop on smaller loop is PERPENDICULAR to the plane of smaller loop hence magnetic flux linked with smaller coil can be written as follows: `phi= Bpir^(2)= (mu_(0)IR^(2)pir^(2))/(2x^(3))`. Hence, option (a) is correct. |
|