Saved Bookmarks
| 1. |
Two circular coils of radii a and 2a having a common centre, carry identical current I but in opposite directions. Number of turns of the second conductor is 8. show that magnetic field intensity at the centre is 3 times that due to the smaller one. Also find out the change in the previous when current flow in the same direction through both the coils. |
|
Answer» Solution :Mangneticfield intensity at the centre O due to the smaller loop is, `B_(1)=(mu_(0)i)/(2a)` (upwards) SIMILARLY magnetic FIELD intensity at the centredue to the bigger loop, `B_(2)=8xx(mu_(0)i)/(2(2a))=(2mu_(0)i)/(a)` (upwards)  `:.` Net magnetic field at O `B=B_(2)-B_(1)` `=(mu_(o)i)/(a)(2-(1)/(2))` `=(3mu_(0)i)/(2a)=3B_(1)[ :. B_(1)=(mu_(0)i)/(2a)]` Hence resultant field is 3 times that due to the smaller lopp. Now if the direction of the CURRENT is same for both the LOOPS, the resultant field will be, `B=B_(1)+B_(2)=(mu_(0)i)/(2a)+(2mu_(0)i)/(a)=5(mu_(0)i)/(2a)=5B_(1)` Hence the resultant field will be 5 times that due to the smaller loop is current flows in the same direction. |
|