Saved Bookmarks
| 1. |
e Example 26.11 Find the magnitude of magnetic moment of the currentcarrying loop ABCDEFA. Each side of the loop is 10 cm long and current intloop is i - 2.0 Aro |
|
Answer» By assuming two equal and opposite currents in BE, two current carrying loop (ABEFA and BCDEB) are formed. Their magnetic moments are equal in magnitude but perpendicular to each other. Hence Mnet= √M2+ M2=√2 M Where M=ia=(2.0)(0.1)(0.1)=0.02A-m2 Mnet=(√2) (0.02) A – m2=0.028 A-m2 |
|