Saved Bookmarks
| 1. |
2 A current is passed through a solenoid which has 5 cm diameter, 10 turns and 10 cm length. Now, near to one of its two ends, a circular loop of radius 2 cm is kept perpendicular to axis of solenoid. Then magnetic flux passing through this loop is ...... Wb. (mu_0=4pixx10^(-7)TmA^(-1) and take pi^@=10) |
|
Answer» `3.2xx10^(-8)` `B=mu_0nI` `B=mu_0 (N/l)I` Now, magnetic flux linked with the loop , mentioned in the statement is : `phi=BA cos theta` `=(mu_0NI)/l xx pir^2xxcos0^@` `=(4pixx10^(-7)xx10xx2xxpixx4xx10^(-4))/0.1` `=32pi^2xx10^(-9)` `=3.20xx10^(-7)` `=3.2xx10^(-7)` WB |
|