Saved Bookmarks
| 1. |
A closely wound solenoid of 1000 tuins and area of Cross-section 2.0 xx 10^(-4) in- carries a current of 2.0 A. It is placed with its horizontal axis al 30^(@) with the direction of a uniform horizontal magnetie field of 0.16 T as shown in the figure. What is the magnitude of the restoring torque produced by the lield on the solenoid! |
|
Answer» 0.016 Nm `M = nIA = 1000xx2xx2xx10^(-4)` `therefore M= 0.4 m^(2)` The direction of M is along the axis of the solenoid i.e., in the direction PQ. It makes an angle of `30^(@)` with the direction of the field. A restorings torqua `TAU` act on the solenoid, to bring the axis of the solenoid N-S direction. The magnitude of torqua, `tau = MB sintheta` `= 0.4 xx0.16 XX sin30^(@) =0.064 xx 1//2 =0.032 Nm` |
|