Saved Bookmarks
| 1. |
If the solenoid in is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25T is applied , what is the magnitude of torque on the solenoid when its axis makes an angle of 30^@ with the direction of applied field ? Given magnetic moment 0.6 JT−1 . |
|
Answer» SOLUTION :B = 0.25 T , `theta =30^@` `taumB sin theta=0.6xx0.25xxsin30=0.6xx0.25xx1/2=0.075Nm ` (or J) |
|