InterviewSolution
Saved Bookmarks
| 1. |
A short bar magnet placed with its axis at `30^@` with a uniform external magnetic field of `0*25T` experiences a torque of magnitude equal to `4*5xx10^-2J`. What is the magnitude of magnetic moment of the magnet? |
|
Answer» Here, `theta=30^@`, `B=0*25T`, `tau=4*5xx10^-2J`, `M=?` As, `tau=MB sin theta` `:. M=(tau)/(B sin theta)=(4*5xx10^-2)/(0*25sin30^@)=0*36JT^-1` |
|