A short bar magnet placed with its axis at `30^(@)` with a uniform external magnetic field of 0.35 T experiences a torque of magnitude equal to `4.5xx
A short bar magnet placed with its axis at `30^(@)` with a uniform external magnetic field of 0.35 T experiences a torque of magnitude equal to `4.5xx10^(-2)Nm`. The magnitude of magnetic moment of the given magnet is
A. 1.36 J/T
B. 0.06 J/T
C. 0.36 J/T
D. 0.60 J/T
5 views
1 Answers
Correct Answer - C
`tau=MB sintheta`
`M=(tau)/(Bsintheta)=(4.5xx10^(-2))/(0.25xxsin30)=0.36 Am^(2)`
5 views
Answered