Question

a bar magnet has a magnetic moment 2.5A m^2 and it is placed in a magnetic field of 0.2T. claculate the work done in turning the magnet from parallel to antiparallel position relative to field direction​

Answer 1

Answer:

PE=MB(1−cosθ)

so PE

1

=MB(1−cosθ)=0

PE

2

=MB(1−cos180

o

)=2MB=2×

7

2.5 J

×0.27

P.E

2

=1 Joule

were=P.E

2

−P.E

1

=1.0=1 Joule