1.

A short bar magnet of amgnetic moment 25 JT^(-1)is placed with its axis perpendicular to earththe resultant field is inclined at 45^(@)with earth fieldif H=0.4xx10^(-4)T

Answer»

5m
0.5 m
2.5 m
0.25 m

Solution :SINCE B and H areto eachother and RESULTANT FIELDIS inclined at an angle`45^(@)`with
so B=H
`therefore (mu_(0))/(4pi)(2M)/(r^(3))=H`
`therefore r^(3)=(mu_(0))/(4pi)(2M)/(H)=0.5 m`


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