1.

For a short bar magnet ("B-axial")/("B-equatorial") is …....

Answer»

`1:2`
`1:1`
`3:2`
`2:1`

Solution :For small BAR magnet `B_("axis")= (mu_0)/( 4pi ) (2m)/( x^3) ( x gt gt gt 2l)`
`B_("equator") = (mu_0)/( 4pi ) (m)/( x^3)` (where `xgt gt gt 2l` )
`THEREFORE (B_("AXIAL"))/( B_("equator")) = (2)/(1) = 2:1`


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