What is the magnitude of the equatorial and axtal feilds due to a bar magnet of length 5.0 cm at a distance of 50 cm from its mid-point? The magnetic moment of the bar magnet is `0.40 A m^2`

Options

A. `b_(eq)=3.2xx10^(-7) T`
B. `B_(ax)=3.2 xx10^(-7) T`
C. `B_(eq)=4T B_(ax)=2T`
D. `B_(ax)=2T B_(eq)=4T`

Answer

A

Explanation

Magnetic length 2l =5.0
`l=2.5 cm xx10^(-2) m`
`d=50 cm =0.5 and M =0.40 A m^(2)`
As `B_(equi)=(mu_(0)M)/(4pid^(3))`
`=(4pixx10^(-7)T mA^(-1)Am^(2))/(4pi(0.5m)^(3)`
`B_("equi")=3.2 xx10^(-7) T`
`B_("axial")=(mu_(0)2M)/(4pid^(3))`
`B_("net")=B_(M_(1))+B_(m_(2))+B_(H)`
`=(mu_(0))/(4pix^(3))`
=`(10)^(-7)/(10^(-3))xx2.2+3.6xx10^(-5)`
`=2.56 xx10^(-4) wb//m^(2)`