Define magnetic field intensity .Derive an expression for magnetic field intensity due to a bar magnet at a point on broadside - on position .

Explanation

The capability of magnetic field to magnetic the substance is emasured in terms of magnetic intensity of the field . The magnitude of magnetic intensity may be define as the number of amper turns flowing roound unit length of toroid to produce the magnetic induction `B_(0)` , in the toroid , H is given by relation
`H = (B_(0))/(mu_(0))....(i)`
In figure , the point P is shown on equatorial line of the same bar magnet , where OP = d . Magnetic field at P due to N pole of magnet `B = (mu_(0))/(4pi) (mxx1)/((NP)^(2)) =(mu_(0))/(4pi) m/((d^(2)+l^(2)))`
along NP produced
Magnetic field at P due to S ploe of magnet ,
`B = (mu_(0))/(4pi) (mxx1)/(SP^(2)) =(mu_(0))/(4pi) m/((d^(2)+l^(2))) , ` along PS.

As B = B in amgnitude , their components B `sin theta ` along OP produced and B `sin theta ` along PO cancel . However , components along PX parallel to NS add. Therefore , magnetic field at P due to the bar magnet ,
`B_(2) = B cos theta + B cos theta = 2 B cos theta ` along PX
`B_(2) 2 (mu_(0))/(4pi) m/((d^(2)+l^(2))) xxl/(sqrt(d^(2)+l^(2)))`
`=(mu_(0))/(4pi) (m xx2l)/((d^(2)+l^(2))^(3//2))`
`B_(2)=(mu_(0))/(4pi) M/((d^(2)+l^(2))^(3//2))`
If the magnet is short , `l^(2) lt lt d^(2)`
`:. " " (mu_(0))/(4pi) M/((d^(2))^(3//2)) =(mu_(0))/(4pi) M/(d^(3))`
`rArr` Magnetic field intensity at point P is
`H = (B_(2))/(mu_(0)) =1/(4pi) M/(d^(3))`
Hence , magnetc field intensity due to a short bar magnet at any point on the axial line of magnet is twice the magnetic field intensity at a point at the same distance on the equatorial line of the magnet .