Two bar magnets A and B , each having a magnetic length of 4 cm are placed along a straight line with their north poles 8 cm apart and facing each other . The neutral point lies on the axis , 2 cm from the north pole of the magnet A . Calculate the ratio of the magnetic moments of A and B . [ Ignore the effect of earth ' s magnetic field ] .

Explanation

O is the neutral point Fig
Given , No = 2 cm
`thereforeN'O=(8-2)=6cm `
`XN=X'N=l=(4)/(2)cm=2xx10^(-2)m`
`d_(1)=XO=(2+2)cm=4xx10^(3)m`
`d_(2)=XO=(6+2)cm=8xx10^(3)m`

`therefore` Magnetic field at point O due to the magnet A ,
`B_(1)=(mu_(0)2p_(m1)d_(1))/(4pi(d_(1)^(2)-l^(2))^(2))`
`"or, " B_(1)=(mu_(0))/(4pi)(2p_(m_(1))xx4xx10^(-2))/([(4xx10^(-2))^(2)-(2xx10^(-2))^(2)]^(2))`
`=(mu_(0))/(4pi)(2p_(m_(1))xx4xx10^(-2))/((16-4)^(2)xx10^(-8))`
`therefore` Magnetic field at O due to the magnet B ,
`B_(2)=(mu_(0))/(4pi)(2p_(m_(2))xx8xx10^(-2))/([(8xx10^(-2))^(2)-(2xx10^(-2))^(2)]^(2))`
`=(mu_(0))/(4pi)(2p_(m_(2))xx8xx10^(-2))/((64-4)^(2)xx10^(-8))`
According to the question , `B_(1)=B_(2)`
`therefore" " (mu_(0))/(4pi).(2p_(m_(1))xx4xx10^(-2))/(144xx10^(-8))= (mu_(0))/(4pi).(2p_(m_(2))xx8xx10^(-2))/(3600xx10^(-8))`
`"or, "(p_(m_(1)))/(p_(m_(2)))=(2xx144)/(3600)=(2)/(25)`
`therefore` Ratio pf the magnetic moments = `p_(m_(1)):p_(m_(2))=2:25`