Two short bar magnets of length `1cm` each have magnetic moments `1.20 Am^(2) and 1.00 Am^(2)` respectively. They are placed on a horizontal table parallel to each other with their `N` poles pointing towards the south. They have a common magnetic equator and are separted by a distance of `20.0 cm`. The value of the resultant horizontal magnetic induction at the mid - point `O` of the line joining their centres is close to (Horizontal component of earths magnetic induction is `3.6 xx 10.5 Wh//m^(2)`

Options

A. `3.6xx10^(-5)Wb//m^(2)`
B. `2.56xx10^(-4)Wb//m^(2)`
C. `3.50xx10^(-4) Wb//m^(2)`
D. `5.80xx10^(-4)Wb//m^(2)`

Answer

b

Explanation

(b) Given:
`M_(1)=1.20Am^(2)`
`M_(2)=1.00Am^(2)`
`r=(20)/(2) cm =0.1m`
From figure `B_("net")=B_(1)+B_(2)+B_(H)`
`B_("net")=(mu_(0))/(4pi)((M_(1)+M_(2)))/(r^(3))+B_(H)`
`=(10^(-7)(1.2+1))/((0.1)^(3))+3.6xx10^(-5)=2.56xx10^(-4)wb//m^(2)`