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. `2.56 xx 10^(-4) Wb//m^2`
B. `3.50 xx 10^(-4) Wb//m^2`
C. `5.80 xx 10^(-4) Wb//m^2`
D. `3.6 xx 10^(-5) Wb//m^2`

Answer

A

Explanation

We have short magnets
`M_1 = 1.20 Am^2 " and " M_2 = 1.00 Am^2`

` d= 0.1 m ,M_1 = 1.20 Am^2`
`M_2 = 100 Am^2`
Net magnetic field due to a small bar magnet and Earth.s horizontal magnetic field at centre O is:
`B=B_1+B_2+B_H`
`(mu_0)/(4pi) (M)/(d^3) + (mu_0 M_2)/(4pi d^3) + B_H`
` = (10^(-7) xx 1.20 )/( (0.1)^3) + (10^(-7) xx 1)/((0.1)^3) + 3.6 xx 10^(-5)`
` = 106(-7) xx 10^(-3) xx 2.20 + 3.36 xx 10^(-5)`
` = 2.56 xx 10^(-4) Wh//m^2`