Two identical short bar magnets each of magnetic moment `12*5Am^2` are placed at a separation of `10cm` between their centres, such that their axes are perpendicular to each other. Find the magnetic field at a point midway between the two magnets.

Answer

`2*24xx10^-2T`, `theta=26*6^@`

Explanation

Here, `M_1=M_2=12.5Am^2`.
`O_1O_2=10cm`figure.
P is midway between `O_1` and `O_2`
`:. d=O_1P=O_2 P=5cm=5xx10^-2m`

As P lies an axial line of `N_1S_1`,
`:. B_1=(mu_0)/(4pi)(2M_1)/((d)^3)=(10^-7xx2xx12*5)/((5xx10^-2)^3)`
`=0*02T`, along `O_1PO_2`
Again, P lies on equatorial line of `N_2S_2`
`:. B_2=(mu_0)/(4pi)(M_2)/(d^3)=(10^-7xx12*5)/((5xx10^-2)^3)=0*01`, `_|_PO_2`
Resultant magnetic field of strength at P
`B=sqrt(B_1^2+B_2^2)=sqrt((0*02)^2+(0*01)^2)`
`=2*24xx10^-2T`
If `theta` is the angle which `vecB` makes with `vecB_1`, then
`tan theta=B_2/B_1=(0*01)/(0*02)=0*5`, `theta=26*6^@`