A magnet `15cm` long with poles of strength `250Am` lies on a table. Find the magnitude of the magnetic intensity B at a point P, `20cm` directly above the north pole of the magnet.

Answer

`388*1muT`

Explanation

As is clear from figure.

`B_N=(mu_0)/(4pi)(m_1xxm_2)/((NP)^2)=(10^-7xx250xx1)/((0*2)^2)`
`=625xx10^-6T=625muT`
Now `SP=sqrt((0*2)^2+(0*15)^2)=0*25m`
`B_S=(10^-7xx250xx1)/((0*25)^2)=400muT`,
Vertical component of `B_S` is
`B_Scostheta=400xx(20)/(25)=320muT`, opposite to `B_N`
Horizontal components of `B_S` is
`B_Ssintheta=400xx(15)/(25)=240muT`
Resultant vertical component `=625-320`
`=305muT`
Resultant field at `P=sqrt((305)^2+(240)^2)`
`=388*1muT`