A strong magnet is placed under a horizontal conducting ring of radius r that carries current i as shown in Fig. If the magnetic field makes an angle `theta` with the vertical at the ring's location, what are the magnitude and direction of the resultant force on the ring?

Answer

`vecF=2piriB sin thetahatj`

Explanation

The ring carries current `i` and the magnetic field B make an angle
`theta` with the vertical at the ring's location.

Consider a small element subtending an angle `dphi` at the centre of the ring.
Hence, force on the ring due to horizontal component of the field is
`dF=idl B sin thetahatj`
`dF=iBsin theta rdphi hatj`
`:. F_j=intdF=riBsin thetaint_0^(2pi)dphi hatj`
`vecF_y=2piirB sin theta`

Force on the ring due to vertical component of the field is
`dF=-idlB cos theta hati`
`F_x=intdF=-iint dlBcos thetahati=0`
As the ring lies in a region of constant magnetic field, `Bcos theta bot` to it.
The force on it due to this component is zero as `int dl=0`
`:. vecF=2piriB sin theta hatj`.