In the general case , the unit normal vector `hatn` and magnetic field `vecB` is with an angle `theta` as shown in Figure.
(a) The force on section in PQ
` vec l = a hatj " and " vecB = B hat i`
`vecF_("PQ") = vec(Il) xx vecB = I a B ( hatj xx hati ) = - I a B hat k `
Since , the unit vector normal to the plane `hatn` is along the direction of `hat k`.
(b ) The force on section QR
`vecl = b cos (pi/2 - theta) ĭ sin (pi/2 - theta) k̆ " and " vecB = B ĭ `
`vecF_("QR") = vec(Il) xx vecB = - IbB (pi/2 - theta) hat j `
`vecF_("QR") = - IbB cos theta hat j `
(c) The force on section RS
`vecl = a hatj " and " vecB = B hat i `
` vecF_("RS") = vec(Il) xx vecB = I a B ( hatj xx hati) = - IaB hat k `
Since , the unit vector normal to the plane is along the direction of ` - hat k` .
(d) The force on section SP
`vecl = b cos ( pi/2 - theta ) hati + sin ( pi/2 + theta ) hatk " and " vecB = B hat i`
` vecF_("SP") = vec(Il) xx vecB = IbB sin (pi/2 + theta ) hat j`
` vecF_("SP") = - IbB cos theta hatj`
The net force on the rectangular loop is
`vecF_("net") = vecF_("PQ") + vecF_("QR") + vecF_("RS") + vecF_("SP") `
`vecF_("net") = IaB hatk - IbB cos theta hatj - IaB hatk + Ib B cos theta hatj vecF_("net") = vec0`
Hence, the net force on the rectangular loop in this configuration is also zero.
Notice that the force on section QR and SP are not zero here. But, they have equal and opposite effects, but we assume that the loop to be rigid, so no deformation. So, no torque produced by these tow sections.
Even though the forces PQ and RS also are equal and opposite, they not collinear. So these two forces constitute a couple as shown in Figure (a) . Hence the net torque produced by these two forces about the axis of the rectangular loop is given by `vec tau_("net") = baBI sin theta hatk = ABI sin theta hat k`
From the Figure (c )
`vec(OA) = b/2 cos (pi/2 - theta) ( - hat i) + b/2 sin ( pi/2 - theta) ( - hat k ) `
` = b/2 ( - sin theta hati + cos theta hatk) `
`vec(OB) = b/2 cos ( pi/2 - theta ) ( hat i) + b/2 sin ( pi/2 - theta ) ( - hat k) `
` = b/2 ( - sin theta hati + cos theta hatk) `
`vec(OA) xx vecF_("PQ") = { b/2 ( -sin theta hati + cos theta hatk)} xx { Ia Bhatk}`
` = 1/2 IabB sin theta hatj `
`vec(OB) xx vecF_("RS") = { b/2 ( sin theta hati + cos theta hatk)} xx { - I aB hatk}`
` = 1/2 IabB sin theta hatj`
The net torque `vectau_("net") = I abB sin theta hatj` .....(1)
Note that the net torque is in the positive y direction which tends to rotate the loop in clockwise direction about the y axis. If the current is passed in the other way `(P to S to R to Q to P)` , then total torque will point in the negative y direction which tends to rotate the loop in anticlockwise direction about y axis .
Another important point is to note that the torque is less in this case compared to earlier case (where the `hat n` is perpendicular to the magnetic field `vecB`) . It is becasue the perpendicular distance is reduced between the forces `vecF_("PQ") " and " vecF_("RS") ` in this case.
The equation (1) can also be rewritten in terms of magnetic dipole moment `vecp_(m) = I vecA = I ab hatn vectau_("net") = vecp xx vecB`
