Let ABCD = be a rectangular coil suspended in a uniform magnetic field `(vec(B))`. Let AB=CD= l (length of coil) BC= DA= b (breadth of coil). I = be the current flowing in the direction ABCD. `theta` =be the angle between plane of coil and magnetic field `vec(B)`
a= angle between normal to the coil and magnetic field.
`vec(F)_(1)`= be the force acting on current carrying arm AB, `vec(F)_(2)`= be the force acting on current carrying arm BC, `vec(F)_(3)` = be the force acting on curretn carrying arm CD, `vec(F)_(4)` = be the force acting on current carrying arm DA
The force on arm DA is given by `vec(F)_(4)= I (vec(DA) xx vec(B)) = I (DA) (B) sin (180^(@)- theta)`
`vec(F)_(4)=1 b B sin theta`
The direction of this force is in the direction of `DA xx B` i.e., in the plane of coil directed upwards. The force on the arm BC is given by `vec(F)_(2)= I (vec(BC) xx vec(B)) = I (BC) (B) sin theta`
`F_(2)= I b B sin theta`
The direction of this force is in direction of `vec(BC) xx vec(B)` i.e., in the plane of coil directed downward. Since the forces `vec(F)_(2) and vec(F)_(4)` are the equal in magnitude and acting in opposite direction along the same straight line, they cancel out each other. Now the force on arm AB is given by `vec(F)_(1)= I (vec(AB) xx vec(B))`
`F_(1)= I (AB) (B) sin 90^(@)`
`vec(F)_(1)=I l B`
According to Fleming.s left hand rule, the direction of this force `vec(F)_(1)` is `bot` to the plane of coil directed towards the reader. Force on arm CD is giben by `vec(F)_(3)= I (vec(CD) xx vec(B))`
`F_(3)= I (CD) (B) sin 90^(@)`
`F_(3)= I l B`
According to Fleming.s left hand rule, the direction of this force `vec(F)_(3)` is `bot` to the plane of coil directed away from the reader. The forces acting on the arms AB and CD are equal, Parallel and acting in opposite directions having different lines of action. These forces a couple, the effect of which is to rotate the coil in the anti clockwise direction about the dotted line as axis. Torque in the coil is given by `tau`= either force `xx` (`bot` distance) i.e, (arm of couple)
SInce, `cos theta= (DE)/(AD)`
`cos theta= (DE)/(b)`
`DE = b cos theta`
`therefore tau = I//B xx DE`
`tau = I//B b cos theta`
`tau = IAB cos theta (because I xx b = A)`
It the rectangular coil has n turns then `tau = eta IAB cos theta`
If the normal draw on the plane of coil makes an angle `alpha` with direction of magnetic field then `theta+alpha= 90^(@)`
`theta= 90^(@)-a`
`cos theta= cos (90^@)-a)`
`cos theta= sin a`
`tau = eta IBA sin a`
`tau = MB sin a`
`vec(tau) = vec(M) xx vec(B) ((because [n=1]),(I xx A= M))`
where M is Magnetic dipole moment
Special case: 1. If the coil is set with its plne parallel to magnetic field then `theta = 0^(@) or alpha= 90^(@)`
`tau = eta IBA sin 90^(@)`
`tau = eta IBA` Maximum. This is radial magnetic field
2. If the coil is set with its plane `bot` to magnetic field then `theta= 90^(@), a=0`
`tau = eta IBA cos 90^(@)`
`tau_("min") = 0`
