Torque experienced by a current loop in uniform magnetic field :
Consider a rectangle conducting loop ABCD of a wire of length l and width b carrying current I ( clock wise ) is placed in a uniform magnetic field $\overrightarrow{B}$. The magnetic field $\overrightarrow{B}$ is acting X - axis along X-axis and the Normal of the plane of the loop $\hat{n}$ makes an angle $\theta$ with the magnetic field B.
The magnetic field B is perpendicular to the arms AB and CD of the loop and the angle between the arm BC or AD and the magnetic field is not 90°.
Force acting on the arm AB of the conducting loop carrying current I due to magnetic field is given by :
$$\overrightarrow{F_{1}} = I(\overrightarrow{l}\times \overrightarrow{B})$$
According to Fleming's left hand rule , direction of $\overrightarrow{F_{1}}$ is perpendicular to the length of arm AB and directed into the plane of the paper
The magnitude of force $\overrightarrow{F_{1}}$ is given by
$$\overrightarrow{F_{1}}= I.l.B.sin90^{0}$$
$$\overrightarrow{F_{1}}= I.l.B $$
Force acting on the arm CD of the loop carrying current I due to magnetic field is given by :
$$\overrightarrow{F_{2}} = I(\overrightarrow{l}\times \overrightarrow{B})$$
As per Fleming's left hand rule , the direction of $\overrightarrow{F_{2}}$ is perpendicular to the length of the arm CD and directed outward the plane of the paper.
The magnitude of force $\overrightarrow{F_{2}}$ is given by
$$\overrightarrow{F_{2}}= I.l.B.sin90^{0}$$
$$\overrightarrow{F_{2}}= I.l.B $$
Force acting on the arm AC of the loop is given by
$$\overrightarrow{F_{3}} = I(\overrightarrow{b}\times \overrightarrow{B})$$
The direction of $\overrightarrow{F_{3}}$ is along Y-axis.
Force acting on the arm DA of the loop is given by
$$\overrightarrow{F_{4}} = I(\overrightarrow{b}\times \overrightarrow{B})$$
Direction of force $\overrightarrow{F_{4}}$ is in the direction opposite to the direction of $\overrightarrow{F_{3}}$.
Since
$\overrightarrow{F_{3}}$ =- $\overrightarrow{F_{4}}$
Their lines of action coincide with each other, therefore these forces cancel each other.
Force $\overrightarrow{F_{1}}$ and $\overrightarrow{F_{2}}$ are equal in magnitude but their lines of action do not coincide with each other . Hence , these forces form a couple and tends to rotate the loop clockwise.
Magnitude of the torque acting on the current loop :
The end view of the loop
The moment arm of force $\overrightarrow{F_{1}}$ about point O ( passing through axis of rotation) is $\frac{b}{2} sin \theta$ and moment arm of force $\overrightarrow{F_{2}}$ about point O ( passing through axis of rotation) is $\frac{b}{2} sin \theta$
Magnitude of net torque on the loop about is given by
$\tau = \overrightarrow{F_{1}} \times \frac{b}{2} sin \theta + \overrightarrow{F_{2}} \times \frac{b}{2} sin \theta$
Since
$$\overrightarrow{F_{1}} = \overrightarrow{F_{2}} = BIl$$
$\tau = BIl \times \frac{b}{2} sin \theta + BIl \times \frac{b}{2} sin \theta$
$\tau = BI(lb)sin \theta$
Since lb= A
$\tau = BIAsin \theta$
In vector form ,
$\overrightarrow{\tau} = I (\overrightarrow{A} \times \overrightarrow{B})$
If the loop has N turns , then net torque acting on the loop is given by
$\tau_{N} = N.\tau = NIABsin \theta$
Special cases :
1. If the current carrying loop or coil is aligned in such a way that the plane of the loop or coil is parallel to the direction of the magnetic field , then $\theta = 90^{0}$
Hence , torque acting on the loop or coil is given by
$\tau$ = NIABsin90°
$\tau= NIAB$
Thus , maximum torque acts on the loop or coil . The magnetic field which is parallel to the plane of the loop or coil is called radial magnetic field.
2. If the current carrying loop or coil is aligned in such a way that the plane of the coil is perpendicular to the direction of magnetic field, then $\theta$=0°
Hence , torque acting on the loop or coil , is given by
$\tau = NIAB sin0^{0}$
$\tau = 0$
Thus , no torque acts on the loop and the loop will be in the position of equilibrium.