## Torque Experienced by A Current Loop in Uniform Magnetic Field

Consider a rectangular loop PQRS carrying current I placed within a uniform magnetic field of strength \overrightarrow{B}.

**IMAGE – 1**

Let,

l = Length of loop = PQ = RS

b = Breadth of loop = QR = SP

Let at any instant the plane of loop makes angle \Theta with the direction of the magnetic field.

Since current carrying conductor placed in a magnetic field experienced force.

Now the forces on the side SP is given by

\overrightarrow{F_{4}} = I (\overrightarrow{b}\times\overrightarrow{B})…….(1)According to **Right Hand Screw Rule**, The direction of \overrightarrow{F_{4}} is vertically upwards in the plane of coil.

Hence, force on side QR is given by

\overrightarrow{F_{2}} = I (\overrightarrow{b}\times\overrightarrow{B})…….(2)The direction of force \overrightarrow{F_{2}} is vertically downwards in the plane of coil.

Since \overrightarrow{F_{2}} and \overrightarrow{F_{4}} are equal in magnitude and act in opposite direction. So that they cancel out each other.

Therefore net force by the torque on the loop is **ZERO**.

Now, Force on side PQ is given by

\overrightarrow{F_{1}} = I (\overrightarrow{l}\times\overrightarrow{B}) \Rightarrow \left |\overrightarrow{F_{1}} \right | = IlB\sin\Theta.\widehat{n} \Rightarrow \left |\overrightarrow{F_{1}} \right | = IlB\sin 90^{0}.\left |\widehat{n}\right | {F_{1}} = IlB.......(3)Similarly, Force on side RS is given by

\overrightarrow{F_{3}} = I (\overrightarrow{l}\times\overrightarrow{B}) \Rightarrow \left |\overrightarrow{F_{3}} \right | = IlB\sin\Theta.\widehat{n} \Rightarrow \left |\overrightarrow{F_{3}} \right | = IlB\sin 90^{0}.\left |\widehat{n}\right | {F_{3}} = IlB.......(4)Since, the forces F_{1} and F_{3} are equal in magnitude and opposite in direction. But they act on the different line of action.

So, they constitute a couple, which rotates the loop in the anti-clockwise direction about the vertical axis.

**IMAGE – 2**

Now, torque on the loop is given by

\tau = either force x perpendicular distance between two forces

\Rightarrow \tau = (IlB) \times b \cos \Theta \Rightarrow \tau = I(lb) B \cos \ThetaBut lb = A = Area of the loop

\Rightarrow \tau = IAB \cos \Theta.......(5)If, the loop has “N” number of turns, then

\tau = NIAB \cos \Theta.......(6)**Special Cases**

#### Case – 1

When \Theta = 0^{0}, the plane of the coil is parallel to the direction of the magnetic field.

\tau = NIAB \cos 0^{0}\tau = NIBA……. (Maximum value)

#### Case – 2

When \Theta = 90^{0}, the plane of the coil is perpendicualr too the direction of the magnetic field.

\tau = NIAB \cos 90^{0} \tau = NIAB \cos 90^{0} \tau = 0### Conclusion

Hence, when the plane of the coil is parallel to the direction of magnetic field, torque will be maximum.

*Thanks for Today And Meet You On The Next….*

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