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 \Theta

But $lb$ = 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….

Torque Experienced by A Current Loop in Uniform Magnetic Field