## Find torque on current loop placed in a Magnetic Field.

*When a current carrying loop or coil is placed in a magnetic field, it will experience a magnetic force which produces a torque in the current loop.*

- The magnitude of the magnetic force developed on the current loop is according to the Lorentz Law.
- The direction of the magnetic force and so the torque developed on the current loop is according to the Fleming’s Left Hand Rule.
- Most of the electrical instruments like
*Galvanometer*, Voltmeter, Ammeter etc. work on this principle. - Therefore, measurement of torque acting on a current loop is most important for accuracy of electrical instruments.

### Expression for Torque on Current loop

Consider about a rectangular loop ABCD of a conducting material placed in a magnetic field as shown in figure. Let –

- Length of the loop is ( l ) .
- Breadth of the loop is ( b ) .
- Loop is pivoted at the mid points of arms AD \ \text {and} \ BC .
- Intensity of uniform
*magnetic field*( \vec {B} ) . - Current flowing in clockwise direction in the loop is ( I ) .
- The angle between the normal ( \hat {n} ) to the plane of the loop and the direction of magnetic field is ( \theta ) .

From *Lorentz force*, we know that net force acting on a conductor carrying current will be –

\vec {F} = I \left ( \vec {l} \times \vec {B} \right )

#### Force acting on arms AB & CD

Magnetic force acting on the arm AB of the loop will be –

\vec {F_1} = B I l ………. (1)

According to *Fleming’s left hand rule*, direction of force ( \vec F_1 ) is perpendicular to the length of the arm AB and directed penetrating inside the plane of paper.

Magnetic force ( F_2 ) acting on the arm CD of the loop will be –

\vec {F_2} = B I l ………. (2)

According to *Fleming’s left hand rule*, direction of force ( \vec {F_2} ) is perpendicular to the length of arm CD and is directed coming up from the plane of the paper. Therefore, direction of ( \vec F_2 ) is opposite to the direction of ( \vec F_1 ) .

#### Force acting on arms BC & AD

*Magnetic force* acting on the arm BC of the loop will be –

\vec {F_3} = B I b ………. (3)

According to *Fleming’s left hand rule*, direction of force ( \vec F_3 ) is perpendicular to the length of the arm BC and directed emerging up from the plane of paper. Magnetic force acting on the arm AD of the loop will be –

\vec {F_4} = B I b ………. (4)

According to *Fleming’s left hand rule*, direction of force ( \vec F_4 ) is perpendicular to the length of the arm AD and directed penetrating inside the plane of paper. Therefore, direction of ( \vec F_4 ) is opposite to the direction of ( \vec F_3 ) .

### Torque on loop

Forces ( \vec {F_1} ), \ ( \vec {F_2} ), \ ( \vec {F_3} ), \ \text {and} \ ( \vec {F_4} ) , all are acting at the mid points of their respective arms.

- Forces ( \vec {F_3} ) and ( \vec {F_4} ) are equal, opposite in direction and parallel. But their line of action are passing through the pivot point. Hence, they will cancel to each other.
- Forces ( \vec {F_1} ) and ( \vec {F_2} ) are equal, opposite in direction and their line of action is different. Being opposite in direction, they will form a
*couple*and try to rotate the loop in clockwise direction as shown in figure.

The magnitude of the *torque* ( \tau ) due to *forces* ( \vec {F_1} ) and ( \vec {F_2} ) will be –

\text {Torque} = \text {Force} \ \times \ \text {Moment arm}

Therefore, \quad \tau = F_1 \ \times \ DN

= B I l \ \times \ DN

= I \ \times \ \left ( \vec {l} \times \vec {B} \right ) \ \times \ DN

Vectors ( \vec {l} ) \ \text {and} \ ( \vec {B} ) are mutually perpendicular to each other and ( DN = b \sin \theta )

Therefore, \quad \tau = I ( l B \sin 90 \degree ) \ \times \ ( b \sin \theta )

= I ( l b ) B \sin \theta = I A B \sin \theta ………. (3)

### Torque on a coil

Now consider that, the loop has ( N ) number of turns. Then net torque acting on the loop will be –

\tau_N = N I A B \sin \theta ……….. (4)

If the plane of loop makes an angle ( \alpha ) with the magnetic field ( \vec {B} ) . Then, \quad ( \theta + \alpha ) = 90 \degree

Hence, equation (3) becomes –

\tau_N = N I B \sin \left ( 90 \degree - \alpha \right ) = N I B \cos \theta

## Radial Magnetic Field

Torque developed in a current loop or coil is given by the equation –

\tau_N = N I A B \sin \theta …….. (4)

If the value of ( \sin \theta ) attains the maximum value then maximum torque will develop in the loop. But maximum value of ( \sin \theta = 1 ) .

Therefore, when maximum torque is acting on the loop, then ( \theta = 90 \degree ) . Therefore, maximum value of *torque* will be –

\tau_{Max} = N I A B \sin 90 \degree = N I A B …….. (5)

*Therefore, magnetic field which is parallel to the plane of the loop or coil is called radial magnetic field. This magnetic field develops maximum torque in the loop or coil.*

## Dipole Moment of current loop

Torque developed in a current loop is given by the equation –

\tau = I A B \sin \theta ………. (3)

*The product ( I A ) in equation (3) is known as magnetic dipole moment of the current loop. It is denoted by ( m ) .*

Therefore, \quad \text {Magnetic dipole moment of current loop} ( m ) = I A

If the loop has* * ( N ) number of turns,

Then,* * \quad m = N I A