Torque Current Loop: Torque in Magnetic Field & Solved Examples

Torque is a kind of twisting force. It is mainly meant to provide rotational movement to any object. For instance, a door rotates around its attached hinge or a spinning wheel which moves on its own axis; all these rotations work on the concept of torque. Torque is given by the formula,

T = F × r

Where,

F → Force

r → Distance between the centre of the axis of rotation and to the point of force.

However, in the case of the torque current loop, the magnetic field is used to accelerate the force on the rectangular loop. In this article, we will discuss the meaning and the details of the torque current loop along with the diagrams and other related concepts.

Key Terms: Torque, Current, Magnetic Dipole, Magnetic Moment, Rotational Motion, Force, Area, Rotation, Current loop

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What is Torque?

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Torque is a term which is used to define a kind of force that supports twisting movement causing rotation. This rotation takes place on the fixed axis of rotation. ‘T’ is a symbol which is used to denote torque. Mathematically,

T = F × r

Here,

‘F’ represents the amount of force applied and ‘r’ represents the distance between the centre of the axis of rotation and to the point where force is applied.

Torque

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Torque on Current Loop - Magnetic Dipole 

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A current loop starts behaving like a magnetic dipole when it is surrounded by a magnetic field. In that case, the magnetic dipole has certain limitations and can look towards only either the north or the south poles. This choice of a pair of poles remains as soon as the dimensions of the source are reduced to zero but the magnetic moment is kept constant. Therefore, through this, now we shall show a uniform magnetic field and a rectangular loop that has a steady current passing through it. This passing current shall further experience a kind of rotational force known as torque. However, there shall not be any net force but the behaviour is believed to be similar to that of an electric dipole in a uniform electric field. Now, let us see two different cases.

Torque on Current Loop - Magnetic Dipole

Case 1: In this case, let us assume a plane of the loop as uniform magnetic field B. This rectangular loop is placed such that the uniform magnetic field B exerts no force on the arms of the loops, i.e. PS and QR. However, on the other hand, the plane lies perpendicular to the arm PQ of the loop and exerts a force F₁ on it. And this force F₁ is directly exerted into the plane of the loop. The magnitude of the same can be represented as the following,

F₁ = IzB

Simultaneously, it also exerts a force F₂ on the arm RS and now, this force F₂ is directly exerted outwards the plane of the loop. Hence, the next equation can be written as,

F₂ = IzB = F₁

From the above equation, as we can see that the forces F₁ and F₂ neutralise each other, therefore, the net force exerted on the planes of the loop is zero. As a result, the loop shall experience a torque which creates a rotation on the loop in an anti-clockwise direction. Therefore,

T = F₁ (y / 2) + F₂ (y / 2)

= IzB (y / 2) + IzB (y / 2)

= I (y × z) B

T = IAB ………….(1)

Here, A = y × z is the area of the rectangle.

Case 2: In this case, let us assume that the rectangular loop has a plane which makes an angle between the field and the normal to the coil and that is angle Θ. From the above diagram, we could see that the forces on the arms QR and SP are equal and opposite. And simultaneously, this force acts along the axis of the coil, which connects the centres of mass of QR and SP. Further, since both the forces lie in the same line to that of the axis, that is, they are collinear along the axis, therefore, both shall cancel each other or neutralise their effects. Hence, no net force or torque is there. The two arms of the loop, PQ and RS have the force as F₁ and F₂ which are also equal and opposite, with magnitude as,

F₁ = F₂ = IzB

So, now the magnitude of the torque on the loop is following,

τ = F₁ (y / 2) sinθ + F₂ (y / 2) sinθ

τ = I (y × z) B sinθ

τ = IAB sinθ …...(2)

Now, from the equations (1) and (2), the torques can be expressed as the vector product of the magnetic moment of the coil and the magnetic field. Therefore, the magnetic moment of the current loop can be as the following,

m = IA

here,

A represents the direction of the area vector and θ represents the angle between m and B.

τ = m×B

here,

m represents the magnetic moment and

B represents the uniform magnetic field.

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Torque on a Current Loop Equation 

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Generally, in terms of the equation,

F = IBL

where I represents the width of the loop, L represents the Height of the loop (the length of the wire feeling the force), and B represents the arms.

So, an equation for torque shall be as the following,

T = Fd

(d = distance of the wire from the axis of rotation)

d = W / 2

As,

F = IBL,

So,

T_Left = (IBL) W/2

T_Right = (IBL) W/2

and when the loop is in the middle, no torque will act on it.

Therefore,

T_Net = T_Left + 0 + T_Right = BIA

A is the area of the loop

Here, area vector A points outward in the middle.

So,

T = IABSinθ

If there are N number of loops inside the field, we have:

T = NIABSinθ……(3)

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Torque on Current Loop Due to the Magnetic Moment 

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From equation (3), NIA is the magnetic moment. Therefore, torque on any current-carrying loop is the magnetic moment times the magnetic field.

Torque on Current Loop Due to the Magnetic Moment

Case 1: Here, let us consider a situation when the angle between two arms M and B is 0°

It means both M and B are pointing inwards.

F_net = FQR + FRS + FSP + FPQ

F_Net = Force on FQR and FSP will cancel the force on FRS and FPQ

So,

F_net = 0 (Always zero in loop on uniform magnetic field)

Torque = F x ⊥d

Here, Torque_net = 0 as net forces are zero, won’t rotate the loop.

Case 2: Here, the angle between the vectors M and B = θ.

And the wires RS and PQ are still perpendicular to the magnetic field while QR and SP are making an angle θ with the magnetic field. So, forces on the wires RS and PQ will be:

FPQ = FRS = IBL (Sin 90° =1) for each

Since, these two acting forces don’t have the same line of action, they will rotate the loop about an axis and hence generate the torque.

Fnet = zero.

However, Torque ≠ 0 due to horizontal forces not having the same line of action. Hence, from the top-view of the loop,

Torque_net = F x ⊥d + F x ⊥d

= FBW / 2 LSinθ + FBW / 2 LSinθ

= FBWLSinθ = FIABSinθ (Torque × M = IA)

Hence, the torque on the loop in a uniform magnetic field shall be Torque_net = MBSinθ.

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Things to Remember

• The concept of torque or torque originated in the US with the work of Archimedes on levers.
• A torque is a force which is known to create rotation on a body experiencing the force. The fundamental law of the lever is used for the measurement of the torque.
• The units which are used for expressing the torque are in. lbs. (inch pounds), in. ozs. (inch ounces), ft. lbs. (foot pounds), Nm (Newton metre) and so on.
• Torque helps in determining the direction and the magnitude of the rotational force acting on a current loop which further guides in evaluating other outcomes.
• Torque also helps in determining the acceleration or speed as happens in the case of a car where the speed generated by the piston produces the torque.

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