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Let us consider a current-carrying rectangular loop inside an electric motor.

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When the current is passed through this loop, a magnetic field is produced which exerts a torque on the loops, rotating the shaft.

Here, the magnetic field is uniform all around it.

A current-carrying loop of wire in the above arrangement is attached to a vertical rotating shaft that feels magnetic forces that produce a clockwise torque as viewed from above.

This is how electrical energy is transformed into mechanical work.

If you look at Fig.1, four wires are joined to form a loop. They are placed inside the magnetic field.

When the current is passed through this loop, the magnetic field of lines crosses the loop. It experiences a force, which in turn generates a torque due to the force as shown in Fig.2.

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Now, the loop will experience a force due to the magnetic field.

Here, the current is flowing from D to C and the magnetic field in the opposite direction.

To find the direction of the force in the wire CD.

Let’s Apply Fleming’s Right-Hand Rule

It states that if we stretch our index finger, middle finger and the thumb in such a way that they are mutually perpendicular to each other where the index finger indicates the direction of the magnetic field, middle finger, the direction of an induced current, while the thumb represents the direction of motion.

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As we align our fingers in this way, we would observe that the current and the magnetic field are in opposite directions.

So, they are parallel, i.e. Sin 0° = 0.

Therefore, no force is acting on the wire CD.

Similarly, if we look at wire AB, the direction of current is opposite to that in the wire CD, however in the same direction to that of the magnetic field.

Here also Sin 0° = 0, no force is acting on wire AB.

For wire CA,

The current is flowing in an upward direction, which means the electrons are flowing in the opposite direction.

Applying Fleming’s right-hand rule, the magnetic field is in a direction perpendicular to that of current. The force is acting inwards.

While for the wire BD, the direction of the current is downward and the direction of force is outward.

A torque is exerted on the loop about an axis, making the loop rotate.

Now, we’ll deduce the equation for the same.

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Here, we would define some quantities:

W = width of the loop

L = Height of the loop, which is the length of the wire feeling the force.

F = IBL

We took F = IBL because in maximum cases, F and B are perpendicular to each other and Sin 90° = 1.

So, an equation for torque:

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

d = W/2

As F = IBL

So, てleft (torque from the left) = (IBL) W/2 and てright = (IBL) W/2, and when the loop is in the middle, no torque will act on it.

てnet = てleft + 0 + てright = BIHW

∵ HW = A (area of the loop)

て = BIA

Here, area vector A points outward in the middle.

So, て = IABSinӨ

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

て = NIABSinӨ …(1)

From equation (1),

Here, NIA is called the magnetic moment.

So, torque on any current-carrying loop is the magnetic moment times the magnetic field.

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Force is acting outwards whether the current is flowing inside or outside the loop, this magnetic force will continue spinning the loop.

We know that the current loop when placed inside the magnetic field, it behaves like a magnetic dipole where it has a North and the South pole.

So, magnetic moment, M = IA.

Now, consider a rectangular loop placed inside the magnetic field.

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Case 1: When the angle between M and B is 0°

It means both M and B are pointing inwards.

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Fnet = FQR + FRS + FSP + FPQ

= Force on FQR and FSP on FRS and FPQ cancel out each other

So, Fnet = 0 (Always zero in loop on uniform magnetic field)

∵ て = F x 丄d

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

Case 2: When the angle between the vectors M and B = Ө

Wires RS and PQ are still perpendicular to the magnetic field while QR and SP are making an angle Ө with the magnetic field.

So, forces on the wires RS and PQ will be:

FPQ = FRS = IBL (Sin 90° =1) for each

∵ These two forces don’t have the same line of action, so they will rotate the loop about an axis and would hence generate the torque.

Fnet = zero.

However, て ≠ 0 due to horizontal forces not having the same line of action.

Hence, from the top-view of the loop,

てnet = F x 丄d + F x 丄d

= FBW/2 LSinӨ + FBW/2 LSinӨ

= FBWLSinӨ =FIABSinӨ (∵ M = IA)

It is the torque on the loop in a uniform magnetic field.

FAQ (Frequently Asked Questions)

Q1: How do you get Maximum Torque?

Ans: Since we know that, て = MBSinӨ. We can obtain maximum torque only when Ө = 90°. It is possible only when F and B are perpendicular to each other.

Q2: Is there a Net Force on the Loop of Wire From the Magnetic Field?

Ans: In a uniform magnetic field, F_{net} = 0, because the force acting on the left and right arm of the loop, i.e., F = IBL, would cancel out each other.

Q3: What is the Position of the Loop When the Current is Maximum?

Ans: The current is maximum only when the loop is at the beginning, half-way, and at the end of the region of the field.

Q4: Can a Torque be Balanced by a Single Force?

Ans: No, because torque can be balanced by two forces. If one force rotates the loop in a clockwise direction, then another force would rotate it in an anticlockwise direction. Thus, no toque will be acting on the body.