Lesson Explainer: The Torque on a Current-Carrying Rectangular Loop of Wire in a Magnetic Field | Nagwa Lesson Explainer: The Torque on a Current-Carrying Rectangular Loop of Wire in a Magnetic Field | Nagwa

Lesson Explainer: The Torque on a Current-Carrying Rectangular Loop of Wire in a Magnetic Field Physics • Third Year of Secondary School

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In this explainer, we will learn how to calculate the torque on a current-carrying rectangular loop of wire in a uniform magnetic field.

Let us start by considering a rectangular loop of wire in a horizontal plane, with current 𝐼 through it. A magnetic field, with strength 𝐡, exists, also in the horizontal plane. The following diagram shows this loop of wire.

Let us look at the rectangular loop in more detail. The following diagram shows each side of the loop labeled numerically.

As we can see, the rectangular loop has two pieces of wire providing current from the middle of the base of the rectangle. The gap between these two pieces of wire is π‘‘οŠ©, which is very small: π‘‘β‰ˆ0.

We will also say that the length of the two pieces of wire supplying current is small. This means we only need to consider the rectangular section (from 2 to 5) of wire.

Recall that the force, 𝐹, on a wire of length 𝑙, carrying a current 𝐼, perpendicular to a magnetic field 𝐡, is equal to 𝐹=𝐡𝐼𝑙.

If we look at each section of wire, we will see that there is only force acting on sections 3 and 5. Taking vertically upward as positive, the force acting on each of these is equal to 𝐹=βˆ’π΅πΌπ‘‘,𝐹=𝐡𝐼𝑑.

So, the two sections of wire have forces of equal magnitude acting on them, but in opposite directions. The direction of the force can be found using Fleming’s left-hand rule.

Fleming’s left-hand rule is a convention used to find the direction of the force acting on a wire from the direction of the magnetic field the wire is in and the direction of current through the wire. The following diagram shows Fleming’s left-hand rule and the axes made from the magnetic field, force, and current.

In Fleming’s left-hand rule, the thumb points in the direction of the force, the index finger points in the direction of the magnetic field, and the middle finger points in the direction of the current through the wire.

We can use Fleming’s left-hand rule to find the direction of the force acting on sections 3 and 5 of the loop of wire. The following diagram illustrates this.

As we can see, the force acting on section 3 of the wire acts downward, and the force acting on section 5 of the wire acts upward. The following diagram shows the forces acting on the rectangular loop of wire; 𝐹 acts downward, and 𝐹 acts upward.

Now, let us consider an axis through the center of the rectangle, represented by the dashed line in the following diagram.

The torque around this axis can be calculated by multiplying each force by the perpendicular distance to the axis: 𝜏=πΉο€½βˆ’π‘‘2+𝐹𝑑2.

Substituting in expressions for 𝐹 and 𝐹, we get 𝜏=βˆ’π΅πΌπ‘‘ο€½βˆ’π‘‘2+𝐡𝐼𝑑𝑑2𝜏=𝐡𝐼𝑑𝑑.

It can be seen that this contains π‘‘οŠ§ multiplied by π‘‘οŠ¨, which is the area of the rectangle, 𝐴. The torque can, therefore, be written as 𝜏=𝐡𝐼𝐴.

We can also calculate the magnetic dipole moment of the loop of wire. The magnetic dipole moment is defined as the torque on the wire divided by the magnetic field strength: π‘š=𝜏𝐡.

Definition: Magnetic Dipole Moment

The magnetic dipole moment of a loop of wire carrying a current in a magnetic field is defined as the torque acting on the loop of wire divided by the magnetic field strength: π‘š=𝜏𝐡.

For the rectangular wire, we can substitute the expression for torque we calculated previously: π‘š=π΅πΌπ΄π΅π‘š=𝐼𝐴.

We will now work through an example question calculating the torque and magnetic dipole moment of a rectangular wire.

Example 1: Calculating the Torque and Magnetic Dipole Moment of a Rectangular Loop of Wire Carrying a Current in a Magnetic Field

The diagram shows a rectangular loop of current-carrying wire between the poles of a magnet. The sides of the loop parallel to line π‘‘οŠ§ are parallel to the magnetic field, and the sides of the loop parallel to line π‘‘οŠ¨ are perpendicular to the magnetic field. The current in the loop is 350 mA, and the magnetic field strength is 0.12 T. The length of 𝑑=0.025m and the length of 𝑑=0.015m.

  1. Find the torque acting on the loop to the nearest micronewton-metre.
  2. Find the magnetic dipole moment of the loop to the nearest micronewton-metre per tesla.

Answer

Part 1

The torque acting on the loop of wire, 𝜏, can be found using the formula 𝜏=𝐡𝐼𝐴, where 𝐡 is the magnetic field strength, 𝐼 is the current in the loop of wire, and 𝐴 is the area of the loop of wire.

First, we will convert the current in the wire to SI units of amperes: 𝐼=350𝐼=0.35.mAA

The loop of wire is rectangular, so the area of the loop of wire is simply 𝐴=𝑑𝑑, where 𝑑=0.025m and 𝑑=0.015m, as stated in the question. This gives an area of 𝐴=3.75Γ—10.οŠͺm

Substituting the values of 𝐼 and 𝐴 into the equation for torque, along with the magnetic field strength 𝐡=0.12T, gives 𝜏=0.12Γ—0.35Γ—3.75Γ—10𝜏=1.575Γ—10β‹….TAmNmοŠͺ

In micronewton-metres, this is equal to 𝜏=15.75β‹….ΞΌNm

To the nearest micronewton-metre, the torque acting on the loop of wire is equal to 16 ΞΌNβ‹…m.

Part 2

The magnetic dipole moment, π‘šοŒ½, of a loop of wire carrying a current in a magnetic field is equal to the torque acting on the loop of wire, 𝜏, divided by the magnetic field strength, 𝐡: π‘š=𝜏𝐡.

Substituting in the previously calculated value of 𝜏=1.575Γ—10β‹…οŠ±οŠ«Nm and the given value of 𝐡=0.12T gives π‘š=1.575Γ—10β‹…0.12π‘š=1.313Γ—10β‹…/.οŠͺNmTNmT

In micronewton-metres per tesla, this is π‘š=131.3β‹…/.μNmT

The magnetic dipole moment of the loop to the nearest micronewton-metre per tesla is equal to 131 ΞΌNβ‹…m/T.

Now, we will consider what will happen if there are multiple loops, or turns, of rectangular wire. The following diagram shows the same rectangular loop as before, except there are now many turns to the wire.

Each piece of wire contributes the same amount of force. If there are 𝑁 wires, the torque on the axis is equal to 𝜏=𝐡𝐼𝐴𝑁.

We will now work through an example question where the rectangular loop of wire has multiple turns.

Example 2: Calculating the Torque Acting on a Rectangular Loop of Wire with Multiple Turns Carrying a Current in a Magnetic Field

The diagram shows a rectangular conducting coil with 3 turns that is in a magnetic field. There is a current of 8.5 A in the coil. The sides of the loop parallel to line π‘‘οŠ§ are parallel to the magnetic field, and the sides of the loop parallel to line π‘‘οŠ¨ are perpendicular to the magnetic field. The length of 𝑑=0.035m and the length of 𝑑=0.025m. The torque on the loop is 15 mNβ‹…m. Find the magnitude of the magnetic field to the nearest millitesla.

Answer

The formula to calculate the torque on a rectangular loop of wire with multiple turns carrying a current in a magnetic field is 𝜏=𝐡𝐼𝐴𝑁.

We can rearrange this equation for the magnetic field strength, 𝐡: 𝐡=πœπΌπ΄π‘.

First, the area of the rectangular loop of wire, 𝐴, can be calculated by multiplying π‘‘οŠ§ and π‘‘οŠ¨: 𝐴=𝑑×𝑑𝐴=0.035Γ—0.025𝐴=0.000875.mmm

The torque on the loop can be converted to SI units of newton-metres: 𝜏=15β‹…=0.015β‹….mNmNm

These values can be substituted into the equation for magnetic field strength, along with the given values of 𝐼=8.5A and 𝑁=3: 𝐡=0.015β‹…8.5Γ—0.000875Γ—3𝐡=0.672.NmAmT

To the nearest millitesla, this is 𝐡=672.mT

Now, let’s consider what happens when the loop of wire has rotated about the axis. We will measure the angle, πœƒ, from the direction perpendicular to the loop of wire (the direction normal to the rectangle) to the magnetic field.

The forces acting on the wire are now slightly different.

The forces acting on sections 3 and 5 still have the same magnitude and are still vertical. However, sections 2 and 4 are now no longer parallel to the magnetic field. This means that there will be some force perpendicular to the direction of the magnetic field and the current direction in those wires.

This means that the force on these sections of wire is along the axis of rotation, in opposite directions, so there is no torque about the axis from these forces.

We can label these forces on the diagram of the wire, as shown.

The torque on the wire can be calculated by multiplying each force by its perpendicular distance to the axis. For the wire rotated by an angle πœƒ, the perpendicular distance, π‘₯, from section 3 to the axis is π‘₯=βˆ’π‘‘2πœƒ.sin

Similarly, for the perpendicular distance, π‘₯, from section 5 to the axis: π‘₯=𝑑2πœƒ.sin

We can use these values to calculate the torque on the axis: 𝜏=𝐹π‘₯+𝐹π‘₯.

Substituting in the expressions for π‘₯ and π‘₯, this is equal to 𝜏=𝐹𝑑2πœƒο‰+πΉο€½βˆ’π‘‘2πœƒο‰.sinsin

Substituting in known expressions 𝐹=βˆ’π΅πΌπ‘‘οŠ©οŠ§ and 𝐹=π΅πΌπ‘‘οŠ«οŠ§, this is equal to 𝜏=βˆ’π΅πΌπ‘‘ο€½βˆ’π‘‘2πœƒο‰βˆ’π΅πΌπ‘‘ο€½π‘‘2πœƒο‰πœ=π΅πΌπ‘‘π‘‘πœƒ.sinsinsin

Again, this can be written in terms of the area of the rectangle, 𝐴: 𝜏=π΅πΌπ΄πœƒ.sin

When there are 𝑁 turns of wire, the torque is equal to 𝜏=π΅πΌπ΄π‘πœƒ.sin

Definition: The Torque on a Current-Carrying Rectangular Loop of Wire in a Magnetic Field

The torque, 𝜏, on a rectangular loop of wire with area 𝐴 and number of loops 𝑁, carrying a current 𝐼 in a magnetic field of strength 𝐡, at an angle πœƒ from the perpendicular to the magnetic field is 𝜏=π΅πΌπ΄π‘πœƒ.sin

We will now work through an example question where the rectangular loop of wire is at an angle to the magnetic field.

Example 3: Calculating the Torque Acting on a Rectangular Loop of Wire Carrying a Current in a Magnetic Field at an Angle

The diagram shows a rectangular loop of current-carrying wire between the poles of a magnet. The sections of the loop ab and dc are perpendicular to the magnetic field. The diagonal lines bc and ad are aligned at an angle πœƒ=33∘ from the direction of the magnetic field. The current in the loop is 1.75 A, and the magnetic field strength is 0.15 T. Length acm=0.065 and length abm=0.045. Find the torque acting on the loop to the nearest micronewton-metre.

Answer

This question asks us to calculate the torque acting on a rectangular loop of wire carrying a current in a magnetic field at an angle.

First, we can calculate the area, 𝐴, of the rectangle by multiplying the lengths of the two sides, ab and ac: 𝐴=×𝐴=0.045Γ—0.065𝐴=0.0029.abacmmm

Next, we can calculate the angle, πœ™, that the normal of the rectangle makes with the magnetic field. The angle given in the question, πœƒ, is the angle the rectangle makes with the magnetic field, so πœ™=90βˆ’πœƒπœ™=90βˆ’33πœ™=57.∘

The torque on the loop of wire can then be calculated with the formula 𝜏=π΅πΌπ΄πœ™,sin where 𝐡=0.15T and 𝐼=1.75A, as stated in the question.

So, 𝜏=0.15Γ—1.75Γ—0.0029Γ—57𝜏=0.000644β‹….TAmsinNm

In micronewton-metres, this is 𝜏=644β‹….ΞΌNm

We will now work through a question looking at how the angle of the the loop of wire affects the torque acting on the loop.

Example 4: The Effect of Angle on Torque Acting on a Rectangular Loop of Wire Carrying a Current in a Magnetic Field

The diagram shows a rectangular loop of current-carrying wire between the poles of a magnet. The longer sides of the loop are initially parallel to the magnetic field, and the shorter sides of the loop are initially perpendicular to the magnetic field. The loop then rotates through 90∘ so that all its sides are perpendicular to the magnetic field. Which of the lines on the graph correctly represents the change in the torque acting on the loop as the angle its longest sides make with the magnetic field direction varies from 0∘ to 90∘?

  1. Blue
  2. Green
  3. Red
  4. Orange
  5. None of these lines

Answer

In this question, we must consider the torque on a loop of wire as the angle of the loop changes.

Recall that the torque, 𝜏, on a rectangular loop of wire carrying a current in a magnetic field is equal to 𝜏=π΅πΌπ΄πœ™,sin where 𝐡 is the magnetic field strength, 𝐼 is the current the wire is carrying, 𝐴 is the area of the rectangular loop, and πœ™ is the angle the normal to the rectangle makes with the magnetic field.

In this question, however, the angle is measured from the sides of the rectangle to the magnetic field. This angle is written as πœƒ: πœ™=90βˆ’πœƒ.

So, the equation for torque can be written as 𝜏=𝐡𝐼𝐴(90βˆ’πœƒ).sin

Noting that sincos(90βˆ’πœƒ)=πœƒ, this can be written as 𝜏=π΅πΌπ΄πœƒ.cos

So, when πœƒ=0, the torque will be at its maximum, and when πœƒ=90∘, the torque will be zero. The line on the graph that matches this is the red line.

The answer is option C, β€œRed.”

Finally, we will work through an example question combining everything we have learned.

Example 5: Calculating the Torque Acting on a Rectangular Loop of Wire Carrying a Current in a Magnetic Field from Magnetic Dipole Moment

The diagram shows a rectangular loop of current-carrying wire between the poles of a magnet that produces a field with a magnitude of 250 mT. The longer sides of the loop are initially parallel to the magnetic field, and the shorter sides of the loop are initially perpendicular to the magnetic field. The loop has a magnetic dipole moment of 500 ΞΌNβ‹…m/T. The loop is then rotated by an external torque through 90∘ so that all its sides are perpendicular to the magnetic field.

  1. How much does the torque on the loop change by due to its rotation? Answer to the nearest micronewton-metre.
  2. As the loop increases its rotation angle to values greater than 90∘ but less than 180∘, how does the direction of the applied torque on the loop compare to the direction of the magnetic torque acting on it?
    1. The direction of the applied torque on the loop is opposite to the direction of the magnetic torque on the loop.
    2. The direction of the applied torque on the loop is the same as the direction of the magnetic torque on the loop.

Answer

Part 1

This part of the question asks us to calculate the torque on the loop of wire before and after a 90∘ rotation.

Recall that the torque, 𝜏, acting on a loop of wire is equal to 𝜏=π΅πΌπ΄π‘πœƒ,sin where the magnetic field strength 𝐡=250mT, 𝐼 is the current the wire is carrying, 𝐴 is the area of the rectangular loop of wire, 𝑁 is the number of turns of wire, and πœƒ is the angle the normal of the loop makes with the magnetic field.

Before the rotation, the normal of the loop of wire is perpendicular to the magnetic field, so πœƒ=90∘. After the loop has rotated by 90∘, the normal of the loop of wire will be parallel to the magnetic field, so the torque at this point is zero.

The change in torque due to the rotation is, therefore, just equal to the torque before the rotation.

In this case, we are given the magnetic dipole moment of the loop before the rotation and the magnetic field strength. Recall that the magnetic dipole moment, π‘šοŒ½, is related to the torque on the loop of wire, 𝜏, and the magnetic field strength, 𝐡, by the following equation: π‘š=𝜏𝐡.

We can rearrange this for the torque on the loop of wire: 𝜏=π‘šπ΅.

We are given values of π‘š=500β‹…/μNmT and 𝐡=250mT. Converting these to SI units, we get π‘š=500β‹…/π‘š=0.0005β‹…/,𝐡=250𝐡=0.25.μNmTNmTmTT

We can now calculate the torque on the loop of wire: 𝜏=π‘šπ΅πœ=0.0005β‹…/Γ—0.25𝜏=0.000125β‹….NmTTNm

In micronewton-metres, this is 𝜏=125β‹….ΞΌNm

Part 2

This part of the question asks us to consider what happens to the torque acting on the loop of wire when it rotates more than 90∘ from its initial angle.

We can draw a diagram of this configuration.

As seen, there is a force acting downward on the right-hand section of the loop, and there is a force acting upward on the left-hand section of the loop.

This means that there will be a magnetic torque acting clockwise on the loop when the loop has been rotated between 90∘ and 180∘.

The applied torque to the loop to rotate it from 90∘ to 180∘ is also clockwise.

This means that the direction of the applied torque on the loop is the same as the direction of the magnetic torque on the loop. The answer is option B.

We can summarize what we have learned in this explainer in the following key points.

Key Points

  • The magnetic dipole moment, π‘šοŒ½, of a loop of wire carrying a current in a magnetic field is equal to the torque acting on the loop of wire, 𝜏, divided by the magnetic field strength the wire is in, 𝐡: π‘š=𝜏𝐡.
  • The torque, 𝜏, acting on a rectangular loop of wire with area 𝐴, carrying a current, 𝐼, in a magnetic field of magnitude 𝐡 when the loop of wire is parallel to the magnetic field is equal to 𝜏=𝐡𝐼𝐴.
  • When the rectangular loop is comprised of 𝑁 turns of wire, the torque acting on the loop is equal to 𝜏=𝐡𝐼𝐴𝑁.
  • If the normal to the rectangular loop is at an angle πœƒ to the magnetic field, the torque acting on the loop is equal to 𝜏=π΅πΌπ΄π‘πœƒ.sin

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