Question

A rectangular coil with 20 windings carries a current of 2.00 mA flowing in the counterclockwise direction. It has two sides that are parallel to the \(y\) -axis and have length \(8.00 \mathrm{~cm}\) and two sides that are parallel to the \(x\) -axis and have length \(6.00 \mathrm{~cm} .\) A uniform magnetic field of \(50.0 \mu \mathrm{T}\) acts in the positive \(x\) -direction. What torque must be applied to the loop to hold it steady?

Short answer

Based on the given information, the torque that must be applied to hold the rectangular coil steady in the given uniform magnetic field is: $$\tau_{total} = 9.6\times10^{-7}\,\text{N}\cdot\text{m}$$

Step-by-step solution

Identify the relevant quantities

Given: - Number of windings: \(n = 20\) - Current: \(I = 2.00 \,\text{mA} = 2.00\times10^{-3}\,\text{A}\) - Length of sides parallel to the \(y\)-axis: \(l_y = 8.00 \,\text{cm} = 0.08 \,\text{m}\) - Length of sides parallel to the \(x\)-axis: \(l_x = 6.00 \,\text{cm} = 0.06 \,\text{m}\) - Uniform magnetic field strength: \(B = 50.0\,\mu\text{T} = 50.0\times10^{-6}\,\text{T}\)

Find the magnetic force on the sides parallel to the \(y\)-axis

Since the magnetic field is parallel to the \(x\)-axis, there will be no force exerted on the sides parallel to the \(y\)-axis (the force will be perpendicular to length and field). Therefore, we can move on to the next step.

Find the magnetic force on the sides parallel to the \(x\)-axis

The magnetic force on a current-carrying wire can be calculated using the formula: $$F = nIBL\sin\theta$$ As the magnetic field is parallel to the \(x\)-axis, it will be perpendicular to the sides of the loop parallel to the \(x\)-axis, making the angle \(\theta = 90^\circ\). Therefore, the magnetic force on these sides is: $$F_{x} = nIBl_y\sin(90^\circ) = 20 \times 2.00\times10^{-3} \text{A} \times 50.0\times10^{-6} \text{T} \times 0.08 \text{m}$$ $$F_{x} = 1.6\times10^{-5} \,\text{N}$$

Calculate the net torque on the loop

To find the net torque on the loop, which must be counteracted to hold the loop steady, we need to calculate the torque on both sides parallel to the \(x\)-axis and add them up. The torque is given by the cross product: $$\tau = r \times F_{x}$$ Since \(F_{x}\) acts equally on both sides, we can calculate the torque on one side and simply double it. The torque on one side is: $$\tau_{one-side} = \frac{l_x}{2} \times F_{x}$$ $$\tau_{one-side} = 0.5 \times 0.06\,\text{m} \times 1.6\times10^{-5}\,\text{N} = 4.8\times10^{-7} \,\text{N}\cdot\text{m}$$ Now, double the torque to account for both sides: $$\tau_{total} = 2 \times \tau_{one-side} = 2 \times 4.8\times10^{-7}\,\text{N}\cdot\text{m} = 9.6\times10^{-7} \,\text{N}\cdot\text{m}$$

Final answer

The torque that must be applied to hold the rectangular coil steady in the given uniform magnetic field is: $$\tau_{total} = 9.6\times10^{-7}\,\text{N}\cdot\text{m}$$

Key concepts

Current-Carrying Wire

When we speak of a current-carrying wire, we refer to the flow of electric charge through a conductor such as copper or aluminum. In this exercise, the wire is wound to form a rectangular coil. This coil consists of 20 loops, or windings, through which a current of 2.00 mA (milliamperes) is passed.

Electric current in a wire is measured in amperes (A), which denotes the flow of charge per unit time. With this specific problem, multiplying the current with the number of windings is essential since each loop contributes to the magnetic force. As a result, the entire coil experiences a magnetic force, which becomes significant when calculating torque. Understanding the current flow is key to predicting electromagnetic interactions such as magnetic force and torque.

Magnetic Field

The magnetic field plays a crucial role in the dynamics of the coil. In terms of physics, a magnetic field is a vector field that exerts force on moving electric charges, which in turn influences a current-carrying wire.

In our particular problem, the magnetic field is uniform and acts in the positive x-direction with a strength of 50.0 μT (microteslas). This means that the magnetic field's strength is constant and aligned along the x-axis. This alignment is crucial because it affects how the forces are applied to the coil. When the field is perpendicular to the current, it results in maximum force, which is core to the calculation of torque. Understanding the direction and magnitude of the magnetic field helps in determining how it interacts with the wire's current to produce rotational effects, or torque.

Rectangular Coil

A rectangular coil is essentially a loop of wire in a rectangular shape, often used to illustrate electromagnetic principles. In this exercise, our coil features two sides that are parallel to the y-axis and two sides parallel to the x-axis, with lengths of 8.00 cm and 6.00 cm respectively.

Rectangular coils are ideal for educational demonstrations because their geometry allows straightforward calculations and predictable behavior in magnetic fields. In this example, it's vital to recognize that the coil's sides parallel to the x-axis will encounter different forces compared to those parallel to the y-axis, primarily due to their orientation relative to the magnetic field. Understanding this geometry helps simplify complex electromagnetic calculations by focusing on the active components of the coil in the presence of the field.

Torque Calculation

Calculating torque in this scenario involves understanding how forces interact with the coil. Torque is a measure of the rotational force acting on an object and is calculated as the product of force and the distance from the pivot point.

In our situation, the torque on the coil arises from the magnetic force exerted on the sides of the coil that are parallel to the x-axis. The formula for calculating torque (τ) is given by \( au = r \times F\), where \(r\) is the distance from the pivot (in this case, half the side length of the rectangular coil), and \(F\) is the force calculated using \(F = nIBl_y\sin \theta\).
In this problem, because \( heta = 90^{\circ} \), the sine component equals 1, leading to the maximum possible force. Finally, since both sides parallel to x-axis contribute to torque, we double the calculated torque for one side to find the total, arriving at a torque value of \(9.6 \times 10^{-7} \,\text{N}\cdot\text{m}\). This value tells us the torque required to hold the coil steady in the magnetic field.