A 50-turn rectangular coil of wire of dimensions \(10.0 \mathrm{~cm}\) by \(20.0 \mathrm{~cm}\) lies in a horizontal plane, as shown in the figure. The axis of rotation of the coil is aligned north and south. It carries a current \(i=1.00 \mathrm{~A}\), and is in a magnetic field pointing from west to east. A mass of \(50.0 \mathrm{~g}\) hangs from one side of the loop. Determine the strength the magnetic field has to have to keep the loop in the horizontal orientation.

Short Answer

Expert verified
Answer: The required strength of the magnetic field is approximately 0.0245 T.

Step by step solution

01

Calculate the magnetic torque

The magnetic torque acting on the rectangular coil can be found using the formula: $$\tau_{m} = NiAB\sin\theta,$$ where \(\tau_{m}\) is the magnetic torque, \(N\) is the number of turns, \(i\) is the current, \(A\) is the area of the coil, \(B\) is the magnetic field strength, and \(\theta\) is the angle between the magnetic field and the normal to the plane of the coil. In this problem, the magnetic field is aligned with the plane of the coil, so the angle between them is 0 degrees, thus, \(\sin\theta = \sin0°= 0.\) Since \(\tau_{m} = 0,\) the magnetic torque cannot solely maintain the horizontal orientation.
02

Calculate the gravitational torque

Let's now calculate the gravitational torque acting on the mass hanging from one side of the loop. The gravitational torque can be found using the formula: $$\tau_{g} = mgL,$$ where \(\tau_{g}\) is the gravitational torque, \(m\) is the mass, \(g\) is the acceleration due to gravity, and \(L\) is the length from the axis of rotation to the point where the mass is hanging (half the length of the shorter side of the rectangular coil). Express \(m\) in kg and substitute the given values and \(g=9.81 \mathrm{~m/s^2}\) to find \(\tau_{g}\): $$\tau_{g} = (0.050 \mathrm{~kg})(9.81 \mathrm{~m/s^2})(0.1/2\mathrm{~m}) = 0.024525\mathrm{~N \cdot m}$$
03

Balance torques and solve for the magnetic field strength

Since the magnetic torque is 0, we need a torque that will balance the gravitational torque to keep the loop in a horizontal orientation. This can be achieved by having another current carrying loop on the opposite side of the coil (mirror image). The magnitude of the torque in this loop will be equal and opposite to that of \(\tau_{g}\). Therefore, we can write the following equation: $$NiAB = \tau_{g}.$$ Now, rearrange the equation to solve for the magnetic field strength \(B\): $$B = \frac{\tau_{g}}{NiA}.$$ Substitute the known values to find B: $$B = \frac{0.024525\mathrm{~N \cdot m}}{(50\ \mathrm{turns})(1.00\ \mathrm{A})(0.1\mathrm{~m} \times 0.2\mathrm{~m})} \approx 0.0245\ \mathrm{T},$$ So, the required strength of the magnetic field to keep the loop in a horizontal orientation is approximately \(0.0245\ \mathrm{T}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Magnetic Torque
Magnetic torque is a force that arises within a magnetic field when a current-carrying conductor, such as a coil of wire, is placed within it. This torque tends to rotate the conductor around its axis. Mathematically, it is expressed as \(\tau_m = NiAB\sin\theta\), where \(N\) represents the number of turns in the coil, \(i\) the current flowing through it, \(A\) the area of the coil, \(B\) the magnetic field strength, and \(\theta\) the angle between the field and the normal to the coil.
With a 90-degree angle, \(\sin\theta = 1\), resulting in maximum torque; however, if the field and the planar coil are parallel (\theta = 0 or 180 degrees), the sine term becomes zero, and so does the magnetic torque. This is an essential concept because if the coil in an exercise is perfectly aligned with the magnetic field, as in the given problem, the magnetic torque would not be able to support the coil against other forces, like gravity.
Gravitational Torque
Gravitational torque occurs when a force, specifically gravity, acts on a mass at some distance from an axis of rotation, causing a rotational effect. The formula to find the gravitational torque is \(\tau_g = mgL\), where \(m\) is the mass of the object, \(g\) is the gravitational acceleration (typically \(9.81 \mathrm{m/s^2}\) on Earth), and \(L\) is the lever arm, the distance from the rotation axis to where the force is applied. It's important to note that this calculation requires the mass in kilograms, not grams as often provided in exercises.
In our sample problem, a mass hangs from one side of the loop, generating torque due to gravity. To maintain horizontal stability, this gravitational torque must be countered by another force, such as an opposing magnetic torque, generated by another loop positioned symmetrically.
Torque Balance
Torque balance is the condition where the sum of all torques acting on a system is zero. In this balanced state, the system remains static or rotates with constant angular velocity. When dealing with exercises that involve torque balance, it's crucial to ensure that all forces and their respective distances from the pivot point are considered to accurately find the conditions for balance. If a coil is to remain in a horizontal position, as with our coil in the exercise, the torques produced by magnetic forces must exactly counteract the torques from gravitational forces. The resulting equation typically takes the form \(\tau_m = \tau_g\), guiding us to the conditions that will maintain the coil's position.
Rectangular Coil
A rectangular coil in physics problems is often a simplified model used to calculate magnetic and gravitational torques. When evaluating the forces acting on a rectangular coil, it is essential to understand its dimensions and orientation relative to external fields. The coil's area, given by multiplying its length and width, feeds into the formulas for both magnetic and gravitational torques. The coil's placement in a magnetic field - whether at an angle or aligned with the field lines - and the location of masses attached to it determine the type and magnitude of torques we calculate.
In the scenario provided, our coil is lying in a horizontal plane. When visualizing such problems, imagining the coil from a top-down perspective can help determine the direction and effectiveness of the forces acting upon it. Ultimately, knowing how to manipulate these variables can allow students to solve for unknown quantities, such as the needed magnetic field strength to achieve a certain equilibrium.

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Most popular questions from this chapter

A current element produces a magnetic field in the region surrounding it. At any point in space, the magnetic field produced by this current element points in a direction that is a) radial from the current element to the point in space. b) parallel to the current element. c) perpendicular to the current element and to the radial direction.

A long solenoid (diameter of \(6.00 \mathrm{~cm}\) ) is wound with 1000 turns per meter of thin wire through which a current of 0.250 A is maintained. A wire carrying a current of 10.0 A is inserted along the axis of the solenoid. What is the magnitude of the magnetic field at a point \(1.00 \mathrm{~cm}\) from the axis?

Consider a model of the hydrogen atom in which an electron orbits a proton in the plane perpendicular to the proton's spin angular momentum (and magnetic dipole moment) at a distance equal to the Bohr radius, \(a_{0}=5.292 \cdot 10^{-11} \mathrm{~m} .\) (This is an oversimplified classical model.) The spin of the electron is allowed to be either parallel to the proton's spin or antiparallel to it; the orbit is the same in either case. But since the proton produces a magnetic field at the electron's location, and the electron has its own intrinsic magnetic dipole moment, the energy of the electron differs depending on its spin. The magnetic field produced by the proton's spin may be modeled as a dipole field, like the electric field due to an electric dipole discussed in Chapter 22 Calculate the energy difference between the two electronspin configurations. Consider only the interaction between the magnetic dipole moment associated with the electron's spin and the field produced by the proton's spin.

Exposed to sufficiently high magnetic fields, materials saturate, or approach a maximum magnetization. Would you expect the saturation (maximum) magnetization of paramagnetic materials to be much less than, roughly the same as, or much greater than that of ferromagnetic materials? Explain why.

An electron has a spin magnetic moment of magnitude \(\mu=9.285 \cdot 10^{-24} \mathrm{~A} \mathrm{~m}^{2}\). Consequently, it has energy associated with its orientation in a magnetic field. If the difference between the energy of an electron that is "spin up" in a magnetic field of magnitude \(B\) and the energy of one that is "spin down" in the same magnetic field (where "up" and "down" refer to the direction of the magnetic field) is \(9.460 \cdot 10^{-25} \mathrm{~J}\), what is the field magnitude, \(B\) ?

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