Question

A circular coil with a radius of \(10.0 \mathrm{~cm}\) has 100 turns of wire and carries a current, \(i=100 . \mathrm{mA} .\) It is free to rotate in a region with a constant horizontal magnetic field given by \(\vec{B}=(0.0100 \mathrm{~T}) \hat{x}\). If the unit normal vector to the plane of the coil makes an angle of \(30.0^{\circ}\) with the horizontal, what is the magnitude of the net magnetic torque acting on the coil? 27.61 At \(t=0\) an electron crosses the positive \(y\) -axis (so \(x=0\) ) at \(60.0 \mathrm{~cm}\) from the origin with velocity \(2.00 \cdot 10^{5} \mathrm{~m} / \mathrm{s}\) in the positive \(x\) -direction. It is in a uniform magnetic field. a) Find the magnitude and the direction of the magnetic field that will cause the electron to cross the \(x\) -axis at \(x=60.0 \mathrm{~cm}\). b) What work is done on the electron during this motion? c) How long will the trip take from \(y\) -axis to \(x\) -axis?

Short answer

#Answer#: To solve Part 1, follow these steps: 1. Calculate the area of the circular coil: Area = πr² = π (0.10 m)² ≈ 0.0314 m² 2. Calculate the magnetic field component in the plane of the coil: \(B_{plane} = B \times \cos{θ}\) = (5 T) × cos(30.0°) ≈ 4.33 T 3. Calculate the magnetic flux passing through the coil: Magnetic flux = n × Area × \(B_{plane}\) = 100 × 0.0314 m² × 4.33 T ≈ 13.6 Wb 4. Calculate the net magnetic torque on the coil: Torque = nABisini = (100)(0.0314 m²)(5 T)(0.1 A)sin(30.0°) ≈ 7.86 Nm The magnetic torque on the circular coil is approximately 7.86 Nm.

Step-by-step solution

Determine the magnetic torque on the circular coil

First, we need to find the area of the circular coil: Area = \(πr^2\), where r = 0.10 m Now, calculate the magnetic field component in the plane of the coil: \(B_{plane} = B \times \cos{θ}\), where θ = 30.0° After determining \(B_{plane}\), find the magnetic flux passing through the coil: Magnetic flux = \(n \times Area \times B_{plane}\), where n = 100 The net magnetic torque on the coil is determined by the formula: Torque = nABisini, B is magnetic field, i is current We already have the \(B_{plane}\), and the current i = 100 mA. Calculate the magnetic torque using the given values.

Key concepts

Circular Coil

A circular coil is essentially made up of numerous loops of wire, each neatly wound in a circle. The coil referenced in our exercise is particularly interesting because it has 100 turns, which means the wire has been looped 100 times. This configuration increases its ability to generate magnetic effects due to the more concentrated field created.
Coils can have various uses in physics and engineering. They are often integrated into electrical circuits to shape magnetic fields or even convert energy from one form to another.
The circular coil's radius affects its overall area, which is calculated using the formula for the area of a circle: \[ \text{Area} = \pi r^2 \], where \( r \) is the coil's radius. If the radius is 0.10 meters, then the area is 0.0314 square meters. This area contributes directly to determining the magnetic flux and torque in this scenario.

Magnetic Field

A magnetic field, represented by the symbol \( \vec{B} \), describes the magnetic influence produced by electric currents and magnetic materials. In our example, the magnetic field is \(0.0100 \text{ T} \) and directed along the x-axis.
The strength and direction of the magnetic field are crucial in determining how a coil or a charged particle, like an electron, is affected.
Some important aspects of magnetic fields include:

• Magnetic fields exert forces on charges moving through them.
• The direction and magnitude of forces can be computed using the right-hand rule and formulas such as the Lorentz force law.

In this exercise, the magnetic field helps determine the torque applied to the coil, depending on how it interacts with the coil's plane. Additionally, it influences the trajectory of electrons passing through, impacting our calculations for electron motion.

Electron Motion

Magnetic fields significantly affect electron motion. Electrons, being negatively charged particles, react to magnetic fields in predictable ways.
When an electron moves through a uniform magnetic field, it experiences a force perpendicular to both its velocity and the magnetic field direction, described by the formula \( F = qvB \sin(\theta) \), where:

• \( q \) is the charge of the electron (approximately \( -1.6 \times 10^{-19} \) C)
• \( v \) is the velocity of the electron
• \( B \) is the magnetic field strength
• \( \theta \) is the angle between the velocity and the magnetic field direction

In our setup, the electron initially moves along the y-axis and is aimed to cross the x-axis. The magnetic field's role is crucial as it bends the electron’s path by exerting a force that alters the electron's direction, causing it to cross the x-axis at the desired point. This alteration requires careful determination of the magnetic field's strength and orientation. The work done by the magnetic field is noted to be zero since it acts perpendicular to the direction of motion, changing the electron's path without affecting its speed.

Magnetic Flux

Magnetic flux quantifies the amount of magnetic field passing through a given area, like the loops of a coil. It provides insight into how much of the magnetic field interacts with the coil, influencing both the induced current and torque.
The formula for magnetic flux \( \Phi \) is:\[ \Phi = n \, \cdot \, A \, \cdot \, B_{plane} \]where:

• \( n \) is the number of turns in the coil
• \( A \) is the area of the coil
• \( B_{plane} \) is the component of the magnetic field perpendicular to the coil

Since the coil in our situation is tilted, the effective magnetic field component in its plane is calculated as \( B_{plane} = B \, \cos(\theta) \), where \( \theta \) is 30 degrees. By calculating the magnetic flux, we can better understand the induced forces and the resultant torque acting on the coil.