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Chapter 28: Problem 14

Figure \(28-27\) shows a wire ring of radius \(a=1.8 \mathrm{~cm}\) that is perpendicular to the general direction of a radially symmetric, diverging magnetic fleld. The magnetic field at the ring is everywhere of the same magnitude \(B=3.4 \mathrm{mT}\), and its direction at the ring everywhere makes an angle \(\theta=15^{\circ}\) with a normal to the plane of the ring. The twisted lead wires have no effect on the problem. Find the magnitude of the force the field exerts on the ring if the ring carries a current \(i=4.6 \mathrm{~mA}\).

Short Answer

Expert verified
The force is approximately \(3.2 \times 10^{-6} \text{ N}\).

Step by step solution

01

Understanding the Magnetic Force on the Current

The force on a current-carrying wire in a magnetic field is given by the formula: \[ F = iL(B \sin \theta) \] where \( i \) is the current, \( L \) is the length of the wire, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the wire and the magnetic field.
02

Calculate the Length of the Wire

The ring is circular, so its length (circumference) is given by: \[ L = 2 \pi a \] where \( a \) is the radius of the ring. Substitute \( a = 1.8 \text{ cm} = 0.018 \text{ m} \):\[ L = 2 \pi \times 0.018 \approx 0.1131 \text{ m} \]
03

Calculate the Magnetic Force

Substitute the values into the force equation: - Current \( i = 4.6 \text{ mA} = 0.0046 \text{ A} \)- Magnetic field \( B = 3.4 \text{ mT} = 0.0034 \text{ T} \)- Angle \( \theta = 15^{\circ} \)Calculate the force:\[ F = 0.0046 \times 0.1131 \times 0.0034 \times \sin (15^{\circ}) \approx 3.2 \times 10^{-6} \text{ N} \]

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Key Concepts

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

Current-Carrying Wire
A wire carrying an electric current generates a magnetic field around it. This is due to the movement of electrons through the wire, creating a circular magnetic field pattern. In our context, the wire forms a circular loop and carries a specified current, which flows uniformly throughout the circle.
  • The current (\(i\)) in this problem is given as \(4.6\) mA, which converts to \(0.0046\) A (since 1 mA = 0.001 A).
  • The direction and magnitude of this current affect how the wire interacts with external magnetic fields.
This interaction is crucial for calculating the magnetic force that acts on the wire. When placed within an external magnetic field, the current-carrying wire experiences a force. The magnitude of this force depends not only on the current but also on the geometry and orientation concerning the field.
Magnetic Field Strength
Understanding magnetic field strength (\(B\)) is key to solving the given problem. It represents the intensity of the magnetic field at a certain point. Measured in Tesla (T), it shows how much force a magnetic field can exert on a current-carrying wire. In this exercise:
  • We have a magnetic field strength of \(3.4\) mT, which equates to \(0.0034\) T (since 1 mT = 0.001 T).
  • The field's uniform direction ensures a consistent interaction across the wire loop.
  • The angle \(\theta = 15^{\circ}\) between the field direction and the loop's plane is critical to determining the force component along the wire.
In the presence of this magnetic field, the force on the wire is proportional to \(B\) and depends on its orientation, as captured by \(\sin \theta\).
Circular Ring Geometry
The wire ring has a specific circular geometry, defined by its radius (\(a\)). This affects both the length of the wire and the force it experiences in a magnetic field.
  • The ring's radius is \(1.8\) cm, which converts to \(0.018\) m for calculations.
  • The circumference of the ring, which acts as the length \(L\) of the wire in our calculations, is derived using the formula: \[L = 2\pi a\]
  • Substituting the given radius, we calculate \(L\approx 0.1131\) m.
This length plays a direct role in computing the force experienced by the wire, as it defines how much of the wire is interacting with the field.
Force Calculation
To find the magnetic force acting on the wire ring, we use the formula: \[F = iL(B \sin \theta)\]Here's the step-by-step approach:
  • The current is \(0.0046\) A, the length of the wire is \(0.1131\) m, the magnetic field strength is \(0.0034\) T, and the angle is \(15^{\circ}\).
  • First, find the sine of the angle: \(\sin(15^{\circ})\).
  • Plug these values into the force equation to compute:\[F = 0.0046 \times 0.1131 \times 0.0034 \times \sin(15^{\circ})\]
  • The result of the computation gives \(F \approx 3.2 \times 10^{-6}\) N.
This result indicates the force in newtons, underscoring the influence of current, wire length, field strength, and angle on the magnetic force experienced by the wire.

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Most popular questions from this chapter

A current loop, carrying a current of \(7.5 \mathrm{~A}\), is in the shape of a right triangle with sides 30,40 , and \(50 \mathrm{~cm}\). The loop is in a uniform magnetic field of magnitude \(120 \mathrm{mT}\) whose direction is parallel to the current in the \(50 \mathrm{~cm}\) side of the loop. Find the magnitude of (a) the magnetic dipole moment of the loop and (b) the torque on the loop.

A conducting rectangular solid of dimensions \(d_{x}=5.00 \mathrm{~m}, d_{y}=3.00 \mathrm{~m}\), and \(d_{z}=2.00 \mathrm{~m}\) moves with a constant velocity \(\vec{v}=(20.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}\) through a uniform magnetic field \(\vec{B}=(40.0 \mathrm{mT}) \hat{\mathrm{j}}\) (Fig. 28-22). What are the resulting (a) electric field within the solid, in unit- vector notation, and (b) potential difference across the solid? (c) Which face becomes negatively charged?

A positron with kinetic energy \(950 \mathrm{eV}\) is projected into a uniform magnetic field \(\vec{B}\) of magnitude \(0.732 \mathrm{~T}\), with its velocity vector making an angle of \(89.0^{\circ}\) with \(\vec{B}\). Find (a) the period, (b) the pitch \(p\), and (c) the radius \(r\) of its helical path.

A long, rigid conductor, lying along an \(x\) axis, carries a current of \(7.0 \mathrm{~A}\) in the negative \(x\) direction. A magnetic field \(\vec{B}\) is present, given by \(\vec{B}=3.0 \mathrm{i}+8.0 x^{2} \hat{\mathrm{j}}\), with \(x\) in meters and \(\vec{B}\) in milliteslas. Find, in unit-vector notation, the force on the \(2.0 \mathrm{~m}\) segment of the conductor that lies between \(x=1.0 \mathrm{~m}\) and \(x=3.0 \mathrm{~m}\).

A strip of copper \(75.0 \mu \mathrm{m}\) thick and \(4.5 \mathrm{~mm}\) wide is placed in a uniform magnetic field \(\vec{B}\) of magnitude \(0.65 \mathrm{~T}\), with \(\vec{B}\) perpendicular to the strip. A current \(i=57 \mathrm{~A}\) is then sent through the strip such that a Hall potential difference \(V\) appears across the width of the strip. Calculate \(V\). (The number of charge carriers per unit volume for copper is \(8.47 \times 10^{28}\) electrons \(/ \mathrm{m}^{3}\).)
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