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Chapter 27: Problem 5

(II) The force on a wire is a maximum of \(7.50 \times 10^{-2} \mathrm{N}\) when placed between the pole faces of a magnet. The current flows horizontally to the right and the magnetic field is vertical. The wire is observed to "jump" toward the observer when the current is turned on. (a) What type of magnetic pole is the top pole face? (b) If the pole faces have a diameter of \(10.0 \mathrm{cm},\) estimate the current in the wire if the field is 0.220 \(\mathrm{T}\) . (c) If the wire is tipped so that it makes an angle of \(10.0^{\circ}\) with the horizontal, what force will it now feel?

Short Answer

Expert verified
(a) South Pole, (b) 3.41 A, (c) 7.38 脳 10鈦宦� N

Step by step solution

01

Understanding the Problem

We are given that the force on a wire is maximum when it is placed in a magnetic field, and the current flows horizontally where the wire moves toward the observer. We need to determine the polarity of the magnetic field, estimate the current required for this force, and calculate the force with a tilted wire.
02

Identify Pole Type

According to the right-hand rule for forces in magnetic fields, if the current is moving to the right and the wire jumps towards the observer, the magnetic field direction is from the top to the bottom. This points towards identifying the top pole as the South Pole, since the force is directed outward.
03

Calculate Maximum Force Equation

The maximum force on the wire can be expressed as: \[ F = I L B \]where \( F = 7.50 \times 10^{-2} \text{ N} \), \( B = 0.220 \text{ T} \), and \( L \) is the diameter of the pole faces, thus the length of the wire within the field, which is \( 0.10 \text{ m} \). We need to solve for the current \( I \).
04

Solve for Current

Using the formula and solving for \( I \):\[I = \frac{F}{L B} = \frac{7.50 \times 10^{-2}}{0.10 \times 0.220} \approx 3.41 \text{ A}\]
05

Force with Tilted Wire

When the wire is tipped at an angle \( \theta = 10.0^{\circ} \), the effective component of the current along the length of the magnetic field changes. The new force is:\[ F_{\text{new}} = I L B \cos(\theta) \]Substitute in the known values: \[F_{\text{new}} = 3.41 \times 0.10 \times 0.220 \times \cos(10.0^{\circ}) \approx 7.38 \times 10^{-2} \text{ N}\]

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Key Concepts

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

Right-Hand Rule
It鈥檚 essential to grasp how the right-hand rule works. It helps in determining the direction of the magnetic force on a current-carrying wire. Here is how you can apply it:
  • Point your thumb in the direction of the conventional current (positive to negative).
  • Extend your fingers in the direction of the magnetic field lines.
  • The force exerted on the wire pushes perpendicular from your palm.
In this scenario, because the wire "jumps" toward the observer and the current flows horizontally to the right, the magnetic force pushes outward. Following the right-hand rule with the magnetic field going from top to bottom (vertical), aligns with the scenario where the force pushes toward you.
Magnetic Poles
Understanding magnetic poles helps clarify which pole affects the wire. If the wire jumps toward the observer, the force resulting from a vertically oriented field suggests a specific pole configuration. In this case, the top pole must be the South Pole.
  • Magnetic field lines emerge from the North Pole and enter the South Pole.
  • Since the field direction is downward in this context (from top to bottom), the right-hand rule implies the top is South.
Keep in mind that the poles of a magnet do not physically push or pull; instead, they create a field interacting with charges moving perpendicular to the field.
Current Calculation
To calculate current in the wire, we use the equation for magnetic force: \[ F = I L B \] Where:
  • \( F \) is the force exerted on the wire, given as \( 7.50 \times 10^{-2} \text{ N} \).
  • \( L \) is the length of the wire within the magnetic field, here \( 0.10 \text{ m} \).
  • \( B \) is the magnetic field strength, \( 0.220 \text{ T} \).
Rearrange to solve for current \( I \):\[ I = \frac{F}{L B} = \frac{7.50 \times 10^{-2}}{0.10 \times 0.220} \approx 3.41 \text{ A} \]This calculation gives the magnitude of the current needed to produce the specified force under the given field conditions.
Inclined Wire in Magnetic Field
When the wire is inclined, not all of the current contributes effectively to generate a magnetic force. We must consider the angle between the wire and the magnetic field. The effective component of the force is determined by the cosine of this angle. The formula for this scenario is:\[ F_{\text{new}} = I L B \cos(\theta) \]Substitute the values: \[ F_{\text{new}} = 3.41 \times 0.10 \times 0.220 \times \cos(10.0^{\circ}) \approx 7.38 \times 10^{-2} \text{ N} \]Here, \( \theta \) represents the angle, in this case, \( 10^{\circ} \), which modifies the effective magnetic force felt by the wire. The calculation shows how the tilt reduces the force compared to when the wire is completely horizontal.

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Most popular questions from this chapter

(I) An electron is projected vertically upward with a speed of \(1.70 \times 10^{6} \mathrm{m} / \mathrm{s}\) into a uniform magnetic field of 0.480 \(\mathrm{T}\) that is directed horizontally away from the observer. Describe the electron's path in this field.

(I) Alpha particles of charge \(q=+2 e\) and mass \(m=6.6 \times 10^{-27} \mathrm{kg}\) are emitted from a radioactive source at a speed of \(1.6 \times 10^{7} \mathrm{m} / \mathrm{s}\) . What magnetic field strength would be required to bend them into a circular path of radius \(r=0.18 \mathrm{m} ?\)

The net force on a current loop whose face is perpendicular to a uniform magnetic field is zero, since contributions to the net force from opposite sides of the loop cancel. However, if the field varies in magnitude from one side of the loop to the other, then there can be a net force on the loop. Consider a square loop with sides whose length is \(a\), located with one side at \(x=b\) in the \(x y\) plane (Fig. 27-55). A magnetic field is directed along \(z\), with a magnitude that varies with \(x\) according to $$ B=B_{0}\left(1-\frac{x}{b}\right) $$ If the current in the loop circulates counterclockwise (that is, the magnetic dipole moment of the loop is along the \(z\) axis), find an expression for the net force on the loop.

An electron experiences a force \(\overrightarrow{\mathbf{F}}=(3.8 \hat{\mathbf{i}}-2.7 \hat{\mathbf{j}}) \times 10^{-13} \mathbf{N}\) when passing through a magnetic field \(\overrightarrow{\mathbf{B}}=(0.85 \mathrm{~T}) \hat{\mathbf{k}}\). Determine the electron's velocity.

A Hall probe used to measure magnetic field strengths consists of a rectangular slab of material (free-electron density \(n\) ) with width \(d\) and thickness \(t,\) carrying a current \(I\) along its length \(\ell\). The slab is immersed in a magnetic field of magnitude \(B\) oriented perpendicular to its rectangular face (of area \(\ell d\) ), so that a Hall emf \(\mathscr{E}_{\mathrm{H}}\) is produced across its width \(d\). The probe's magnetic sensitivity, defined as \(K_{\mathrm{H}}=\mathscr{E}_{\mathrm{H}} / I B,\) indicates the magnitude of the Hall emf achieved for a given applied magnetic field and current. A slab with a large \(K_{\mathrm{H}}\) is a good candidate for use as a Hall probe. (a) Show that \(K_{\mathrm{H}}=1\) /ent. Thus, a good Hall probe has small values for both \(n\) and \(t .(b)\) As possible candidates for the material used in a Hall probe, consider (i) a typical metal \(\left(n \approx 1 \times 10^{29} / \mathrm{m}^{3}\right)\) and (ii) a (doped) semiconductor \(\left(n \approx 3 \times 10^{22} / \mathrm{m}^{3}\right)\). Given that a semiconductor slab can be manufactured with a thickness of \(0.15 \mathrm{~mm}\), how thin (nm) should a metal slab be to yield a \(K_{H}\) value equal to that of the semiconductor slab? Compare this metal slab thickness with the 0.3 -nm size of a typical metal atom. (c) For the typical semiconductor slab described in part \((b),\) what is the expected value for \(\mathscr{E}_{\mathrm{H}}\) when \(I=100 \mathrm{~mA}\) and \(B=0.1 \mathrm{~T} ?\)
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