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Chapter 29: Problem 27

A \(1.41-\mathrm{m}\) bar moves through a uniform, 1.20 . \(T\) magnetic field with a speed of 2.50 \(\mathrm{m} / \mathrm{s}\) (Fig, 29.40\()\) . In cach case, find the emf induced between the ends of this bar and identify which, if any, end \((a \text { or } b)\) is at the higher potential. The bar moves in the direction of (a) the \(+x\) -axis; (b) the \(-y\) -axis; (c) the \(+z\) -axis. (d) How should this bar move so that the emf across its ends has the greatest possible value with \(b\) at a higher potential than \(a\) , and what is this maximum emf?

Short Answer

Expert verified
(a) 0 V; (b) 4.23 V, b higher; (c) 4.23 V, b higher; (d) 4.23 V, direction perpendicular to the magnetic field.

Step by step solution

01

Understand the Problem and Variables

We are asked to find the induced electromotive force (emf) between the ends of a moving bar and determine which end is at a higher potential. The bar has a length of 1.41 m, moves in a magnetic field of 1.20 T, and at a speed of 2.50 m/s. We need to analyze three different movement directions: the +x-axis, -y-axis, and +z-axis.
02

Use the Formula for Induced EMF

The formula for the emf induced in a bar moving in a magnetic field is \( \text{emf} = B \cdot L \cdot v \cdot \sin(\theta) \), where \( B \) is the magnetic field strength, \( L \) is the length of the bar, \( v \) is the velocity, and \( \theta \) is the angle between the direction of the velocity and the magnetic field. Since the magnetic field, assumed default by convention, is along one of the coordinate axes, judging by given cases (a, b, c), we will assume it's aligned with the +x-axis.
03

Solve for Direction (a) +x-axis

For the bar moving along the +x-axis, the velocity and magnetic field are parallel. Here \( \theta = 0 \), which makes \( \sin(\theta) = 0 \). Therefore, the induced emf is 0 V, as the velocity is parallel to the magnetic field.
04

Solve for Direction (b) -y-axis

For movement along the -y-axis, the velocity is perpendicular to the magnetic field (\( \theta = 90^\circ \)), thus \( \sin(\theta) = 1 \). The induced emf is calculated as: \[ \text{emf} = 1.20 \times 1.41 \times 2.50 = 4.23 \text{ V}. \] By the right-hand rule, if we assume the magnetic field is along the +x-axis, end b would be at a higher potential than end a.
05

Solve for Direction (c) +z-axis

For movement along the +z-axis, the velocity is again perpendicular to the magnetic field (\( \theta = 90^\circ \)), so \( \sin(\theta) = 1 \). The induced emf is: \[ \text{emf} = 1.20 \times 1.41 \times 2.50 = 4.23 \text{ V}. \] Similar reasoning using the right-hand rule shows that it will be equivalent to case (b).
06

Determine Maximum EMF and Condition (d)

To achieve the maximum emf with end b at a higher potential than end a, the velocity must be perpendicular to both the bar and the magnetic field. The maximum emf is the same as in cases where \( \theta = 90^\circ \), which is 4.23 V. Movement in any direction with angle 90 degrees to the magnetic field would achieve this emf, as shown in scenarios (b) and (c).

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Key Concepts

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

Faraday's Law
Faraday's Law of electromagnetic induction is a fundamental principle that describes how a changing magnetic field can induce an electromotive force (emf) in a conductor. In simpler terms, when you move a conductor through a magnetic field, or the magnetic field around a conductor changes, an emf is induced, giving rise to an electric current if the circuit is closed.

The mathematical expression of Faraday's Law is given by the formula:
  • \[\text{emf} = -\frac{d\Phi}{dt}\]
Here, \( \Phi \) represents the magnetic flux, and \( \frac{d\Phi}{dt} \) indicates the rate of change of magnetic flux.

In the context of our exercise, Faraday's Law helps explain why an emf is induced when the bar moves through the magnetic field. The motion of the bar changes the magnetic flux through the bar, thus inducing the emf.
magnetic fields
Magnetic fields are invisible fields that exert a force on moving charges and magnetic materials. The strength of a magnetic field is measured in teslas (T), and it defines how much force a magnetic field can exert on a moving charge.

In this exercise, a uniform magnetic field of 1.20 T is present. When the bar moves through this magnetic field, it cuts through the field lines, which is what induces an emf according to Faraday's Law. Understandably, the orientation between the motion of the conductor and magnetic field lines is crucial to determine if and how much emf will be induced.

The force experienced by charges in the bar is related to both the quantity of the magnetic field (B) and the velocity of the charge carriers (v), coupled with the length (L) of the bar moving through the field, calculated using the formula:
  • \[\text{emf} = B \cdot L \cdot v \cdot \sin(\theta)\]
where \( \theta \) is the angle between the magnetic field and the direction of movement of the bar.
right-hand rule
The right-hand rule is a method used to determine the direction of the induced current or magnetic force in a scenario involving a conductor in a magnetic field. For the specific orientation of forces and fields, the right hand's thumb points in the direction of the current or velocity (the movement of a positively charged particle), and the fingers point in the direction of the magnetic field.

In our exercise, using this rule helps determine which end of the bar (end a or end b) is at higher potential. For instance, if the velocity is in a direction perpendicular to the magnetic field, say the -y-axis, and assuming the magnetic field is in the +x-axis, the palm facing direction tells us that end b will be at a higher potential due to the leftward direction of the induced emf.
induced current
Induced current is the result of an induced emf in a closed circuit. When a conductor like the bar in our exercise moves through a magnetic field and induces an emf, it can drive a current if the circuit is closed.

However, in the exercise discussed, we only consider the induced emf and not the induced current itself, as the question does not specify a complete circuit attached to the moving bar. If there were one, charges would move, creating a current that circulates through the circuit.

The induced emf calculated for different scenarios indicates the potential difference across the bar's ends, and this potential difference would drive the current once the circuit is closed, adhering to Ohm's Law:

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Most popular questions from this chapter

A circular loop of wire with a radius of 12.0 \(\mathrm{cm}\) and oriented in the horizontal \(x y\) -plane is located in a region of uniform magnetic field. A field of 1.5 \(\mathrm{T}\) is directed along the positive z-direction, which is upward. (a) If the loop is removed from the field region in a time interval of 2.0 \(\mathrm{ms}\) , find the average emf that will be induced in the wire loop during the extraction process. (b) If the coil is viewed looking down on it from above, is the induced current in the loop clockwise or counterclockwise?

A capacitor has two parallel plates with area \(A\) separated by a distance \(d\) . The space between plates is filled with a material having dielectric constant \(K\) . The material is not a perfect insulator but has resistivity \(\rho\) . The capacitor is initially charged with charge of magnitude \(Q_{0}\) on each plate that gradually discharges by conduction through the dielectric. (a) Calculate the conduction current density \(j_{\mathrm{C}}(t)\) in the dielectric. (b) Show that at any instant the dis-placement current density in the diclectric is equal in magnitude to the oonduotion current density but opposite in direction, so the total current density is zero at every instant.

A long, straight solenoid with a cross-sectional area of 8.00 \(\mathrm{cm}^{2}\) is wound with 90 turns of wire per centimeter, and the windings carry a current of 0.350 \(\mathrm{A}\) . A second winding of 12 tums encircles the solenoid at its center. The current in the solenoid is turned off such that the magnetic field of the solenoid becomes zero in 0.0400 s. What is the average indnced emf in the second winding?

A long, straight wire made of a type-I superconductor carries a constant current \(I\) along its length. Show that the current cannot be uniformly spread over the wire's cross section but instead must all be at the surface.

A rod of pure silicon (resistivity \(\rho=2300 \Omega \cdot \mathrm{m} )\) is carry-ing a current. The electric field varies sinusoidally with time according to \(E=E_{0} \sin \omega t,\) where \(E_{0}=0.450 \mathrm{V} / \mathrm{m}, \omega=2 \pi f,\) and the frequency \(f=120 \mathrm{Hz}\) (a) Find the magnitude of the maximum conduction current density in the wire. (b) Assuming \(\epsilon=\epsilon_{0}\) , find the maximum displacement current density in the wire, and compare with the result of part (a). (c) At what frequency \(f\) would the maximum conduction and displacement densitics become equal if \(\epsilon=\epsilon_{0}\) (which is not actually the case)? (d) At the frequency determined in part (c), what is the relative phase of the conduction and displacement currents?
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