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Chapter 22: Problem 1

A \(0.80-\mathrm{m}\) aluminum bar is held with its length parallel to the east- west direction and dropped from a bridge. Just before the bar hits the river below, its speed is \(22 \mathrm{m} / \mathrm{s},\) and the emf induced across its length is \(6.5 \times 10^{-4} \mathrm{V} .\) Assuming the horizontal component of the earth's magnetic field at the location of the bar points directly north, (a) determine the magnitude of the horizontal component of the earth's magnetic field, and (b) state whether the east end or the west end of the bar is positive.

Short Answer

Expert verified
(a) 3.69 脳 10鈦烩伒 T; (b) West end is positive.

Step by step solution

01

Understand the context and setup the problem

An aluminum bar is dropped while oriented in the east-west direction, being affected by the Earth's magnetic field, which is directed north. We need to use Faraday's law of induction to compute the horizontal component of the magnetic field as there is a change in magnetic flux as the bar moves.
02

Use the formula for motional emf

The formula for the motional emf induced along a length of a conductor moving through a magnetic field is given by:\[\text{emf} = B \cdot L \cdot v \cdot \sin(\theta)\]Where:- \(B\) is the magnetic field- \(L\) is the length of the rod- \(v\) is the velocity of the rod- \(\theta\) is the angle between the velocity and the magnetic field.Since the velocity direction is vertical and the magnetic field is horizontal (90 degrees btw them), \(\sin(\theta) = 1\).
03

Plug in values to find the magnetic field

Given:- \(\text{emf} = 6.5 \times 10^{-4} \text{ V}\)- \(L = 0.80 \text{ m}\)- \(v = 22 \text{ m/s}\)Substitute these values into the equation to solve for \(B\):\[6.5 \times 10^{-4} = B \times 0.80 \times 22\]Solving for \(B\), we have:\[\begin{aligned} B &= \frac{6.5 \times 10^{-4}}{0.80 \times 22} \ &= \frac{6.5 \times 10^{-4}}{17.6} \ &= 3.69 \times 10^{-5} \text{ T (tesla)} \end{aligned}\]
04

Determine the polarity of the bar

Using the right-hand rule for the direction of force on a moving conductor in a magnetic field, point your thumb in the direction of the rod's velocity (down), the fingers in the direction of the magnetic field (north), and your palm will face the force direction, showing electron flow. This indicates current moving towards the east end, making it negatively charged, thus the induced emf makes the west end positive.

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Key Concepts

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

Motional EMF
Motional EMF is a fascinating phenomenon that occurs when a conductor moves through a magnetic field. This results in an induced electromotive force (emf) across the conductor's length due to the movement relative to the magnetic field.
The magnitude of the motional emf can be calculated using the formula, \[\text{emf} = B \cdot L \cdot v \cdot \sin(\theta) \]
  • \(B\) represents the magnetic field strength.
  • \(L\) is the length of the conductor.
  • \(v\) is the velocity of the conductor.
  • \(\theta\) is the angle between the direction of the velocity and the magnetic field.
In most practical scenarios, the angle \(\theta\) is either \(0^\circ\) or \(90^\circ\), simplifying the calculation of \(\sin(\theta)\) to either 0 or 1.
In the given problem, because the velocity of the bar is perpendicular to the magnetic field, \(\theta = 90^\circ\), which means \(\sin(\theta) = 1\). Hence, the formula simplifies to \[\text{emf} = B \cdot L \cdot v \]which is straightforward for computation when the bar moves at right angles to the magnetic field.
Magnetic Field
The magnetic field is an invisible field that exerts magnetic forces on moving charges and magnetic materials in its vicinity. In this exercise, the Earth's magnetic field plays a crucial role. Its horizontal component is necessary to induce a motional emf in a falling aluminum bar.
To find the magnitude of this horizontal magnetic field (denoted as \(B\)), we use the formula provided for the motional emf:\[B = \frac{\text{emf}}{L \cdot v} \]where the emf has been provided as \(6.5 \times 10^{-4} \text{ V}\), the length of the bar \(L\) is \(0.80 \text{ m}\), and the speed of the bar \(v\) is \(22 \text{ m/s}\).
Substituting the given values, you calculate:\[B = \frac{6.5 \times 10^{-4}}{0.80 \times 22} = 3.69 \times 10^{-5} \text{ T}\]
The above result gives us the strength of the horizontal component of the Earth's magnetic field in Tesla, indicating the power of magnetic forces in inducing an electromotive force.
Right-Hand Rule
The right-hand rule is an intuitive tool used to determine the direction of the induced current in a conductor moving through a magnetic field. This rule can be a bit tricky at first, but visualizing it helps immensely.
To apply the right-hand rule, follow these steps:
  • Point your thumb in the direction of the conductor's velocity (in this exercise, this is vertically down since the bar is falling).
  • Extend your fingers in the direction of the magnetic field (pointing north as per the problem statement).
  • Your palm will naturally face the direction of the force experienced by positive charges (or the direction electrons move due to the force).
In the context of the aluminum bar problem, this means the electrons will experience a force towards the east end, making it negatively charged. Conversely, this means the west end of the bar will be positively charged.
This rule is crucial in determining polarity, as understanding the internal flow of charge carriers in moving conductors affects circuits and real-life magnetic applications.

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Most popular questions from this chapter

Two flat surfaces are exposed to a uniform, horizontal magnetic field of magnitude 0.47 T. When viewed edge-on, the first surface is tilted at an angle of \(12^{\circ}\) from the horizontal, and a net magnetic flux of \(8.4 \times 10^{-3} \mathrm{Wb}\) passes through it. The same net magnetic flux passes through the second surface. (a) Determine the area of the first surface. (b) Find the smallest possible value for the area of the second surface.

The rechargeable batteries for a laptop computer need a much smaller voltage than what a wall socket provides. Therefore, a transformer is plugged into the wall socket and produces the necessary voltage for charging the batteries. The batteries are rated at \(9.0 \mathrm{V},\) and a current of \(225 \mathrm{mA}\) is used to charge them. The wall socket provides a voltage of \(120 \mathrm{V}\). (a) Determine the turns ratio of the transformer. (b) What is the current coming from the wall socket? (c) Find the average power delivered by the wall socket and the average power sent to the batteries.

A square loop of wire consisting of a single turn is perpendicular to a uniform magnetic field. The square loop is then re-formed into a circular loop, which also consists of a single turn and is also perpendicular to the same magnetic field. The magnetic flux that passes through the square loop is \(7.0 \times 10^{-3}\) Wb. What is the flux that passes through the circular loop?

A flat coil of wire has an area \(A, N\) turns, and a resistance \(R\). It is situated in a magnetic field, such that the normal to the coil is parallel to the magnetic field. The coil is then rotated through an angle of \(90^{\circ},\) so that the normal becomes perpendicular to the magnetic field. The coil has an area of \(1.5 \times 10^{-3} \mathrm{m}^{2}, 50\) turns, and a resistance of \(140 \Omega .\) During the time while it is rotating, a charge of \(8.5 \times 10^{-5} \mathrm{C}\) flows in the coil. What is the magnitude of the magnetic field?

The drawing shows a straight wire carrying a current \(I\). Above the wire is a rectangular loop that contains a resistor \(R\). If the current \(I\) is decreasing in time, what is the direction of the induced current through the resistor \(R-\) left-to-right or right-to-left?
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