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CALC A conducting rod with length \(L=0.200 \mathrm{m},\) mass \(m=0.120 \mathrm{kg},\) and resistance \(R=80.0 \Omega\) moves without friction on metal rails as shown in Fig. \(29.11 .\) A uniform magnetic field with magnitude \(B=1.50 \mathrm{T}\) is directed into the plane of the figure. The rod is initially at rest, and then a constant force with magnitude \(F=1.90 \mathrm{N}\) and directed to the right is applied to the bar. How many seconds after the force is applied does the bar reach a speed of \(25,0 \mathrm{m} / \mathrm{s} ?\)

Short Answer

Expert verified
The time is approximately 1.76 seconds.

Step by step solution

01

Understand the Physics of the Problem

The rod is moving in a uniform magnetic field, and a force is applied horizontally. Due to the motion of the rod within the magnetic field, a current will be induced in the rod. This situation can be described by Faraday's Law of Induction and Lenz's Law.
02

Calculate the Induced EMF

The electromotive force (EMF) induced in the circuit can be calculated using the formula: \[ \text{EMF} = B \cdot L \cdot v \]where \(B = 1.50 \text{ T}\), \(L = 0.200 \text{ m}\), and \(v\) is the velocity of the rod.
03

Calculate the Induced Current

Using Ohm's Law, the induced current \(I\) is given by:\[ I = \frac{\text{EMF}}{R} = \frac{B \cdot L \cdot v}{R} \]where \(R = 80.0 \Omega\).
04

Determine the Magnetic Damping Force

The force due to the induced current interacting with the magnetic field is given by:\[ F_{\text{damping}} = I \cdot L \cdot B = \frac{(B \cdot L \cdot v)^2}{R} \]
05

Use Newton’s Second Law to Set Up the Equation of Motion

According to Newton’s Second Law:\[ F - F_{\text{damping}} = m \cdot a \]where \(a\) is the acceleration of the rod and \(m = 0.120 \text{ kg}\). Solve for \(a\).
06

Solve the Motion Equation for Time

You need to find the time \(t\) when the rod reaches a speed of \(v_f = 25.0 \text{ m/s}\):\[ a = \frac{dv}{dt} \implies v_f = at \]Rearrange to solve for \(t\). Substitute the value of \(a\) from Step 5 and solve for \(t\).
07

Final Calculations and Solve for Time

Substitute the known values into the equations derived. The final time can be computed by plugging in all the values and solving for \(t\), using the formula:\[ t = \frac{m \cdot v_f}{F - \frac{(B \cdot L \cdot v)^2}{R}} \]Solving will give the time it takes for the rod to reach a velocity of 25.0 m/s.

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Key Concepts

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

Faraday's Law of Induction
Faraday's Law of Induction is fundamental to understanding how a changing magnetic environment can generate an electric current. This principle is at the core of many devices like generators and transformers. In this context, when the conducting rod moves through a magnetic field, it experiences a change in magnetic flux. This change induces an electromotive force (EMF) in the rod. The formula to calculate the EMF is given by: \[ \text{EMF} = B \cdot L \cdot v \]where
  • \(B\) is the magnetic field strength in tesla (T),
  • \(L\) is the length of the rod in meters (m),
  • \(v\) is the velocity of the rod in meters per second (m/s).
This relationship demonstrates that the induced voltage (EMF) is dependent on how fast the magnetic field is cut, the strength of the magnetic field, and the length of the rod. As the rod in our problem starts moving, the velocity increases, thus causing an increase in the induced EMF. Hence, as per Faraday's Law, the faster the rod moves, the greater the induced EMF.
Lenz's Law
Lenz's Law is an extension of the principles explained by Faraday's Law. It predicts the direction of the induced current. According to Lenz's Law, the induced current creates a magnetic field that opposes the change in the original magnetic field. This is expressed in saying "induced currents counter the change that produced them." In our scenario with the rod, the motion through the magnetic field generates a current in a direction that opposes the motion of the rod.
Here's how it works:
  • As the rod moves to the right, an upward force acts on it due to the induced current.
  • This creates a magnetic field that opposes the motion of the rod.
  • The result is a damping force that tries to slow down the rod's movement.
This opposing induced force is why it's sometimes challenging to keep objects like this rod moving through a magnetic field; the induced fields respond to resist any change in motion.
Ohm's Law
Ohm's Law is a staple in the study of electricity and magnetism, describing the relationship between current, voltage, and resistance. It is succinctly expressed as \[ I = \frac{\text{EMF}}{R} \]where
  • \(I\) is the current in amperes (A),
  • \(\text{EMF}\) is the electromotive force in volts (V),
  • \(R\) is the resistance in ohms (Ω).
In the exercise, once the EMF has been induced by the rod's motion in the magnetic field, Ohm's Law allows us to calculate the induced current through the rod. Since the resistance in the rod is constant (80.0 Ω), the induced current depends directly on the EMF. Hence, a greater velocity not only increases the EMF but also proportionally increases the current flowing through the rod. This relationship is crucial to solve for the damping force as part of the motion equation.
Newton's Second Law
Newton's Second Law of Motion, typically referred to as \( F = m \cdot a \), expresses that the force acting on an object is equal to the mass of the object multiplied by its acceleration.
In the context of the exercise, the rod is subjected to two main forces:
  • The applied force \(F\), which aims to move the rod to the right.
  • The magnetic damping force \(F_{\text{damping}}\), which opposes its motion.
The net force acting on the rod is thus the difference: \[ F_{\text{net}} = F - F_{\text{damping}} \]Using Newton's Second Law, we write \[ m \cdot a = F_{\text{net}} \]This relationship allows us to solve for the acceleration \(a\), knowing the masses and forces involved. Finally, to find the time \(t\) when the rod reaches a specified speed, we rearrange the equation to \[ t = \frac{v_f}{a} \]where \(v_f\) is the final velocity. This connection helps us understand how the interplay of forces influences the motion of the rod over time.

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Most popular questions from this chapter

A long, thin solenoid has 900 turns per meter and radius 2.50 \(\mathrm{cm} .\) The current in the solenoid is increasing at a uniform rate of 60.0 \(\mathrm{A} / \mathrm{s}\) . What is the magnitude of the induced electric field at a point near the center of the solenoid and (a) 0.500 \(\mathrm{cm}\) from the axis of the solenoid; (b) 1.00 \(\mathrm{cm}\) from the axis of the solenoid?

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CP CALC A square, conducting, wire loop of side \(L\) , total mass \(m,\) and total resistance \(R\) initially lies in the horizontal \(x y\) -plane, with corners at \((x, y, z)=(0,0,0),(0, L, 0),(L, 0,0)\) and \((L, L, 0) .\) There is a uniform, upward magnetic field \(\vec{\boldsymbol{B}}=\hat{B hat{\boldsymbol{k}}}\) in the space within and around the loop. The side of the loop that extends from \((0,0,0)\) to \((L, 0,0)\) is held in place on the \(x\) -axis; the rest of the loop is free to pivot around this axis. When the loop is released, it begins to rotate due to the gravitational torque. (a) Find the net torque (magnitude and direction) that acts on the loop when it has rotated through an angle \(\phi\) from its original orientation and is rotating downward at an angular speed \(\omega\) . (b) Find the angular acceleration of the loop at the instant described in part (a). (c) Compared to the case with zero magnetic field, does it take the loop a longer or shorter time to rotate through \(90^{\circ} ?\) Explain. (d) Is mechanical energy conserved as the loop rotates downward? Explain.

CALC In a region of space, a magnetic field points in the \(+x\) -direction (toward the right). Its magnitude varies with position according to the formula \(B_{x}=B_{0}+b x,\) where \(B_{0}\) and \(b\) are positive constants, for \(x \geq 0 .\) A flat coil of area \(A\) moves with uniform speed \(v\) from right to left with the plane of its area always perpendicular to this field. (a) What is the emf induced in this coil while it is to the right of the origin? (b) As viewed from the origin, what is the direction (clockwise or counterclockwise) of the current induced in the coil? (c) If instead the coil moved from left to right, what would be the answers to parts (a) and (b)?

A rectangle measuring 30.0 \(\mathrm{cm}\) by 40.0 \(\mathrm{cm}\) is located inside a region of a spatially uniform magnetic field of 1.25 \(\mathrm{T}\) , with the field perpendicular to the plane of the coil (Fig. E29.24). The coil is pulled out at a steady rate of 2.00 \(\mathrm{cm} / \mathrm{s}\) traveling perpendicular to the field lines. The region of the field ends abruptly as shown. Find the emf induced in this coil when it is (a) all inside the field; (b) partly inside the field; (c) all outside the field.

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